3.386 \(\int \frac {\sec (c+d x)}{a+b \sin ^3(c+d x)} \, dx\)

Optimal. Leaf size=290 \[ -\frac {b \log \left (a+b \sin ^3(c+d x)\right )}{3 d \left (a^2-b^2\right )}+\frac {\sqrt [3]{b} \left (a^{4/3}+b^{4/3}\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{6 a^{2/3} d \left (a^2-b^2\right )}-\frac {\sqrt [3]{b} \left (a^{4/3}+b^{4/3}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{3 a^{2/3} d \left (a^2-b^2\right )}-\frac {\sqrt [3]{b} \left (a^{4/3}-b^{4/3}\right ) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sin (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} d \left (a^2-b^2\right )}-\frac {\log (1-\sin (c+d x))}{2 d (a+b)}+\frac {\log (\sin (c+d x)+1)}{2 d (a-b)} \]

[Out]

-1/2*ln(1-sin(d*x+c))/(a+b)/d+1/2*ln(1+sin(d*x+c))/(a-b)/d-1/3*b^(1/3)*(a^(4/3)+b^(4/3))*ln(a^(1/3)+b^(1/3)*si
n(d*x+c))/a^(2/3)/(a^2-b^2)/d+1/6*b^(1/3)*(a^(4/3)+b^(4/3))*ln(a^(2/3)-a^(1/3)*b^(1/3)*sin(d*x+c)+b^(2/3)*sin(
d*x+c)^2)/a^(2/3)/(a^2-b^2)/d-1/3*b*ln(a+b*sin(d*x+c)^3)/(a^2-b^2)/d-1/3*b^(1/3)*(a^(4/3)-b^(4/3))*arctan(1/3*
(a^(1/3)-2*b^(1/3)*sin(d*x+c))/a^(1/3)*3^(1/2))/a^(2/3)/(a^2-b^2)/d*3^(1/2)

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Rubi [A]  time = 0.33, antiderivative size = 290, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3223, 2074, 1871, 1860, 31, 634, 617, 204, 628, 260} \[ \frac {\sqrt [3]{b} \left (a^{4/3}+b^{4/3}\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{6 a^{2/3} d \left (a^2-b^2\right )}-\frac {b \log \left (a+b \sin ^3(c+d x)\right )}{3 d \left (a^2-b^2\right )}-\frac {\sqrt [3]{b} \left (a^{4/3}+b^{4/3}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{3 a^{2/3} d \left (a^2-b^2\right )}-\frac {\sqrt [3]{b} \left (a^{4/3}-b^{4/3}\right ) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sin (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} d \left (a^2-b^2\right )}-\frac {\log (1-\sin (c+d x))}{2 d (a+b)}+\frac {\log (\sin (c+d x)+1)}{2 d (a-b)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]/(a + b*Sin[c + d*x]^3),x]

[Out]

-((b^(1/3)*(a^(4/3) - b^(4/3))*ArcTan[(a^(1/3) - 2*b^(1/3)*Sin[c + d*x])/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(2/3)*
(a^2 - b^2)*d)) - Log[1 - Sin[c + d*x]]/(2*(a + b)*d) + Log[1 + Sin[c + d*x]]/(2*(a - b)*d) - (b^(1/3)*(a^(4/3
) + b^(4/3))*Log[a^(1/3) + b^(1/3)*Sin[c + d*x]])/(3*a^(2/3)*(a^2 - b^2)*d) + (b^(1/3)*(a^(4/3) + b^(4/3))*Log
[a^(2/3) - a^(1/3)*b^(1/3)*Sin[c + d*x] + b^(2/3)*Sin[c + d*x]^2])/(6*a^(2/3)*(a^2 - b^2)*d) - (b*Log[a + b*Si
n[c + d*x]^3])/(3*(a^2 - b^2)*d)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 1860

Int[((A_) + (B_.)*(x_))/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{r = Numerator[Rt[a/b, 3]], s = Denominator[R
t[a/b, 3]]}, -Dist[(r*(B*r - A*s))/(3*a*s), Int[1/(r + s*x), x], x] + Dist[r/(3*a*s), Int[(r*(B*r + 2*A*s) + s
*(B*r - A*s)*x)/(r^2 - r*s*x + s^2*x^2), x], x]] /; FreeQ[{a, b, A, B}, x] && NeQ[a*B^3 - b*A^3, 0] && PosQ[a/
b]

Rule 1871

Int[(P2_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{A = Coeff[P2, x, 0], B = Coeff[P2, x, 1], C = Coeff[P2, x,
2]}, Int[(A + B*x)/(a + b*x^3), x] + Dist[C, Int[x^2/(a + b*x^3), x], x] /; EqQ[a*B^3 - b*A^3, 0] ||  !Ration
alQ[a/b]] /; FreeQ[{a, b}, x] && PolyQ[P2, x, 2]

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rule 3223

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With
[{ff = FreeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p, x]
, x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (EqQ[n, 4] || GtQ[m, 0
] || IGtQ[p, 0] || IntegersQ[m, p])

Rubi steps

\begin {align*} \int \frac {\sec (c+d x)}{a+b \sin ^3(c+d x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \left (a+b x^3\right )} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-\frac {1}{2 (a+b) (-1+x)}+\frac {1}{2 (a-b) (1+x)}+\frac {b \left (b-a x+b x^2\right )}{\left (-a^2+b^2\right ) \left (a+b x^3\right )}\right ) \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac {\log (1-\sin (c+d x))}{2 (a+b) d}+\frac {\log (1+\sin (c+d x))}{2 (a-b) d}-\frac {b \operatorname {Subst}\left (\int \frac {b-a x+b x^2}{a+b x^3} \, dx,x,\sin (c+d x)\right )}{\left (a^2-b^2\right ) d}\\ &=-\frac {\log (1-\sin (c+d x))}{2 (a+b) d}+\frac {\log (1+\sin (c+d x))}{2 (a-b) d}-\frac {b \operatorname {Subst}\left (\int \frac {b-a x}{a+b x^3} \, dx,x,\sin (c+d x)\right )}{\left (a^2-b^2\right ) d}-\frac {b^2 \operatorname {Subst}\left (\int \frac {x^2}{a+b x^3} \, dx,x,\sin (c+d x)\right )}{\left (a^2-b^2\right ) d}\\ &=-\frac {\log (1-\sin (c+d x))}{2 (a+b) d}+\frac {\log (1+\sin (c+d x))}{2 (a-b) d}-\frac {b \log \left (a+b \sin ^3(c+d x)\right )}{3 \left (a^2-b^2\right ) d}-\frac {b^{2/3} \operatorname {Subst}\left (\int \frac {\sqrt [3]{a} \left (-a^{4/3}+2 b^{4/3}\right )+\sqrt [3]{b} \left (-a^{4/3}-b^{4/3}\right ) x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\sin (c+d x)\right )}{3 a^{2/3} \left (a^2-b^2\right ) d}-\frac {\left (b^{2/3} \left (a^{4/3}+b^{4/3}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,\sin (c+d x)\right )}{3 a^{2/3} \left (a^2-b^2\right ) d}\\ &=-\frac {\log (1-\sin (c+d x))}{2 (a+b) d}+\frac {\log (1+\sin (c+d x))}{2 (a-b) d}-\frac {\sqrt [3]{b} \left (a^{4/3}+b^{4/3}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{3 a^{2/3} \left (a^2-b^2\right ) d}-\frac {b \log \left (a+b \sin ^3(c+d x)\right )}{3 \left (a^2-b^2\right ) d}+\frac {\left (b^{2/3} \left (a^{4/3}-b^{4/3}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\sin (c+d x)\right )}{2 \sqrt [3]{a} \left (a^2-b^2\right ) d}+\frac {\left (\sqrt [3]{b} \left (a^{4/3}+b^{4/3}\right )\right ) \operatorname {Subst}\left (\int \frac {-\sqrt [3]{a} \sqrt [3]{b}+2 b^{2/3} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\sin (c+d x)\right )}{6 a^{2/3} \left (a^2-b^2\right ) d}\\ &=-\frac {\log (1-\sin (c+d x))}{2 (a+b) d}+\frac {\log (1+\sin (c+d x))}{2 (a-b) d}-\frac {\sqrt [3]{b} \left (a^{4/3}+b^{4/3}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{3 a^{2/3} \left (a^2-b^2\right ) d}+\frac {\sqrt [3]{b} \left (a^{4/3}+b^{4/3}\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{6 a^{2/3} \left (a^2-b^2\right ) d}-\frac {b \log \left (a+b \sin ^3(c+d x)\right )}{3 \left (a^2-b^2\right ) d}+\frac {\left (\sqrt [3]{b} \left (a^{4/3}-b^{4/3}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} \sin (c+d x)}{\sqrt [3]{a}}\right )}{a^{2/3} \left (a^2-b^2\right ) d}\\ &=-\frac {\sqrt [3]{b} \left (a^{4/3}-b^{4/3}\right ) \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} \sin (c+d x)}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt {3} a^{2/3} \left (a^2-b^2\right ) d}-\frac {\log (1-\sin (c+d x))}{2 (a+b) d}+\frac {\log (1+\sin (c+d x))}{2 (a-b) d}-\frac {\sqrt [3]{b} \left (a^{4/3}+b^{4/3}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{3 a^{2/3} \left (a^2-b^2\right ) d}+\frac {\sqrt [3]{b} \left (a^{4/3}+b^{4/3}\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{6 a^{2/3} \left (a^2-b^2\right ) d}-\frac {b \log \left (a+b \sin ^3(c+d x)\right )}{3 \left (a^2-b^2\right ) d}\\ \end {align*}

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Mathematica [C]  time = 0.20, size = 268, normalized size = 0.92 \[ \frac {b^{5/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )+3 a^{2/3} b \sin ^2(c+d x) \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};-\frac {b \sin ^3(c+d x)}{a}\right )-2 a^{2/3} b \log \left (a+b \sin ^3(c+d x)\right )+3 a^{2/3} b \log (1-\sin (c+d x))+3 a^{2/3} b \log (\sin (c+d x)+1)-3 a^{5/3} \log (1-\sin (c+d x))+3 a^{5/3} \log (\sin (c+d x)+1)-2 b^{5/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )+2 \sqrt {3} b^{5/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sin (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{6 a^{2/3} d (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]/(a + b*Sin[c + d*x]^3),x]

[Out]

(2*Sqrt[3]*b^(5/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*Sin[c + d*x])/(Sqrt[3]*a^(1/3))] - 3*a^(5/3)*Log[1 - Sin[c + d*
x]] + 3*a^(2/3)*b*Log[1 - Sin[c + d*x]] + 3*a^(5/3)*Log[1 + Sin[c + d*x]] + 3*a^(2/3)*b*Log[1 + Sin[c + d*x]]
- 2*b^(5/3)*Log[a^(1/3) + b^(1/3)*Sin[c + d*x]] + b^(5/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*Sin[c + d*x] + b^(2/3)
*Sin[c + d*x]^2] - 2*a^(2/3)*b*Log[a + b*Sin[c + d*x]^3] + 3*a^(2/3)*b*Hypergeometric2F1[2/3, 1, 5/3, -((b*Sin
[c + d*x]^3)/a)]*Sin[c + d*x]^2)/(6*a^(2/3)*(a - b)*(a + b)*d)

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fricas [C]  time = 1.48, size = 4396, normalized size = 15.16 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a+b*sin(d*x+c)^3),x, algorithm="fricas")

[Out]

-1/36*(2*(a^2 - b^2)*(9*(I*sqrt(3) + 1)*(-1/54*b/(a^4*d^3 - a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(
a^2 + b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3) + b^2*(-I*sqrt(3) + 1)/((a^2*d - b^2*d)^2*(-1/54*b/(a^4*d^3 - a^2*
b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 + b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3)) + 6*b/(a^2*d - b^2*
d))*d*log(-1/36*(a^5 - a^3*b^2)*(9*(I*sqrt(3) + 1)*(-1/54*b/(a^4*d^3 - a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)
^3 + 1/54*(a^2 + b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3) + b^2*(-I*sqrt(3) + 1)/((a^2*d - b^2*d)^2*(-1/54*b/(a^4
*d^3 - a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 + b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3)) + 6*b/(a
^2*d - b^2*d))^2*d^2 + a*b^2 + 1/6*(2*a^3*b + a*b^3)*(9*(I*sqrt(3) + 1)*(-1/54*b/(a^4*d^3 - a^2*b^2*d^3) - 1/2
7*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 + b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3) + b^2*(-I*sqrt(3) + 1)/((a^2*d - b
^2*d)^2*(-1/54*b/(a^4*d^3 - a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 + b^2)*b/((a^2 - b^2)^2*a^2*
d^3))^(1/3)) + 6*b/(a^2*d - b^2*d))*d - (a^2*b + b^3)*sin(d*x + c)) - ((a^2 - b^2)*(9*(I*sqrt(3) + 1)*(-1/54*b
/(a^4*d^3 - a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 + b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3) + b^
2*(-I*sqrt(3) + 1)/((a^2*d - b^2*d)^2*(-1/54*b/(a^4*d^3 - a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^
2 + b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3)) + 6*b/(a^2*d - b^2*d))*d - 3*sqrt(1/3)*(a^2 - b^2)*d*sqrt(-((a^4 -
2*a^2*b^2 + b^4)*(9*(I*sqrt(3) + 1)*(-1/54*b/(a^4*d^3 - a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2
+ b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3) + b^2*(-I*sqrt(3) + 1)/((a^2*d - b^2*d)^2*(-1/54*b/(a^4*d^3 - a^2*b^2*
d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 + b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3)) + 6*b/(a^2*d - b^2*d))^
2*d^2 - 12*(a^2*b - b^3)*(9*(I*sqrt(3) + 1)*(-1/54*b/(a^4*d^3 - a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/
54*(a^2 + b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3) + b^2*(-I*sqrt(3) + 1)/((a^2*d - b^2*d)^2*(-1/54*b/(a^4*d^3 -
a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 + b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3)) + 6*b/(a^2*d -
b^2*d))*d - 108*b^2)/((a^4 - 2*a^2*b^2 + b^4)*d^2)) - 18*b)*log(1/36*(a^5 - a^3*b^2)*(9*(I*sqrt(3) + 1)*(-1/54
*b/(a^4*d^3 - a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 + b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3) +
b^2*(-I*sqrt(3) + 1)/((a^2*d - b^2*d)^2*(-1/54*b/(a^4*d^3 - a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(
a^2 + b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3)) + 6*b/(a^2*d - b^2*d))^2*d^2 - a*b^2 - 1/6*(2*a^3*b + a*b^3)*(9*(
I*sqrt(3) + 1)*(-1/54*b/(a^4*d^3 - a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 + b^2)*b/((a^2 - b^2)
^2*a^2*d^3))^(1/3) + b^2*(-I*sqrt(3) + 1)/((a^2*d - b^2*d)^2*(-1/54*b/(a^4*d^3 - a^2*b^2*d^3) - 1/27*b^3/(a^2*
d - b^2*d)^3 + 1/54*(a^2 + b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3)) + 6*b/(a^2*d - b^2*d))*d + 1/12*sqrt(1/3)*((
a^5 - a^3*b^2)*(9*(I*sqrt(3) + 1)*(-1/54*b/(a^4*d^3 - a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 +
b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3) + b^2*(-I*sqrt(3) + 1)/((a^2*d - b^2*d)^2*(-1/54*b/(a^4*d^3 - a^2*b^2*d^
3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 + b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3)) + 6*b/(a^2*d - b^2*d))*d^
2 - 6*(a^3*b - a*b^3)*d)*sqrt(-((a^4 - 2*a^2*b^2 + b^4)*(9*(I*sqrt(3) + 1)*(-1/54*b/(a^4*d^3 - a^2*b^2*d^3) -
1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 + b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3) + b^2*(-I*sqrt(3) + 1)/((a^2*d
- b^2*d)^2*(-1/54*b/(a^4*d^3 - a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 + b^2)*b/((a^2 - b^2)^2*a
^2*d^3))^(1/3)) + 6*b/(a^2*d - b^2*d))^2*d^2 - 12*(a^2*b - b^3)*(9*(I*sqrt(3) + 1)*(-1/54*b/(a^4*d^3 - a^2*b^2
*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 + b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3) + b^2*(-I*sqrt(3) + 1)/
((a^2*d - b^2*d)^2*(-1/54*b/(a^4*d^3 - a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 + b^2)*b/((a^2 -
b^2)^2*a^2*d^3))^(1/3)) + 6*b/(a^2*d - b^2*d))*d - 108*b^2)/((a^4 - 2*a^2*b^2 + b^4)*d^2)) - 2*(a^2*b + b^3)*s
in(d*x + c)) - ((a^2 - b^2)*(9*(I*sqrt(3) + 1)*(-1/54*b/(a^4*d^3 - a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 +
1/54*(a^2 + b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3) + b^2*(-I*sqrt(3) + 1)/((a^2*d - b^2*d)^2*(-1/54*b/(a^4*d^3
- a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 + b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3)) + 6*b/(a^2*d
- b^2*d))*d + 3*sqrt(1/3)*(a^2 - b^2)*d*sqrt(-((a^4 - 2*a^2*b^2 + b^4)*(9*(I*sqrt(3) + 1)*(-1/54*b/(a^4*d^3 -
a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 + b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3) + b^2*(-I*sqrt(
3) + 1)/((a^2*d - b^2*d)^2*(-1/54*b/(a^4*d^3 - a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 + b^2)*b/
((a^2 - b^2)^2*a^2*d^3))^(1/3)) + 6*b/(a^2*d - b^2*d))^2*d^2 - 12*(a^2*b - b^3)*(9*(I*sqrt(3) + 1)*(-1/54*b/(a
^4*d^3 - a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 + b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3) + b^2*(
-I*sqrt(3) + 1)/((a^2*d - b^2*d)^2*(-1/54*b/(a^4*d^3 - a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 +
b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3)) + 6*b/(a^2*d - b^2*d))*d - 108*b^2)/((a^4 - 2*a^2*b^2 + b^4)*d^2)) - 1
8*b)*log(-1/36*(a^5 - a^3*b^2)*(9*(I*sqrt(3) + 1)*(-1/54*b/(a^4*d^3 - a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^
3 + 1/54*(a^2 + b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3) + b^2*(-I*sqrt(3) + 1)/((a^2*d - b^2*d)^2*(-1/54*b/(a^4*
d^3 - a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 + b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3)) + 6*b/(a^
2*d - b^2*d))^2*d^2 + a*b^2 + 1/6*(2*a^3*b + a*b^3)*(9*(I*sqrt(3) + 1)*(-1/54*b/(a^4*d^3 - a^2*b^2*d^3) - 1/27
*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 + b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3) + b^2*(-I*sqrt(3) + 1)/((a^2*d - b^
2*d)^2*(-1/54*b/(a^4*d^3 - a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 + b^2)*b/((a^2 - b^2)^2*a^2*d
^3))^(1/3)) + 6*b/(a^2*d - b^2*d))*d + 1/12*sqrt(1/3)*((a^5 - a^3*b^2)*(9*(I*sqrt(3) + 1)*(-1/54*b/(a^4*d^3 -
a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 + b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3) + b^2*(-I*sqrt(3
) + 1)/((a^2*d - b^2*d)^2*(-1/54*b/(a^4*d^3 - a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 + b^2)*b/(
(a^2 - b^2)^2*a^2*d^3))^(1/3)) + 6*b/(a^2*d - b^2*d))*d^2 - 6*(a^3*b - a*b^3)*d)*sqrt(-((a^4 - 2*a^2*b^2 + b^4
)*(9*(I*sqrt(3) + 1)*(-1/54*b/(a^4*d^3 - a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 + b^2)*b/((a^2
- b^2)^2*a^2*d^3))^(1/3) + b^2*(-I*sqrt(3) + 1)/((a^2*d - b^2*d)^2*(-1/54*b/(a^4*d^3 - a^2*b^2*d^3) - 1/27*b^3
/(a^2*d - b^2*d)^3 + 1/54*(a^2 + b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3)) + 6*b/(a^2*d - b^2*d))^2*d^2 - 12*(a^2
*b - b^3)*(9*(I*sqrt(3) + 1)*(-1/54*b/(a^4*d^3 - a^2*b^2*d^3) - 1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 + b^2)*
b/((a^2 - b^2)^2*a^2*d^3))^(1/3) + b^2*(-I*sqrt(3) + 1)/((a^2*d - b^2*d)^2*(-1/54*b/(a^4*d^3 - a^2*b^2*d^3) -
1/27*b^3/(a^2*d - b^2*d)^3 + 1/54*(a^2 + b^2)*b/((a^2 - b^2)^2*a^2*d^3))^(1/3)) + 6*b/(a^2*d - b^2*d))*d - 108
*b^2)/((a^4 - 2*a^2*b^2 + b^4)*d^2)) + 2*(a^2*b + b^3)*sin(d*x + c)) - 18*(a + b)*log(sin(d*x + c) + 1) + 18*(
a - b)*log(-sin(d*x + c) + 1))/((a^2 - b^2)*d)

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giac [A]  time = 0.33, size = 309, normalized size = 1.07 \[ -\frac {\frac {2 \, {\left (a^{3} b^{2} \left (-\frac {a}{b}\right )^{\frac {1}{3}} - a b^{4} \left (-\frac {a}{b}\right )^{\frac {1}{3}} - a^{2} b^{3} + b^{5}\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | -\left (-\frac {a}{b}\right )^{\frac {1}{3}} + \sin \left (d x + c\right ) \right |}\right )}{a^{5} b - 2 \, a^{3} b^{3} + a b^{5}} + \frac {2 \, {\left (\sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} b^{2} + \sqrt {3} \left (-a b^{2}\right )^{\frac {2}{3}} a\right )} \arctan \left (\frac {\sqrt {3} {\left (\left (-\frac {a}{b}\right )^{\frac {1}{3}} + 2 \, \sin \left (d x + c\right )\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{a^{3} b - a b^{3}} + \frac {{\left (\left (-a b^{2}\right )^{\frac {1}{3}} b^{2} - \left (-a b^{2}\right )^{\frac {2}{3}} a\right )} \log \left (\sin \left (d x + c\right )^{2} + \left (-\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x + c\right ) + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{a^{3} b - a b^{3}} + \frac {2 \, b \log \left ({\left | b \sin \left (d x + c\right )^{3} + a \right |}\right )}{a^{2} - b^{2}} - \frac {3 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a - b} + \frac {3 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a + b}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a+b*sin(d*x+c)^3),x, algorithm="giac")

[Out]

-1/6*(2*(a^3*b^2*(-a/b)^(1/3) - a*b^4*(-a/b)^(1/3) - a^2*b^3 + b^5)*(-a/b)^(1/3)*log(abs(-(-a/b)^(1/3) + sin(d
*x + c)))/(a^5*b - 2*a^3*b^3 + a*b^5) + 2*(sqrt(3)*(-a*b^2)^(1/3)*b^2 + sqrt(3)*(-a*b^2)^(2/3)*a)*arctan(1/3*s
qrt(3)*((-a/b)^(1/3) + 2*sin(d*x + c))/(-a/b)^(1/3))/(a^3*b - a*b^3) + ((-a*b^2)^(1/3)*b^2 - (-a*b^2)^(2/3)*a)
*log(sin(d*x + c)^2 + (-a/b)^(1/3)*sin(d*x + c) + (-a/b)^(2/3))/(a^3*b - a*b^3) + 2*b*log(abs(b*sin(d*x + c)^3
+ a))/(a^2 - b^2) - 3*log(abs(sin(d*x + c) + 1))/(a - b) + 3*log(abs(sin(d*x + c) - 1))/(a + b))/d

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maple [A]  time = 0.96, size = 374, normalized size = 1.29 \[ -\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{d \left (2 a +2 b \right )}-\frac {b \ln \left (\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 d \left (a -b \right ) \left (a +b \right ) \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {b \ln \left (\sin ^{2}\left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 d \left (a -b \right ) \left (a +b \right ) \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 d \left (a -b \right ) \left (a +b \right ) \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {a \ln \left (\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 d \left (a -b \right ) \left (a +b \right ) \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {a \ln \left (\sin ^{2}\left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 d \left (a -b \right ) \left (a +b \right ) \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {a \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 d \left (a -b \right ) \left (a +b \right ) \left (\frac {a}{b}\right )^{\frac {1}{3}}}-\frac {b \ln \left (a +b \left (\sin ^{3}\left (d x +c \right )\right )\right )}{3 d \left (a -b \right ) \left (a +b \right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{d \left (2 a -2 b \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)/(a+b*sin(d*x+c)^3),x)

[Out]

-1/d/(2*a+2*b)*ln(sin(d*x+c)-1)-1/3/d*b/(a-b)/(a+b)/(a/b)^(2/3)*ln(sin(d*x+c)+(a/b)^(1/3))+1/6/d*b/(a-b)/(a+b)
/(a/b)^(2/3)*ln(sin(d*x+c)^2-(a/b)^(1/3)*sin(d*x+c)+(a/b)^(2/3))-1/3/d*b/(a-b)/(a+b)/(a/b)^(2/3)*3^(1/2)*arcta
n(1/3*3^(1/2)*(2/(a/b)^(1/3)*sin(d*x+c)-1))-1/3/d/(a-b)/(a+b)*a/(a/b)^(1/3)*ln(sin(d*x+c)+(a/b)^(1/3))+1/6/d/(
a-b)/(a+b)*a/(a/b)^(1/3)*ln(sin(d*x+c)^2-(a/b)^(1/3)*sin(d*x+c)+(a/b)^(2/3))+1/3/d/(a-b)/(a+b)*a*3^(1/2)/(a/b)
^(1/3)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*sin(d*x+c)-1))-1/3/d*b/(a-b)/(a+b)*ln(a+b*sin(d*x+c)^3)+1/d/(2*a-2*b)
*ln(1+sin(d*x+c))

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maxima [A]  time = 0.51, size = 288, normalized size = 0.99 \[ \frac {\frac {2 \, \sqrt {3} {\left (a {\left (3 \, \left (\frac {a}{b}\right )^{\frac {2}{3}} + 2\right )} - b {\left (3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}} + \frac {2 \, a}{b}\right )}\right )} \arctan \left (-\frac {\sqrt {3} {\left (\left (\frac {a}{b}\right )^{\frac {1}{3}} - 2 \, \sin \left (d x + c\right )\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{{\left (a^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}} - b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {3 \, {\left (b {\left (2 \, \left (\frac {a}{b}\right )^{\frac {2}{3}} - 1\right )} - a \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \log \left (\sin \left (d x + c\right )^{2} - \left (\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x + c\right ) + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{a^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}} - b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {6 \, {\left (b {\left (\left (\frac {a}{b}\right )^{\frac {2}{3}} + 1\right )} + a \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \log \left (\left (\frac {a}{b}\right )^{\frac {1}{3}} + \sin \left (d x + c\right )\right )}{a^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}} - b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {9 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a - b} - \frac {9 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a + b}}{18 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a+b*sin(d*x+c)^3),x, algorithm="maxima")

[Out]

1/18*(2*sqrt(3)*(a*(3*(a/b)^(2/3) + 2) - b*(3*(a/b)^(1/3) + 2*a/b))*arctan(-1/3*sqrt(3)*((a/b)^(1/3) - 2*sin(d
*x + c))/(a/b)^(1/3))/((a^2*(a/b)^(2/3) - b^2*(a/b)^(2/3))*(a/b)^(1/3)) - 3*(b*(2*(a/b)^(2/3) - 1) - a*(a/b)^(
1/3))*log(sin(d*x + c)^2 - (a/b)^(1/3)*sin(d*x + c) + (a/b)^(2/3))/(a^2*(a/b)^(2/3) - b^2*(a/b)^(2/3)) - 6*(b*
((a/b)^(2/3) + 1) + a*(a/b)^(1/3))*log((a/b)^(1/3) + sin(d*x + c))/(a^2*(a/b)^(2/3) - b^2*(a/b)^(2/3)) + 9*log
(sin(d*x + c) + 1)/(a - b) - 9*log(sin(d*x + c) - 1)/(a + b))/d

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mupad [B]  time = 0.27, size = 600, normalized size = 2.07 \[ \frac {\left (\sum _{k=1}^3\ln \left (-{\mathrm {root}\left (27\,a^2\,b^2\,z^3-27\,a^4\,z^3-27\,a^2\,b\,z^2-b,z,k\right )}^2\,a\,b^4\,13-{\mathrm {root}\left (27\,a^2\,b^2\,z^3-27\,a^4\,z^3-27\,a^2\,b\,z^2-b,z,k\right )}^3\,a\,b^5\,36-{\mathrm {root}\left (27\,a^2\,b^2\,z^3-27\,a^4\,z^3-27\,a^2\,b\,z^2-b,z,k\right )}^4\,a\,b^6\,36-{\mathrm {root}\left (27\,a^2\,b^2\,z^3-27\,a^4\,z^3-27\,a^2\,b\,z^2-b,z,k\right )}^2\,b^5\,\sin \left (c+d\,x\right )\,16-{\mathrm {root}\left (27\,a^2\,b^2\,z^3-27\,a^4\,z^3-27\,a^2\,b\,z^2-b,z,k\right )}^3\,b^6\,\sin \left (c+d\,x\right )\,12-{\mathrm {root}\left (27\,a^2\,b^2\,z^3-27\,a^4\,z^3-27\,a^2\,b\,z^2-b,z,k\right )}^3\,a^3\,b^3\,27-{\mathrm {root}\left (27\,a^2\,b^2\,z^3-27\,a^4\,z^3-27\,a^2\,b\,z^2-b,z,k\right )}^4\,a^3\,b^4\,180-\mathrm {root}\left (27\,a^2\,b^2\,z^3-27\,a^4\,z^3-27\,a^2\,b\,z^2-b,z,k\right )\,b^4\,\sin \left (c+d\,x\right )\,5-{\mathrm {root}\left (27\,a^2\,b^2\,z^3-27\,a^4\,z^3-27\,a^2\,b\,z^2-b,z,k\right )}^3\,a^2\,b^4\,\sin \left (c+d\,x\right )\,69-{\mathrm {root}\left (27\,a^2\,b^2\,z^3-27\,a^4\,z^3-27\,a^2\,b\,z^2-b,z,k\right )}^4\,a^2\,b^5\,\sin \left (c+d\,x\right )\,162-{\mathrm {root}\left (27\,a^2\,b^2\,z^3-27\,a^4\,z^3-27\,a^2\,b\,z^2-b,z,k\right )}^4\,a^4\,b^3\,\sin \left (c+d\,x\right )\,54\right )\,\mathrm {root}\left (27\,a^2\,b^2\,z^3-27\,a^4\,z^3-27\,a^2\,b\,z^2-b,z,k\right )\right )-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )}{2\,a+2\,b}+\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )}{2\,a-2\,b}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)*(a + b*sin(c + d*x)^3)),x)

[Out]

(symsum(log(- 13*root(27*a^2*b^2*z^3 - 27*a^4*z^3 - 27*a^2*b*z^2 - b, z, k)^2*a*b^4 - 36*root(27*a^2*b^2*z^3 -
27*a^4*z^3 - 27*a^2*b*z^2 - b, z, k)^3*a*b^5 - 36*root(27*a^2*b^2*z^3 - 27*a^4*z^3 - 27*a^2*b*z^2 - b, z, k)^
4*a*b^6 - 16*root(27*a^2*b^2*z^3 - 27*a^4*z^3 - 27*a^2*b*z^2 - b, z, k)^2*b^5*sin(c + d*x) - 12*root(27*a^2*b^
2*z^3 - 27*a^4*z^3 - 27*a^2*b*z^2 - b, z, k)^3*b^6*sin(c + d*x) - 27*root(27*a^2*b^2*z^3 - 27*a^4*z^3 - 27*a^2
*b*z^2 - b, z, k)^3*a^3*b^3 - 180*root(27*a^2*b^2*z^3 - 27*a^4*z^3 - 27*a^2*b*z^2 - b, z, k)^4*a^3*b^4 - 5*roo
t(27*a^2*b^2*z^3 - 27*a^4*z^3 - 27*a^2*b*z^2 - b, z, k)*b^4*sin(c + d*x) - 69*root(27*a^2*b^2*z^3 - 27*a^4*z^3
- 27*a^2*b*z^2 - b, z, k)^3*a^2*b^4*sin(c + d*x) - 162*root(27*a^2*b^2*z^3 - 27*a^4*z^3 - 27*a^2*b*z^2 - b, z
, k)^4*a^2*b^5*sin(c + d*x) - 54*root(27*a^2*b^2*z^3 - 27*a^4*z^3 - 27*a^2*b*z^2 - b, z, k)^4*a^4*b^3*sin(c +
d*x))*root(27*a^2*b^2*z^3 - 27*a^4*z^3 - 27*a^2*b*z^2 - b, z, k), k, 1, 3) - log(sin(c + d*x) - 1)/(2*a + 2*b)
+ log(sin(c + d*x) + 1)/(2*a - 2*b))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec {\left (c + d x \right )}}{a + b \sin ^{3}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a+b*sin(d*x+c)**3),x)

[Out]

Integral(sec(c + d*x)/(a + b*sin(c + d*x)**3), x)

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