3.383 \(\int \frac {\cos ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx\)

Optimal. Leaf size=219 \[ -\frac {\left (a^{4/3}+b^{4/3}\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{6 a^{2/3} b^{5/3} d}+\frac {\left (a^{4/3}+b^{4/3}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{3 a^{2/3} b^{5/3} d}+\frac {\left (a^{4/3}-b^{4/3}\right ) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sin (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} b^{5/3} d}-\frac {2 \log \left (a+b \sin ^3(c+d x)\right )}{3 b d}+\frac {\sin ^2(c+d x)}{2 b d} \]

[Out]

1/3*(a^(4/3)+b^(4/3))*ln(a^(1/3)+b^(1/3)*sin(d*x+c))/a^(2/3)/b^(5/3)/d-1/6*(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)/b^(5/3)/d-2/3*ln(a+b*sin(d*x+c)^3)/b/d+1/2*sin(d*x+c)^2/b/d
+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)/b^(5/3)/d*3^(1/2)

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Rubi [A]  time = 0.29, antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3223, 1887, 1871, 1860, 31, 634, 617, 204, 628, 260} \[ -\frac {\left (a^{4/3}+b^{4/3}\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{6 a^{2/3} b^{5/3} d}+\frac {\left (a^{4/3}+b^{4/3}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{3 a^{2/3} b^{5/3} d}+\frac {\left (a^{4/3}-b^{4/3}\right ) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sin (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} b^{5/3} d}-\frac {2 \log \left (a+b \sin ^3(c+d x)\right )}{3 b d}+\frac {\sin ^2(c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5/(a + b*Sin[c + d*x]^3),x]

[Out]

((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)*b^(5/3)*d)
+ ((a^(4/3) + b^(4/3))*Log[a^(1/3) + b^(1/3)*Sin[c + d*x]])/(3*a^(2/3)*b^(5/3)*d) - ((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)*b^(5/3)*d) - (2*Log[a + b*Sin[c +
d*x]^3])/(3*b*d) + Sin[c + d*x]^2/(2*b*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 1887

Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[Pq/(a + b*x^n), x], x] /; FreeQ[{a, b}, x
] && PolyQ[Pq, x] && IntegerQ[n]

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 {\cos ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^2}{a+b x^3} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {x}{b}+\frac {b-a x-2 b x^2}{b \left (a+b x^3\right )}\right ) \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {\sin ^2(c+d x)}{2 b d}+\frac {\operatorname {Subst}\left (\int \frac {b-a x-2 b x^2}{a+b x^3} \, dx,x,\sin (c+d x)\right )}{b d}\\ &=\frac {\sin ^2(c+d x)}{2 b d}-\frac {2 \operatorname {Subst}\left (\int \frac {x^2}{a+b x^3} \, dx,x,\sin (c+d x)\right )}{d}+\frac {\operatorname {Subst}\left (\int \frac {b-a x}{a+b x^3} \, dx,x,\sin (c+d x)\right )}{b d}\\ &=-\frac {2 \log \left (a+b \sin ^3(c+d x)\right )}{3 b d}+\frac {\sin ^2(c+d x)}{2 b d}+\frac {\left (\frac {1}{a^{2/3}}+\frac {a^{2/3}}{b^{4/3}}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,\sin (c+d x)\right )}{3 d}+\frac {\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} b^{4/3} d}\\ &=\frac {\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} b^{5/3} d}-\frac {2 \log \left (a+b \sin ^3(c+d x)\right )}{3 b d}+\frac {\sin ^2(c+d x)}{2 b d}-\frac {\left (a^{4/3}-b^{4/3}\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} b^{4/3} d}-\frac {\left (a^{4/3}+b^{4/3}\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} b^{5/3} d}\\ &=\frac {\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} b^{5/3} d}-\frac {\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} b^{5/3} d}-\frac {2 \log \left (a+b \sin ^3(c+d x)\right )}{3 b d}+\frac {\sin ^2(c+d x)}{2 b d}-\frac {\left (a^{4/3}-b^{4/3}\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} b^{5/3} d}\\ &=\frac {\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} b^{5/3} d}+\frac {\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} b^{5/3} d}-\frac {\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} b^{5/3} d}-\frac {2 \log \left (a+b \sin ^3(c+d x)\right )}{3 b d}+\frac {\sin ^2(c+d x)}{2 b d}\\ \end {align*}

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Mathematica [C]  time = 0.24, size = 203, normalized size = 0.93 \[ \frac {-b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )-3 a^{2/3} \sin ^2(c+d x) \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};-\frac {b \sin ^3(c+d x)}{a}\right )-4 a^{2/3} \log \left (a+b \sin ^3(c+d x)\right )+3 a^{2/3} \sin ^2(c+d x)+2 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )-2 \sqrt {3} b^{2/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sin (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{6 a^{2/3} b d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5/(a + b*Sin[c + d*x]^3),x]

[Out]

(-2*Sqrt[3]*b^(2/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*Sin[c + d*x])/(Sqrt[3]*a^(1/3))] + 2*b^(2/3)*Log[a^(1/3) + b^(
1/3)*Sin[c + d*x]] - b^(2/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*Sin[c + d*x] + b^(2/3)*Sin[c + d*x]^2] - 4*a^(2/3)*
Log[a + b*Sin[c + d*x]^3] + 3*a^(2/3)*Sin[c + d*x]^2 - 3*a^(2/3)*Hypergeometric2F1[2/3, 1, 5/3, -((b*Sin[c + d
*x]^3)/a)]*Sin[c + d*x]^2)/(6*a^(2/3)*b*d)

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fricas [C]  time = 1.31, size = 3216, normalized size = 14.68 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a+b*sin(d*x+c)^3),x, algorithm="fricas")

[Out]

-1/12*(2*((1/2)^(1/3)*(I*sqrt(3) + 1)*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b^4)/(a^2*
b^5*d^3))^(1/3) + 4/(b*d) + 2*(1/2)^(2/3)*(-I*sqrt(3) + 1)/(b^2*d^2*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a^
2*b^5*d^3) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3)))*b*d*log(1/4*((1/2)^(1/3)*(I*sqrt(3) + 1)*(2/(b^3*d^3) + (a^4 -
2*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3) + 4/(b*d) + 2*(1/2)^(2/3)*(-I*sqrt(3) + 1)/
(b^2*d^2*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3)))^2*a^3*b^3*d
^2 + 2*a^3*b - 2*a*b^3 - 1/2*(4*a^3*b^2 - a*b^4)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2
+ b^4)/(a^2*b^5*d^3) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3) + 4/(b*d) + 2*(1/2)^(2/3)*(-I*sqrt(3) + 1)/(b^2*d^2*(2
/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3)))*d + (a^4 - b^4)*sin(d*
x + c)) + 6*cos(d*x + c)^2 - (((1/2)^(1/3)*(I*sqrt(3) + 1)*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a^2*b^5*d^3
) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3) + 4/(b*d) + 2*(1/2)^(2/3)*(-I*sqrt(3) + 1)/(b^2*d^2*(2/(b^3*d^3) + (a^4 -
2*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3)))*b*d + 3*sqrt(1/3)*b*d*sqrt(-(((1/2)^(1/3)
*(I*sqrt(3) + 1)*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3) + 4/(
b*d) + 2*(1/2)^(2/3)*(-I*sqrt(3) + 1)/(b^2*d^2*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b
^4)/(a^2*b^5*d^3))^(1/3)))^2*b*d - 8*(1/2)^(1/3)*(I*sqrt(3) + 1)*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a^2*b
^5*d^3) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3) - 32/(b*d) - 16*(1/2)^(2/3)*(-I*sqrt(3) + 1)/(b^2*d^2*(2/(b^3*d^3)
+ (a^4 - 2*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3)))/(b*d)) - 12)*log(1/4*((1/2)^(1/3)
*(I*sqrt(3) + 1)*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3) + 4/(
b*d) + 2*(1/2)^(2/3)*(-I*sqrt(3) + 1)/(b^2*d^2*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b
^4)/(a^2*b^5*d^3))^(1/3)))^2*a^3*b^3*d^2 + 2*a^3*b - 2*a*b^3 - 1/2*(4*a^3*b^2 - a*b^4)*((1/2)^(1/3)*(I*sqrt(3)
+ 1)*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3) + 4/(b*d) + 2*(1
/2)^(2/3)*(-I*sqrt(3) + 1)/(b^2*d^2*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b^4)/(a^2*b^
5*d^3))^(1/3)))*d - 3/4*sqrt(1/3)*(((1/2)^(1/3)*(I*sqrt(3) + 1)*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a^2*b^
5*d^3) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3) + 4/(b*d) + 2*(1/2)^(2/3)*(-I*sqrt(3) + 1)/(b^2*d^2*(2/(b^3*d^3) + (
a^4 - 2*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3)))*a^3*b^3*d^2 - 2*(2*a^3*b^2 + a*b^4)*
d)*sqrt(-(((1/2)^(1/3)*(I*sqrt(3) + 1)*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b^4)/(a^2
*b^5*d^3))^(1/3) + 4/(b*d) + 2*(1/2)^(2/3)*(-I*sqrt(3) + 1)/(b^2*d^2*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a
^2*b^5*d^3) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3)))^2*b*d - 8*(1/2)^(1/3)*(I*sqrt(3) + 1)*(2/(b^3*d^3) + (a^4 - 2
*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3) - 32/(b*d) - 16*(1/2)^(2/3)*(-I*sqrt(3) + 1)/
(b^2*d^2*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3)))/(b*d)) - 2*
(a^4 - b^4)*sin(d*x + c)) - (((1/2)^(1/3)*(I*sqrt(3) + 1)*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a^2*b^5*d^3)
- (a^4 - b^4)/(a^2*b^5*d^3))^(1/3) + 4/(b*d) + 2*(1/2)^(2/3)*(-I*sqrt(3) + 1)/(b^2*d^2*(2/(b^3*d^3) + (a^4 -
2*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3)))*b*d - 3*sqrt(1/3)*b*d*sqrt(-(((1/2)^(1/3)*
(I*sqrt(3) + 1)*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3) + 4/(b
*d) + 2*(1/2)^(2/3)*(-I*sqrt(3) + 1)/(b^2*d^2*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b^
4)/(a^2*b^5*d^3))^(1/3)))^2*b*d - 8*(1/2)^(1/3)*(I*sqrt(3) + 1)*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a^2*b^
5*d^3) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3) - 32/(b*d) - 16*(1/2)^(2/3)*(-I*sqrt(3) + 1)/(b^2*d^2*(2/(b^3*d^3) +
(a^4 - 2*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3)))/(b*d)) - 12)*log(-1/4*((1/2)^(1/3)
*(I*sqrt(3) + 1)*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3) + 4/(
b*d) + 2*(1/2)^(2/3)*(-I*sqrt(3) + 1)/(b^2*d^2*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b
^4)/(a^2*b^5*d^3))^(1/3)))^2*a^3*b^3*d^2 - 2*a^3*b + 2*a*b^3 + 1/2*(4*a^3*b^2 - a*b^4)*((1/2)^(1/3)*(I*sqrt(3)
+ 1)*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3) + 4/(b*d) + 2*(1
/2)^(2/3)*(-I*sqrt(3) + 1)/(b^2*d^2*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b^4)/(a^2*b^
5*d^3))^(1/3)))*d - 3/4*sqrt(1/3)*(((1/2)^(1/3)*(I*sqrt(3) + 1)*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a^2*b^
5*d^3) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3) + 4/(b*d) + 2*(1/2)^(2/3)*(-I*sqrt(3) + 1)/(b^2*d^2*(2/(b^3*d^3) + (
a^4 - 2*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3)))*a^3*b^3*d^2 - 2*(2*a^3*b^2 + a*b^4)*
d)*sqrt(-(((1/2)^(1/3)*(I*sqrt(3) + 1)*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b^4)/(a^2
*b^5*d^3))^(1/3) + 4/(b*d) + 2*(1/2)^(2/3)*(-I*sqrt(3) + 1)/(b^2*d^2*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a
^2*b^5*d^3) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3)))^2*b*d - 8*(1/2)^(1/3)*(I*sqrt(3) + 1)*(2/(b^3*d^3) + (a^4 - 2
*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3) - 32/(b*d) - 16*(1/2)^(2/3)*(-I*sqrt(3) + 1)/
(b^2*d^2*(2/(b^3*d^3) + (a^4 - 2*a^2*b^2 + b^4)/(a^2*b^5*d^3) - (a^4 - b^4)/(a^2*b^5*d^3))^(1/3)))/(b*d)) + 2*
(a^4 - b^4)*sin(d*x + c)))/(b*d)

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giac [A]  time = 0.19, size = 221, normalized size = 1.01 \[ \frac {\frac {3 \, \sin \left (d x + c\right )^{2}}{b} - \frac {4 \, \log \left ({\left | b \sin \left (d x + c\right )^{3} + a \right |}\right )}{b} + \frac {2 \, \sqrt {3} {\left (\left (-a b^{2}\right )^{\frac {1}{3}} b^{2} + \left (-a b^{2}\right )^{\frac {2}{3}} a\right )} \arctan \left (\frac {\sqrt {3} {\left (\left (-\frac {a}{b}\right )^{\frac {1}{3}} + 2 \, \sin \left (d x + c\right )\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{a b^{3}} + \frac {{\left (\left (-a b^{2}\right )^{\frac {1}{3}} b^{2} - \left (-a b^{2}\right )^{\frac {2}{3}} a\right )} \log \left (\sin \left (d x + c\right )^{2} + \left (-\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x + c\right ) + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{a b^{3}} + \frac {2 \, {\left (a b^{4} \left (-\frac {a}{b}\right )^{\frac {1}{3}} - b^{5}\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | -\left (-\frac {a}{b}\right )^{\frac {1}{3}} + \sin \left (d x + c\right ) \right |}\right )}{a b^{5}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a+b*sin(d*x+c)^3),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.92, size = 278, normalized size = 1.27 \[ \frac {\sin ^{2}\left (d x +c \right )}{2 b d}+\frac {\ln \left (\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 d b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (\sin ^{2}\left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 d b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 d b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {a \ln \left (\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 d \,b^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}}-\frac {a \ln \left (\sin ^{2}\left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 d \,b^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}}-\frac {a \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 d \,b^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}}-\frac {2 \ln \left (a +b \left (\sin ^{3}\left (d x +c \right )\right )\right )}{3 b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5/(a+b*sin(d*x+c)^3),x)

[Out]

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

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maxima [A]  time = 0.89, size = 210, normalized size = 0.96 \[ \frac {\frac {9 \, \sin \left (d x + c\right )^{2}}{b} - \frac {2 \, \sqrt {3} {\left (a {\left (3 \, \left (\frac {a}{b}\right )^{\frac {2}{3}} - 4\right )} - b {\left (3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}} - \frac {4 \, a}{b}\right )}\right )} \arctan \left (-\frac {\sqrt {3} {\left (\left (\frac {a}{b}\right )^{\frac {1}{3}} - 2 \, \sin \left (d x + c\right )\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{a b} - \frac {3 \, {\left (b {\left (4 \, \left (\frac {a}{b}\right )^{\frac {2}{3}} + 1\right )} + a \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \log \left (\sin \left (d x + c\right )^{2} - \left (\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x + c\right ) + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {6 \, {\left (b {\left (2 \, \left (\frac {a}{b}\right )^{\frac {2}{3}} - 1\right )} - a \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \log \left (\left (\frac {a}{b}\right )^{\frac {1}{3}} + \sin \left (d x + c\right )\right )}{b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}}}{18 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a+b*sin(d*x+c)^3),x, algorithm="maxima")

[Out]

1/18*(9*sin(d*x + c)^2/b - 2*sqrt(3)*(a*(3*(a/b)^(2/3) - 4) - b*(3*(a/b)^(1/3) - 4*a/b))*arctan(-1/3*sqrt(3)*(
(a/b)^(1/3) - 2*sin(d*x + c))/(a/b)^(1/3))/(a*b) - 3*(b*(4*(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))/(b^2*(a/b)^(2/3)) - 6*(b*(2*(a/b)^(2/3) - 1) - a*(a/b)^(1/3))*log(
(a/b)^(1/3) + sin(d*x + c))/(b^2*(a/b)^(2/3)))/d

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mupad [B]  time = 15.02, size = 229, normalized size = 1.05 \[ \frac {\left (\sum _{k=1}^3\ln \left (3\,a+\mathrm {root}\left (27\,a^2\,b^5\,d^3+54\,a^2\,b^4\,d^2+27\,a^2\,b^3\,d+2\,a^2\,b^2-b^4-a^4,d,k\right )\,\left (12\,a\,b+3\,b^2\,\sin \left (c+d\,x\right )+\mathrm {root}\left (27\,a^2\,b^5\,d^3+54\,a^2\,b^4\,d^2+27\,a^2\,b^3\,d+2\,a^2\,b^2-b^4-a^4,d,k\right )\,a\,b^2\,9\right )+\frac {\sin \left (c+d\,x\right )\,\left (a^2+2\,b^2\right )}{b}\right )\,\mathrm {root}\left (27\,a^2\,b^5\,d^3+54\,a^2\,b^4\,d^2+27\,a^2\,b^3\,d+2\,a^2\,b^2-b^4-a^4,d,k\right )\right )+\frac {{\sin \left (c+d\,x\right )}^2}{2\,b}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5/(a + b*sin(c + d*x)^3),x)

[Out]

(symsum(log(3*a + root(27*a^2*b^5*d^3 + 54*a^2*b^4*d^2 + 27*a^2*b^3*d + 2*a^2*b^2 - b^4 - a^4, d, k)*(12*a*b +
3*b^2*sin(c + d*x) + 9*root(27*a^2*b^5*d^3 + 54*a^2*b^4*d^2 + 27*a^2*b^3*d + 2*a^2*b^2 - b^4 - a^4, d, k)*a*b
^2) + (sin(c + d*x)*(a^2 + 2*b^2))/b)*root(27*a^2*b^5*d^3 + 54*a^2*b^4*d^2 + 27*a^2*b^3*d + 2*a^2*b^2 - b^4 -
a^4, d, k), k, 1, 3) + sin(c + d*x)^2/(2*b))/d

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5/(a+b*sin(d*x+c)**3),x)

[Out]

Timed out

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