3.4.95 \(\int \frac {\cos ^3(c+d x)}{(a+b \sin ^3(c+d x))^2} \, dx\) [395]

Optimal. Leaf size=183 \[ -\frac {2 \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sin (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{5/3} \sqrt [3]{b} d}+\frac {2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{9 a^{5/3} \sqrt [3]{b} d}-\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{9 a^{5/3} \sqrt [3]{b} d}+\frac {a+b \sin (c+d x)}{3 a b d \left (a+b \sin ^3(c+d x)\right )} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3302, 1868, 12, 206, 31, 648, 631, 210, 642} \begin {gather*} -\frac {2 \text {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} \sqrt [3]{b} d}-\frac {\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} \sqrt [3]{b} d}+\frac {2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{9 a^{5/3} \sqrt [3]{b} d}+\frac {a+b \sin (c+d x)}{3 a b d \left (a+b \sin ^3(c+d x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 31

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

Rule 206

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], 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 1868

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

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

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Mathematica [A]
time = 0.52, size = 184, normalized size = 1.01 \begin {gather*} \frac {-\frac {2 \sqrt {3} \tan ^{-1}\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 {3 \sin (c+d x)}{a \left (a+b \sin ^3(c+d x)\right )}+\frac {\frac {2 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{a^{5/3}}-\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 )}{a^{5/3}}+\frac {3}{a+b \sin ^3(c+d x)}}{b}}{9 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.96, size = 165, normalized size = 0.90

method	result	size
risch	\(-\frac {4 \left (b \,{\mathrm e}^{4 i \left (d x +c \right )}-b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 i a \,{\mathrm e}^{3 i \left (d x +c \right )}\right )}{3 a b d \left (b \,{\mathrm e}^{6 i \left (d x +c \right )}-3 b \,{\mathrm e}^{4 i \left (d x +c \right )}+3 b \,{\mathrm e}^{2 i \left (d x +c \right )}-8 i a \,{\mathrm e}^{3 i \left (d x +c \right )}-b \right )}+\left (\munderset {\textit {\_R} =\RootOf \left (729 a^{5} b \,d^{3} \textit {\_Z}^{3}-8\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+9 i a^{2} d \textit {\_R} \,{\mathrm e}^{i \left (d x +c \right )}-1\right )\right )\)	\(153\)
derivativedivides	\(\frac {\frac {\sin \left (d x +c \right )}{3 a \left (a +b \left (\sin ^{3}\left (d x +c \right )\right )\right )}+\frac {\frac {2 \ln \left (\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (\sin ^{2}\left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}}{a}+\frac {1}{3 b \left (a +b \left (\sin ^{3}\left (d x +c \right )\right )\right )}}{d}\)	\(165\)
default	\(\frac {\frac {\sin \left (d x +c \right )}{3 a \left (a +b \left (\sin ^{3}\left (d x +c \right )\right )\right )}+\frac {\frac {2 \ln \left (\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (\sin ^{2}\left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}}{a}+\frac {1}{3 b \left (a +b \left (\sin ^{3}\left (d x +c \right )\right )\right )}}{d}\)	\(165\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(1/3*sin(d*x+c)/a/(a+b*sin(d*x+c)^3)+2/3/a*(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)))+1/3/b/(a+b*sin(d*x+c)^3))

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Maxima [A]
time = 0.53, size = 163, normalized size = 0.89 \begin {gather*} \frac {\frac {3 \, {\left (b \sin \left (d x + c\right ) + a\right )}}{a b^{2} \sin \left (d x + c\right )^{3} + a^{2} b} + \frac {2 \, \sqrt {3} \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 \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {\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 \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {2 \, \log \left (\left (\frac {a}{b}\right )^{\frac {1}{3}} + \sin \left (d x + c\right )\right )}{a b \left (\frac {a}{b}\right )^{\frac {2}{3}}}}{9 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.49, size = 665, normalized size = 3.63 \begin {gather*} \left [\frac {3 \, a^{2} b \sin \left (d x + c\right ) + 3 \, a^{3} + 3 \, \sqrt {\frac {1}{3}} {\left (a^{2} b - {\left (a b^{2} \cos \left (d x + c\right )^{2} - a b^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {-\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} \log \left (-\frac {3 \, \left (a^{2} b\right )^{\frac {1}{3}} a \sin \left (d x + c\right ) + a^{2} + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, a b \cos \left (d x + c\right )^{2} - 2 \, a b - \left (a^{2} b\right )^{\frac {2}{3}} \sin \left (d x + c\right ) + \left (a^{2} b\right )^{\frac {1}{3}} a\right )} \sqrt {-\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} + 2 \, {\left (a b \cos \left (d x + c\right )^{2} - a b\right )} \sin \left (d x + c\right )}{{\left (b \cos \left (d x + c\right )^{2} - b\right )} \sin \left (d x + c\right ) - a}\right ) + \left (a^{2} b\right )^{\frac {2}{3}} {\left ({\left (b \cos \left (d x + c\right )^{2} - b\right )} \sin \left (d x + c\right ) - a\right )} \log \left (-a b \cos \left (d x + c\right )^{2} + a b - \left (a^{2} b\right )^{\frac {2}{3}} \sin \left (d x + c\right ) + \left (a^{2} b\right )^{\frac {1}{3}} a\right ) - 2 \, \left (a^{2} b\right )^{\frac {2}{3}} {\left ({\left (b \cos \left (d x + c\right )^{2} - b\right )} \sin \left (d x + c\right ) - a\right )} \log \left (a b \sin \left (d x + c\right ) + \left (a^{2} b\right )^{\frac {2}{3}}\right )}{9 \, {\left (a^{4} b d - {\left (a^{3} b^{2} d \cos \left (d x + c\right )^{2} - a^{3} b^{2} d\right )} \sin \left (d x + c\right )\right )}}, \frac {3 \, a^{2} b \sin \left (d x + c\right ) + 3 \, a^{3} + 6 \, \sqrt {\frac {1}{3}} {\left (a^{2} b - {\left (a b^{2} \cos \left (d x + c\right )^{2} - a b^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, \left (a^{2} b\right )^{\frac {2}{3}} \sin \left (d x + c\right ) - \left (a^{2} b\right )^{\frac {1}{3}} a\right )} \sqrt {\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}}}{a^{2}}\right ) + \left (a^{2} b\right )^{\frac {2}{3}} {\left ({\left (b \cos \left (d x + c\right )^{2} - b\right )} \sin \left (d x + c\right ) - a\right )} \log \left (-a b \cos \left (d x + c\right )^{2} + a b - \left (a^{2} b\right )^{\frac {2}{3}} \sin \left (d x + c\right ) + \left (a^{2} b\right )^{\frac {1}{3}} a\right ) - 2 \, \left (a^{2} b\right )^{\frac {2}{3}} {\left ({\left (b \cos \left (d x + c\right )^{2} - b\right )} \sin \left (d x + c\right ) - a\right )} \log \left (a b \sin \left (d x + c\right ) + \left (a^{2} b\right )^{\frac {2}{3}}\right )}{9 \, {\left (a^{4} b d - {\left (a^{3} b^{2} d \cos \left (d x + c\right )^{2} - a^{3} b^{2} d\right )} \sin \left (d x + c\right )\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/9*(3*a^2*b*sin(d*x + c) + 3*a^3 + 3*sqrt(1/3)*(a^2*b - (a*b^2*cos(d*x + c)^2 - a*b^2)*sin(d*x + c))*sqrt(-(
a^2*b)^(1/3)/b)*log(-(3*(a^2*b)^(1/3)*a*sin(d*x + c) + a^2 + 3*sqrt(1/3)*(2*a*b*cos(d*x + c)^2 - 2*a*b - (a^2*
b)^(2/3)*sin(d*x + c) + (a^2*b)^(1/3)*a)*sqrt(-(a^2*b)^(1/3)/b) + 2*(a*b*cos(d*x + c)^2 - a*b)*sin(d*x + c))/(
(b*cos(d*x + c)^2 - b)*sin(d*x + c) - a)) + (a^2*b)^(2/3)*((b*cos(d*x + c)^2 - b)*sin(d*x + c) - a)*log(-a*b*c
os(d*x + c)^2 + a*b - (a^2*b)^(2/3)*sin(d*x + c) + (a^2*b)^(1/3)*a) - 2*(a^2*b)^(2/3)*((b*cos(d*x + c)^2 - b)*
sin(d*x + c) - a)*log(a*b*sin(d*x + c) + (a^2*b)^(2/3)))/(a^4*b*d - (a^3*b^2*d*cos(d*x + c)^2 - a^3*b^2*d)*sin
(d*x + c)), 1/9*(3*a^2*b*sin(d*x + c) + 3*a^3 + 6*sqrt(1/3)*(a^2*b - (a*b^2*cos(d*x + c)^2 - a*b^2)*sin(d*x +
c))*sqrt((a^2*b)^(1/3)/b)*arctan(sqrt(1/3)*(2*(a^2*b)^(2/3)*sin(d*x + c) - (a^2*b)^(1/3)*a)*sqrt((a^2*b)^(1/3)
/b)/a^2) + (a^2*b)^(2/3)*((b*cos(d*x + c)^2 - b)*sin(d*x + c) - a)*log(-a*b*cos(d*x + c)^2 + a*b - (a^2*b)^(2/
3)*sin(d*x + c) + (a^2*b)^(1/3)*a) - 2*(a^2*b)^(2/3)*((b*cos(d*x + c)^2 - b)*sin(d*x + c) - a)*log(a*b*sin(d*x
+ c) + (a^2*b)^(2/3)))/(a^4*b*d - (a^3*b^2*d*cos(d*x + c)^2 - a^3*b^2*d)*sin(d*x + c))]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 0.48, size = 169, normalized size = 0.92 \begin {gather*} -\frac {\frac {2 \, \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^{2}} - \frac {2 \, \sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} \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^{2} b} - \frac {\left (-a b^{2}\right )^{\frac {1}{3}} \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} b} - \frac {3 \, {\left (b \sin \left (d x + c\right ) + a\right )}}{{\left (b \sin \left (d x + c\right )^{3} + a\right )} a b}}{9 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.38, size = 172, normalized size = 0.94 \begin {gather*} \frac {\frac {\sin \left (c+d\,x\right )}{3\,a}+\frac {1}{3\,b}}{d\,\left (b\,{\sin \left (c+d\,x\right )}^3+a\right )}+\frac {2\,\ln \left (\frac {2\,b^{5/3}}{a^{2/3}}+\frac {2\,b^2\,\sin \left (c+d\,x\right )}{a}\right )}{9\,a^{5/3}\,b^{1/3}\,d}+\frac {\ln \left (\frac {2\,b^2\,\sin \left (c+d\,x\right )}{a}+\frac {b^{5/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{a^{2/3}}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{9\,a^{5/3}\,b^{1/3}\,d}-\frac {\ln \left (\frac {2\,b^2\,\sin \left (c+d\,x\right )}{a}-\frac {b^{5/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{a^{2/3}}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{9\,a^{5/3}\,b^{1/3}\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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