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

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

[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)

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Rubi [A]  time = 0.329331, antiderivative size = 290, normalized size of antiderivative = 1., 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 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])

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 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 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 31

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

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 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 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 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 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]

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*}

Mathematica [C]  time = 0.213135, 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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Maple [A]  time = 0.128, size = 374, normalized size = 1.3 \begin{align*} -{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{d \left ( 2\,a+2\,b \right ) }}-{\frac{b}{3\,d \left ( a-b \right ) \left ( a+b \right ) }\ln \left ( \sin \left ( dx+c \right ) +\sqrt [3]{{\frac{a}{b}}} \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}+{\frac{b}{6\,d \left ( a-b \right ) \left ( a+b \right ) }\ln \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{2}-\sqrt [3]{{\frac{a}{b}}}\sin \left ( dx+c \right ) + \left ({\frac{a}{b}} \right ) ^{{\frac{2}{3}}} \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}-{\frac{b\sqrt{3}}{3\,d \left ( a-b \right ) \left ( a+b \right ) }\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\sin \left ( dx+c \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-1 \right ) } \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}-{\frac{a}{3\,d \left ( a-b \right ) \left ( a+b \right ) }\ln \left ( \sin \left ( dx+c \right ) +\sqrt [3]{{\frac{a}{b}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}+{\frac{a}{6\,d \left ( a-b \right ) \left ( a+b \right ) }\ln \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{2}-\sqrt [3]{{\frac{a}{b}}}\sin \left ( dx+c \right ) + \left ({\frac{a}{b}} \right ) ^{{\frac{2}{3}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}+{\frac{a\sqrt{3}}{3\,d \left ( a-b \right ) \left ( a+b \right ) }\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\sin \left ( dx+c \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-1 \right ) } \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-{\frac{b\ln \left ( a+b \left ( \sin \left ( dx+c \right ) \right ) ^{3} \right ) }{3\,d \left ( a-b \right ) \left ( a+b \right ) }}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{d \left ( 2\,a-2\,b \right ) }} \end{align*}

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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [C]  time = 14.5048, size = 9420, normalized size = 32.48 \begin{align*} \text{result too large to display} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [A]  time = 1.18017, size = 417, normalized size = 1.44 \begin{align*} -\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{6 \,{\left (\left (-a b^{2}\right )^{\frac{1}{3}} b^{2} + \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 )}{\sqrt{3} a^{3} b - \sqrt{3} 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} \end{align*}

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