3.2.85 \(\int \frac {\sin (c+d x)}{a+b \sin ^3(c+d x)} \, dx\) [185]

Optimal. Leaf size=267 \[ \frac {2 (-1)^{2/3} \tan ^{-1}\left (\frac {\sqrt [3]{-1} \sqrt [3]{b}-\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 \sqrt [3]{a} \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}} \sqrt [3]{b} d}-\frac {2 \tan ^{-1}\left (\frac {\sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 \sqrt [3]{a} \sqrt {a^{2/3}-b^{2/3}} \sqrt [3]{b} d}+\frac {2 \sqrt [3]{-1} \tan ^{-1}\left (\frac {(-1)^{2/3} \sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 \sqrt [3]{a} \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}} \sqrt [3]{b} d} \]

[Out]

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

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Rubi [A]
time = 0.19, antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3299, 2739, 632, 210} \begin {gather*} \frac {2 (-1)^{2/3} \text {ArcTan}\left (\frac {\sqrt [3]{-1} \sqrt [3]{b}-\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 \sqrt [3]{a} \sqrt [3]{b} d \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}-\frac {2 \text {ArcTan}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 \sqrt [3]{a} \sqrt [3]{b} d \sqrt {a^{2/3}-b^{2/3}}}+\frac {2 \sqrt [3]{-1} \text {ArcTan}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 \sqrt [3]{a} \sqrt [3]{b} d \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-1)^(2/3)*ArcTan[((-1)^(1/3)*b^(1/3) - a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) - (-1)^(2/3)*b^(2/3)]])/(3*a
^(1/3)*Sqrt[a^(2/3) - (-1)^(2/3)*b^(2/3)]*b^(1/3)*d) - (2*ArcTan[(b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(
2/3) - b^(2/3)]])/(3*a^(1/3)*Sqrt[a^(2/3) - b^(2/3)]*b^(1/3)*d) + (2*(-1)^(1/3)*ArcTan[((-1)^(2/3)*b^(1/3) + a
^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) + (-1)^(1/3)*b^(2/3)]])/(3*a^(1/3)*Sqrt[a^(2/3) + (-1)^(1/3)*b^(2/3)]*b^
(1/3)*d)

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 632

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

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[2*(e/d), Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> Int[ExpandTr
ig[sin[e + f*x]^m*(a + b*sin[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, e, f}, x] && IntegersQ[m, p] && (EqQ[n, 4]
|| GtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 0.12, size = 172, normalized size = 0.64 \begin {gather*} -\frac {\text {RootSum}\left [-i b+3 i b \text {$\#$1}^2+8 a \text {$\#$1}^3-3 i b \text {$\#$1}^4+i b \text {$\#$1}^6\&,\frac {-2 \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )+i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right )+2 \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2}{b-4 i a \text {$\#$1}-2 b \text {$\#$1}^2+b \text {$\#$1}^4}\&\right ]}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*RootSum[(-I)*b + (3*I)*b*#1^2 + 8*a*#1^3 - (3*I)*b*#1^4 + I*b*#1^6 & , (-2*ArcTan[Sin[c + d*x]/(Cos[c + d
*x] - #1)] + I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2] + 2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^2 - I*Log[1 -
2*Cos[c + d*x]*#1 + #1^2]*#1^2)/(b - (4*I)*a*#1 - 2*b*#1^2 + b*#1^4) & ]/d

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.64, size = 78, normalized size = 0.29

method	result	size
derivativedivides	\(\frac {2 \left (\munderset {\textit {\_R} =\RootOf \left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{3}+\textit {\_R} \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 d}\)	\(78\)
default	\(\frac {2 \left (\munderset {\textit {\_R} =\RootOf \left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{3}+\textit {\_R} \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 d}\)	\(78\)
risch	\(-\frac {i \left (\munderset {\textit {\_R} =\RootOf \left (-64+\left (729 a^{4} b^{2} d^{6}-729 a^{2} b^{4} d^{6}\right ) \textit {\_Z}^{6}-972 a^{2} b^{2} d^{4} \textit {\_Z}^{4}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\left (-\frac {243 i d^{5} b^{2} a^{5}}{32 a^{2}+32 b^{2}}+\frac {243 i d^{5} b^{4} a^{3}}{32 a^{2}+32 b^{2}}\right ) \textit {\_R}^{5}+\left (\frac {162 i d^{4} b \,a^{5}}{32 a^{2}+32 b^{2}}-\frac {162 i d^{4} b^{3} a^{3}}{32 a^{2}+32 b^{2}}\right ) \textit {\_R}^{4}+\left (\frac {216 i d^{3} b^{2} a^{3}}{32 a^{2}+32 b^{2}}+\frac {108 i d^{3} b^{4} a}{32 a^{2}+32 b^{2}}\right ) \textit {\_R}^{3}+\left (-\frac {144 i d^{2} b \,a^{3}}{32 a^{2}+32 b^{2}}-\frac {72 i d^{2} b^{3} a}{32 a^{2}+32 b^{2}}\right ) \textit {\_R}^{2}+\left (\frac {48 i d \,a^{3}}{32 a^{2}+32 b^{2}}+\frac {96 i d \,b^{2} a}{32 a^{2}+32 b^{2}}\right ) \textit {\_R} -\frac {32 i a b}{32 a^{2}+32 b^{2}}\right )\right )}{2}\)	\(338\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/d*sum((_R^3+_R)/(_R^5*a+2*_R^3*a+4*_R^2*b+_R*a)*ln(tan(1/2*d*x+1/2*c)-_R),_R=RootOf(_Z^6*a+3*_Z^4*a+8*_Z^3
*b+3*_Z^2*a+a))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains complex when optimal does not.
time = 1.50, size = 18879, normalized size = 70.71 \begin {gather*} \text {too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*sqrt(2/3)*sqrt(1/6)*sqrt(((a^2 - b^2)*(81*(I*sqrt(3) + 1)*(-1/1458/(a^4*b^2*d^6 - a^2*b^4*d^6) - 1/729/(a
^2*d^2 - b^2*d^2)^3 + 1/1458*(a^2 + b^2)/((a^2 - b^2)^2*a^2*b^2*d^6))^(1/3) + 18/(a^2*d^2 - b^2*d^2) + (-I*sqr
t(3) + 1)/((a^2*d^2 - b^2*d^2)^2*(-1/1458/(a^4*b^2*d^6 - a^2*b^4*d^6) - 1/729/(a^2*d^2 - b^2*d^2)^3 + 1/1458*(
a^2 + b^2)/((a^2 - b^2)^2*a^2*b^2*d^6))^(1/3)))*d^2 + 3*sqrt(1/3)*(a^2 - b^2)*d^2*sqrt(-((a^4 - 2*a^2*b^2 + b^
4)*(81*(I*sqrt(3) + 1)*(-1/1458/(a^4*b^2*d^6 - a^2*b^4*d^6) - 1/729/(a^2*d^2 - b^2*d^2)^3 + 1/1458*(a^2 + b^2)
/((a^2 - b^2)^2*a^2*b^2*d^6))^(1/3) + 18/(a^2*d^2 - b^2*d^2) + (-I*sqrt(3) + 1)/((a^2*d^2 - b^2*d^2)^2*(-1/145
8/(a^4*b^2*d^6 - a^2*b^4*d^6) - 1/729/(a^2*d^2 - b^2*d^2)^3 + 1/1458*(a^2 + b^2)/((a^2 - b^2)^2*a^2*b^2*d^6))^
(1/3)))^2*d^4 - 36*(a^2 - b^2)*(81*(I*sqrt(3) + 1)*(-1/1458/(a^4*b^2*d^6 - a^2*b^4*d^6) - 1/729/(a^2*d^2 - b^2
*d^2)^3 + 1/1458*(a^2 + b^2)/((a^2 - b^2)^2*a^2*b^2*d^6))^(1/3) + 18/(a^2*d^2 - b^2*d^2) + (-I*sqrt(3) + 1)/((
a^2*d^2 -  ...

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 16.12, size = 652, normalized size = 2.44 \begin {gather*} \frac {\sum _{k=1}^6\ln \left (-8192\,a^3\,b+{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^2\,a^3\,b^3\,294912+{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^3\,a^4\,b^3\,1548288+{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^4\,a^5\,b^3\,1990656-{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^5\,a^4\,b^5\,7962624+{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^5\,a^6\,b^3\,5971968+65536\,a^2\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )\,a^3\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,196608+{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^2\,a^4\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,294912-{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^3\,a^3\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1769472+{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^3\,a^5\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,221184+{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^4\,a^4\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,2654208-{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^5\,a^5\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1990656\right )\,\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

symsum(log(294912*root(729*a^4*b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^2*d^4 + 1, d, k)^2*a^3*b^3 - 8192*a^3*b +
1548288*root(729*a^4*b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^2*d^4 + 1, d, k)^3*a^4*b^3 + 1990656*root(729*a^4*
b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^2*d^4 + 1, d, k)^4*a^5*b^3 - 7962624*root(729*a^4*b^2*d^6 - 729*a^2*b^4*
d^6 + 243*a^2*b^2*d^4 + 1, d, k)^5*a^4*b^5 + 5971968*root(729*a^4*b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^2*d^4
+ 1, d, k)^5*a^6*b^3 + 65536*a^2*b^2*tan(c/2 + (d*x)/2) + 196608*root(729*a^4*b^2*d^6 - 729*a^2*b^4*d^6 + 243*
a^2*b^2*d^4 + 1, d, k)*a^3*b^2*tan(c/2 + (d*x)/2) + 294912*root(729*a^4*b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^
2*d^4 + 1, d, k)^2*a^4*b^2*tan(c/2 + (d*x)/2) - 1769472*root(729*a^4*b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^2*d
^4 + 1, d, k)^3*a^3*b^4*tan(c/2 + (d*x)/2) + 221184*root(729*a^4*b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^2*d^4 +
1, d, k)^3*a^5*b^2*tan(c/2 + (d*x)/2) + 2654208*root(729*a^4*b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^2*d^4 + 1,
d, k)^4*a^4*b^4*tan(c/2 + (d*x)/2) - 1990656*root(729*a^4*b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^2*d^4 + 1, d,
k)^5*a^5*b^4*tan(c/2 + (d*x)/2))*root(729*a^4*b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^2*d^4 + 1, d, k), k, 1, 6
)/d

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