\(\int \frac {\cos ^5(c+d x)}{(a+b \sin ^3(c+d x))^2} \, dx\) [394]

Optimal result

Integrand size = 23, antiderivative size = 238 \[ \int \frac {\cos ^5(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx=-\frac {2 \left (a^{4/3}+b^{4/3}\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sin (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{5/3} b^{5/3} d}-\frac {2 \left (a^{4/3}-b^{4/3}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{9 a^{5/3} b^{5/3} d}+\frac {\left (a^{4/3}-b^{4/3}\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{9 a^{5/3} b^{5/3} d}+\frac {\sin (c+d x) \left (b-a \sin (c+d x)-2 b \sin ^2(c+d x)\right )}{3 a b d \left (a+b \sin ^3(c+d x)\right )} \]

[Out]

-2/9*(a^(4/3)-b^(4/3))*ln(a^(1/3)+b^(1/3)*sin(d*x+c))/a^(5/3)/b^(5/3)/d+1/9*(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^(5/3)/b^(5/3)/d+1/3*sin(d*x+c)*(b-a*sin(d*x+c)-2*b*sin(d*x+c)^2)
/a/b/d/(a+b*sin(d*x+c)^3)-2/9*(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^(
5/3)/b^(5/3)/d*3^(1/2)

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3302, 1872, 1874, 31, 648, 631, 210, 642} \[ \int \frac {\cos ^5(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx=-\frac {2 \left (a^{4/3}+b^{4/3}\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sin (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{5/3} b^{5/3} d}+\frac {\left (a^{4/3}-b^{4/3}\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{9 a^{5/3} b^{5/3} d}-\frac {2 \left (a^{4/3}-b^{4/3}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{9 a^{5/3} b^{5/3} d}+\frac {\sin (c+d x) \left (-a \sin (c+d x)-2 b \sin ^2(c+d x)+b\right )}{3 a b d \left (a+b \sin ^3(c+d x)\right )} \]

[In]

Int[Cos[c + d*x]^5/(a + b*Sin[c + d*x]^3)^2,x]

[Out]

(-2*(a^(4/3) + b^(4/3))*ArcTan[(a^(1/3) - 2*b^(1/3)*Sin[c + d*x])/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(5/3)*b^(5/
3)*d) - (2*(a^(4/3) - b^(4/3))*Log[a^(1/3) + b^(1/3)*Sin[c + d*x]])/(9*a^(5/3)*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])/(9*a^(5/3)*b^(5/3)*d) + (Sin[c + d*x]
*(b - a*Sin[c + d*x] - 2*b*Sin[c + d*x]^2))/(3*a*b*d*(a + b*Sin[c + d*x]^3))

Rule 31

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

Rule 210

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

Rule 631

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 642

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 648

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 1872

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, Module[{Q = PolynomialQuotient
[b^(Floor[(q - 1)/n] + 1)*Pq, a + b*x^n, x], R = PolynomialRemainder[b^(Floor[(q - 1)/n] + 1)*Pq, a + b*x^n, x
]}, Dist[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), Int[(a + b*x^n)^(p + 1)*ExpandToSum[a*n*(p + 1)*Q + n*(p +
1)*R + D[x*R, x], x], x], x] + Simp[(-x)*R*((a + b*x^n)^(p + 1)/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1))), x]] /
; GeQ[q, n]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1]

Rule 1874

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 3302

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*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (1-x^2\right )^2}{\left (a+b x^3\right )^2} \, dx,x,\sin (c+d x)\right )}{d} \\ & = \frac {\sin (c+d x) \left (b-a \sin (c+d x)-2 b \sin ^2(c+d x)\right )}{3 a b d \left (a+b \sin ^3(c+d x)\right )}-\frac {\text {Subst}\left (\int \frac {-2 b^2-2 a b x}{a+b x^3} \, dx,x,\sin (c+d x)\right )}{3 a b^2 d} \\ & = \frac {\sin (c+d x) \left (b-a \sin (c+d x)-2 b \sin ^2(c+d x)\right )}{3 a b d \left (a+b \sin ^3(c+d x)\right )}-\frac {\text {Subst}\left (\int \frac {\sqrt [3]{a} \left (-2 a^{4/3} b-4 b^{7/3}\right )+\sqrt [3]{b} \left (-2 a^{4/3} b+2 b^{7/3}\right ) x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\sin (c+d x)\right )}{9 a^{5/3} b^{7/3} d}-\frac {\left (2 \left (a^{4/3}-b^{4/3}\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,\sin (c+d x)\right )}{9 a^{5/3} b^{4/3} d} \\ & = -\frac {2 \left (a^{4/3}-b^{4/3}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{9 a^{5/3} b^{5/3} d}+\frac {\sin (c+d x) \left (b-a \sin (c+d x)-2 b \sin ^2(c+d x)\right )}{3 a b d \left (a+b \sin ^3(c+d x)\right )}+\frac {\left (\frac {1}{a^{4/3}}+\frac {1}{b^{4/3}}\right ) \text {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 )}{3 d}+\frac {\left (a^{4/3}-b^{4/3}\right ) \text {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 )}{9 a^{5/3} b^{5/3} d} \\ & = -\frac {2 \left (a^{4/3}-b^{4/3}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{9 a^{5/3} 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 )}{9 a^{5/3} b^{5/3} d}+\frac {\sin (c+d x) \left (b-a \sin (c+d x)-2 b \sin ^2(c+d x)\right )}{3 a b d \left (a+b \sin ^3(c+d x)\right )}+\frac {\left (2 \left (a^{4/3}+b^{4/3}\right )\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} \sin (c+d x)}{\sqrt [3]{a}}\right )}{3 a^{5/3} b^{5/3} d} \\ & = -\frac {2 \left (a^{4/3}+b^{4/3}\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} \sin (c+d x)}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{3 \sqrt {3} a^{5/3} b^{5/3} d}-\frac {2 \left (a^{4/3}-b^{4/3}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{9 a^{5/3} 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 )}{9 a^{5/3} b^{5/3} d}+\frac {\sin (c+d x) \left (b-a \sin (c+d x)-2 b \sin ^2(c+d x)\right )}{3 a b d \left (a+b \sin ^3(c+d x)\right )} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.96 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.08 \[ \int \frac {\cos ^5(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx=\frac {-\frac {4 \sqrt {3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sin (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{a^{5/3} \sqrt [3]{b}}+\frac {4 \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{a^{5/3} \sqrt [3]{b}}-\frac {2 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{a^{5/3} \sqrt [3]{b}}+\frac {9 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},-\frac {b \sin ^3(c+d x)}{a}\right ) \sin ^2(c+d x)}{a b}-\frac {9 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},2,\frac {5}{3},-\frac {b \sin ^3(c+d x)}{a}\right ) \sin ^2(c+d x)}{a b}+\frac {12}{b \left (a+b \sin ^3(c+d x)\right )}+\frac {6 \sin (c+d x)}{a \left (a+b \sin ^3(c+d x)\right )}}{18 d} \]

[In]

Integrate[Cos[c + d*x]^5/(a + b*Sin[c + d*x]^3)^2,x]

[Out]

((-4*Sqrt[3]*ArcTan[(a^(1/3) - 2*b^(1/3)*Sin[c + d*x])/(Sqrt[3]*a^(1/3))])/(a^(5/3)*b^(1/3)) + (4*Log[a^(1/3)
+ b^(1/3)*Sin[c + d*x]])/(a^(5/3)*b^(1/3)) - (2*Log[a^(2/3) - a^(1/3)*b^(1/3)*Sin[c + d*x] + b^(2/3)*Sin[c + d
*x]^2])/(a^(5/3)*b^(1/3)) + (9*Hypergeometric2F1[2/3, 1, 5/3, -((b*Sin[c + d*x]^3)/a)]*Sin[c + d*x]^2)/(a*b) -
(9*Hypergeometric2F1[2/3, 2, 5/3, -((b*Sin[c + d*x]^3)/a)]*Sin[c + d*x]^2)/(a*b) + 12/(b*(a + b*Sin[c + d*x]^
3)) + (6*Sin[c + d*x])/(a*(a + b*Sin[c + d*x]^3)))/(18*d)

Maple [A] (verified)

Time = 1.06 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.19

method	result	size
derivativedivides	\(\frac {\frac {-\frac {\sin ^{2}\left (d x +c \right )}{3 b}+\frac {\sin \left (d x +c \right )}{3 a}+\frac {2}{3 b}}{a +b \left (\sin ^{3}\left (d x +c \right )\right )}+\frac {\frac {2 b \left (\frac {\ln \left (\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (\sin ^{2}\left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )}{3}+\frac {2 a \left (-\frac {\ln \left (\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (\sin ^{2}\left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3}}{a b}}{d}\)	\(284\)
default	\(\frac {\frac {-\frac {\sin ^{2}\left (d x +c \right )}{3 b}+\frac {\sin \left (d x +c \right )}{3 a}+\frac {2}{3 b}}{a +b \left (\sin ^{3}\left (d x +c \right )\right )}+\frac {\frac {2 b \left (\frac {\ln \left (\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (\sin ^{2}\left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )}{3}+\frac {2 a \left (-\frac {\ln \left (\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (\sin ^{2}\left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3}}{a b}}{d}\)	\(284\)
risch	\(-\frac {2 i \left (a \,{\mathrm e}^{5 i \left (d x +c \right )}+6 a \,{\mathrm e}^{3 i \left (d x +c \right )}-2 i b \,{\mathrm e}^{4 i \left (d x +c \right )}+a \,{\mathrm e}^{i \left (d x +c \right )}+2 i b \,{\mathrm e}^{2 i \left (d x +c \right )}\right )}{3 a b d \left (b \,{\mathrm e}^{6 i \left (d x +c \right )}-3 b \,{\mathrm e}^{4 i \left (d x +c \right )}+3 b \,{\mathrm e}^{2 i \left (d x +c \right )}-8 i a \,{\mathrm e}^{3 i \left (d x +c \right )}-b \right )}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (729 a^{5} b^{5} d^{3} \textit {\_Z}^{3}+108 a^{3} b^{3} d \textit {\_Z} +8 a^{4}-8 b^{4}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\left (\frac {81 i a^{5} b^{3} d^{2} \textit {\_R}^{2}}{2 a^{4}+2 b^{4}}+\frac {18 i a^{2} b^{4} d \textit {\_R}}{2 a^{4}+2 b^{4}}+\frac {8 i a^{3} b}{2 a^{4}+2 b^{4}}\right ) {\mathrm e}^{i \left (d x +c \right )}-\frac {2 a^{4}}{2 a^{4}+2 b^{4}}-\frac {2 b^{4}}{2 a^{4}+2 b^{4}}\right )\right )\)	\(300\)

[In]

int(cos(d*x+c)^5/(a+b*sin(d*x+c)^3)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.21 (sec) , antiderivative size = 2255, normalized size of antiderivative = 9.47 \[ \int \frac {\cos ^5(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)^5/(a+b*sin(d*x+c)^3)^2,x, algorithm="fricas")

[Out]

1/36*(12*a*cos(d*x + c)^2 - 2*(a^2*b*d - (a*b^2*d*cos(d*x + c)^2 - a*b^2*d)*sin(d*x + c))*(4^(1/3)*(I*sqrt(3)
+ 1)*((a^4 + b^4)/(a^5*b^5*d^3) - (a^4 - b^4)/(a^5*b^5*d^3))^(1/3) - 4^(2/3)*(-I*sqrt(3) + 1)/(a^2*b^2*d^2*((a
^4 + b^4)/(a^5*b^5*d^3) - (a^4 - b^4)/(a^5*b^5*d^3))^(1/3)))*log(1/4*(4^(1/3)*(I*sqrt(3) + 1)*((a^4 + b^4)/(a^
5*b^5*d^3) - (a^4 - b^4)/(a^5*b^5*d^3))^(1/3) - 4^(2/3)*(-I*sqrt(3) + 1)/(a^2*b^2*d^2*((a^4 + b^4)/(a^5*b^5*d^
3) - (a^4 - b^4)/(a^5*b^5*d^3))^(1/3)))^2*a^5*b^3*d^2 - (4^(1/3)*(I*sqrt(3) + 1)*((a^4 + b^4)/(a^5*b^5*d^3) -
(a^4 - b^4)/(a^5*b^5*d^3))^(1/3) - 4^(2/3)*(-I*sqrt(3) + 1)/(a^2*b^2*d^2*((a^4 + b^4)/(a^5*b^5*d^3) - (a^4 - b
^4)/(a^5*b^5*d^3))^(1/3)))*a^2*b^4*d + 8*a^3*b + 4*(a^4 + b^4)*sin(d*x + c)) + ((a^2*b*d - (a*b^2*d*cos(d*x +
c)^2 - a*b^2*d)*sin(d*x + c))*(4^(1/3)*(I*sqrt(3) + 1)*((a^4 + b^4)/(a^5*b^5*d^3) - (a^4 - b^4)/(a^5*b^5*d^3))
^(1/3) - 4^(2/3)*(-I*sqrt(3) + 1)/(a^2*b^2*d^2*((a^4 + b^4)/(a^5*b^5*d^3) - (a^4 - b^4)/(a^5*b^5*d^3))^(1/3)))
+ 3*sqrt(1/3)*(a^2*b*d - (a*b^2*d*cos(d*x + c)^2 - a*b^2*d)*sin(d*x + c))*sqrt(-((4^(1/3)*(I*sqrt(3) + 1)*((a
^4 + b^4)/(a^5*b^5*d^3) - (a^4 - b^4)/(a^5*b^5*d^3))^(1/3) - 4^(2/3)*(-I*sqrt(3) + 1)/(a^2*b^2*d^2*((a^4 + b^4
)/(a^5*b^5*d^3) - (a^4 - b^4)/(a^5*b^5*d^3))^(1/3)))^2*a^2*b^2*d^2 + 64)/(a^2*b^2*d^2)))*log(1/4*(4^(1/3)*(I*s
qrt(3) + 1)*((a^4 + b^4)/(a^5*b^5*d^3) - (a^4 - b^4)/(a^5*b^5*d^3))^(1/3) - 4^(2/3)*(-I*sqrt(3) + 1)/(a^2*b^2*
d^2*((a^4 + b^4)/(a^5*b^5*d^3) - (a^4 - b^4)/(a^5*b^5*d^3))^(1/3)))^2*a^5*b^3*d^2 - (4^(1/3)*(I*sqrt(3) + 1)*(
(a^4 + b^4)/(a^5*b^5*d^3) - (a^4 - b^4)/(a^5*b^5*d^3))^(1/3) - 4^(2/3)*(-I*sqrt(3) + 1)/(a^2*b^2*d^2*((a^4 + b
^4)/(a^5*b^5*d^3) - (a^4 - b^4)/(a^5*b^5*d^3))^(1/3)))*a^2*b^4*d + 8*a^3*b - 3/4*sqrt(1/3)*((4^(1/3)*(I*sqrt(3
) + 1)*((a^4 + b^4)/(a^5*b^5*d^3) - (a^4 - b^4)/(a^5*b^5*d^3))^(1/3) - 4^(2/3)*(-I*sqrt(3) + 1)/(a^2*b^2*d^2*(
(a^4 + b^4)/(a^5*b^5*d^3) - (a^4 - b^4)/(a^5*b^5*d^3))^(1/3)))*a^5*b^3*d^2 + 4*a^2*b^4*d)*sqrt(-((4^(1/3)*(I*s
qrt(3) + 1)*((a^4 + b^4)/(a^5*b^5*d^3) - (a^4 - b^4)/(a^5*b^5*d^3))^(1/3) - 4^(2/3)*(-I*sqrt(3) + 1)/(a^2*b^2*
d^2*((a^4 + b^4)/(a^5*b^5*d^3) - (a^4 - b^4)/(a^5*b^5*d^3))^(1/3)))^2*a^2*b^2*d^2 + 64)/(a^2*b^2*d^2)) - 8*(a^
4 + b^4)*sin(d*x + c)) + ((a^2*b*d - (a*b^2*d*cos(d*x + c)^2 - a*b^2*d)*sin(d*x + c))*(4^(1/3)*(I*sqrt(3) + 1)
*((a^4 + b^4)/(a^5*b^5*d^3) - (a^4 - b^4)/(a^5*b^5*d^3))^(1/3) - 4^(2/3)*(-I*sqrt(3) + 1)/(a^2*b^2*d^2*((a^4 +
b^4)/(a^5*b^5*d^3) - (a^4 - b^4)/(a^5*b^5*d^3))^(1/3))) - 3*sqrt(1/3)*(a^2*b*d - (a*b^2*d*cos(d*x + c)^2 - a*
b^2*d)*sin(d*x + c))*sqrt(-((4^(1/3)*(I*sqrt(3) + 1)*((a^4 + b^4)/(a^5*b^5*d^3) - (a^4 - b^4)/(a^5*b^5*d^3))^(
1/3) - 4^(2/3)*(-I*sqrt(3) + 1)/(a^2*b^2*d^2*((a^4 + b^4)/(a^5*b^5*d^3) - (a^4 - b^4)/(a^5*b^5*d^3))^(1/3)))^2
*a^2*b^2*d^2 + 64)/(a^2*b^2*d^2)))*log(-1/4*(4^(1/3)*(I*sqrt(3) + 1)*((a^4 + b^4)/(a^5*b^5*d^3) - (a^4 - b^4)/
(a^5*b^5*d^3))^(1/3) - 4^(2/3)*(-I*sqrt(3) + 1)/(a^2*b^2*d^2*((a^4 + b^4)/(a^5*b^5*d^3) - (a^4 - b^4)/(a^5*b^5
*d^3))^(1/3)))^2*a^5*b^3*d^2 + (4^(1/3)*(I*sqrt(3) + 1)*((a^4 + b^4)/(a^5*b^5*d^3) - (a^4 - b^4)/(a^5*b^5*d^3)
)^(1/3) - 4^(2/3)*(-I*sqrt(3) + 1)/(a^2*b^2*d^2*((a^4 + b^4)/(a^5*b^5*d^3) - (a^4 - b^4)/(a^5*b^5*d^3))^(1/3))
)*a^2*b^4*d - 8*a^3*b - 3/4*sqrt(1/3)*((4^(1/3)*(I*sqrt(3) + 1)*((a^4 + b^4)/(a^5*b^5*d^3) - (a^4 - b^4)/(a^5*
b^5*d^3))^(1/3) - 4^(2/3)*(-I*sqrt(3) + 1)/(a^2*b^2*d^2*((a^4 + b^4)/(a^5*b^5*d^3) - (a^4 - b^4)/(a^5*b^5*d^3)
)^(1/3)))*a^5*b^3*d^2 + 4*a^2*b^4*d)*sqrt(-((4^(1/3)*(I*sqrt(3) + 1)*((a^4 + b^4)/(a^5*b^5*d^3) - (a^4 - b^4)/
(a^5*b^5*d^3))^(1/3) - 4^(2/3)*(-I*sqrt(3) + 1)/(a^2*b^2*d^2*((a^4 + b^4)/(a^5*b^5*d^3) - (a^4 - b^4)/(a^5*b^5
*d^3))^(1/3)))^2*a^2*b^2*d^2 + 64)/(a^2*b^2*d^2)) + 8*(a^4 + b^4)*sin(d*x + c)) + 12*b*sin(d*x + c) + 12*a)/(a
^2*b*d - (a*b^2*d*cos(d*x + c)^2 - a*b^2*d)*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^5(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**5/(a+b*sin(d*x+c)**3)**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^5(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx=-\frac {\frac {3 \, {\left (a \sin \left (d x + c\right )^{2} - b \sin \left (d x + c\right ) - 2 \, a\right )}}{a b^{2} \sin \left (d x + c\right )^{3} + a^{2} b} - \frac {2 \, \sqrt {3} {\left (a \left (\frac {a}{b}\right )^{\frac {1}{3}} + b\right )} \arctan \left (-\frac {\sqrt {3} {\left (\left (\frac {a}{b}\right )^{\frac {1}{3}} - 2 \, \sin \left (d x + c\right )\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{a b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {{\left (a \left (\frac {a}{b}\right )^{\frac {1}{3}} - b\right )} \log \left (\sin \left (d x + c\right )^{2} - \left (\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x + c\right ) + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{a b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {2 \, {\left (a \left (\frac {a}{b}\right )^{\frac {1}{3}} - b\right )} \log \left (\left (\frac {a}{b}\right )^{\frac {1}{3}} + \sin \left (d x + c\right )\right )}{a b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}}}{9 \, d} \]

[In]

integrate(cos(d*x+c)^5/(a+b*sin(d*x+c)^3)^2,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {\cos ^5(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx=\int { \frac {\cos \left (d x + c\right )^{5}}{{\left (b \sin \left (d x + c\right )^{3} + a\right )}^{2}} \,d x } \]

[In]

integrate(cos(d*x+c)^5/(a+b*sin(d*x+c)^3)^2,x, algorithm="giac")

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 13.70 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^5(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx=\frac {\sum _{k=1}^3\ln \left (\frac {4\,b+4\,a\,\sin \left (c+d\,x\right )+{\mathrm {root}\left (729\,a^5\,b^5\,d^3+108\,a^3\,b^3\,d-8\,b^4+8\,a^4,d,k\right )}^2\,a^2\,b^3\,81+\mathrm {root}\left (729\,a^5\,b^5\,d^3+108\,a^3\,b^3\,d-8\,b^4+8\,a^4,d,k\right )\,b^3\,\sin \left (c+d\,x\right )\,18}{a\,b\,9}\right )\,\mathrm {root}\left (729\,a^5\,b^5\,d^3+108\,a^3\,b^3\,d-8\,b^4+8\,a^4,d,k\right )}{d}+\frac {\frac {\sin \left (c+d\,x\right )}{3\,a}+\frac {2}{3\,b}-\frac {{\sin \left (c+d\,x\right )}^2}{3\,b}}{d\,\left (b\,{\sin \left (c+d\,x\right )}^3+a\right )} \]

[In]

int(cos(c + d*x)^5/(a + b*sin(c + d*x)^3)^2,x)

[Out]

symsum(log((4*b + 4*a*sin(c + d*x) + 81*root(729*a^5*b^5*d^3 + 108*a^3*b^3*d - 8*b^4 + 8*a^4, d, k)^2*a^2*b^3
+ 18*root(729*a^5*b^5*d^3 + 108*a^3*b^3*d - 8*b^4 + 8*a^4, d, k)*b^3*sin(c + d*x))/(9*a*b))*root(729*a^5*b^5*d
^3 + 108*a^3*b^3*d - 8*b^4 + 8*a^4, d, k), k, 1, 3)/d + (sin(c + d*x)/(3*a) + 2/(3*b) - sin(c + d*x)^2/(3*b))/
(d*(a + b*sin(c + d*x)^3))