3.2.97 \(\int \frac {\sin ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx\) [197]

Optimal. Leaf size=138 \[ -\frac {\sqrt {a} \tan ^{-1}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{2 \sqrt {\sqrt {a}-\sqrt {b}} b^{5/4} d}-\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{2 \sqrt {\sqrt {a}+\sqrt {b}} b^{5/4} d}+\frac {\cos (c+d x)}{b d} \]

[Out]

cos(d*x+c)/b/d-1/2*arctan(b^(1/4)*cos(d*x+c)/(a^(1/2)-b^(1/2))^(1/2))*a^(1/2)/b^(5/4)/d/(a^(1/2)-b^(1/2))^(1/2
)-1/2*arctanh(b^(1/4)*cos(d*x+c)/(a^(1/2)+b^(1/2))^(1/2))*a^(1/2)/b^(5/4)/d/(a^(1/2)+b^(1/2))^(1/2)

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Rubi [A]
time = 0.13, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3294, 1184, 1107, 211, 214} \begin {gather*} -\frac {\sqrt {a} \text {ArcTan}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{2 b^{5/4} d \sqrt {\sqrt {a}-\sqrt {b}}}-\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{2 b^{5/4} d \sqrt {\sqrt {a}+\sqrt {b}}}+\frac {\cos (c+d x)}{b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^5/(a - b*Sin[c + d*x]^4),x]

[Out]

-1/2*(Sqrt[a]*ArcTan[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] - Sqrt[b]]])/(Sqrt[Sqrt[a] - Sqrt[b]]*b^(5/4)*d) - (S
qrt[a]*ArcTanh[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] + Sqrt[b]]])/(2*Sqrt[Sqrt[a] + Sqrt[b]]*b^(5/4)*d) + Cos[c
+ d*x]/(b*d)

Rule 211

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

Rule 214

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

Rule 1107

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

Rule 1184

Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[(d + e*x
^2)^q/(a + b*x^2 + c*x^4), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a
*e^2, 0] && IntegerQ[q]

Rule 3294

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, Dist[-ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - 2*b*ff^2*x^2 + b*ff^4*x^4
)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {\sin ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx &=-\frac {\text {Subst}\left (\int \frac {\left (1-x^2\right )^2}{a-b+2 b x^2-b x^4} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {\text {Subst}\left (\int \left (-\frac {1}{b}+\frac {a}{b \left (a-b+2 b x^2-b x^4\right )}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {\cos (c+d x)}{b d}-\frac {a \text {Subst}\left (\int \frac {1}{a-b+2 b x^2-b x^4} \, dx,x,\cos (c+d x)\right )}{b d}\\ &=\frac {\cos (c+d x)}{b d}+\frac {\sqrt {a} \text {Subst}\left (\int \frac {1}{-\sqrt {a} \sqrt {b}+b-b x^2} \, dx,x,\cos (c+d x)\right )}{2 \sqrt {b} d}-\frac {\sqrt {a} \text {Subst}\left (\int \frac {1}{\sqrt {a} \sqrt {b}+b-b x^2} \, dx,x,\cos (c+d x)\right )}{2 \sqrt {b} d}\\ &=-\frac {\sqrt {a} \tan ^{-1}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{2 \sqrt {\sqrt {a}-\sqrt {b}} b^{5/4} d}-\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{2 \sqrt {\sqrt {a}+\sqrt {b}} b^{5/4} d}+\frac {\cos (c+d x)}{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 = 0.17, size = 198, normalized size = 1.43 \begin {gather*} \frac {2 \cos (c+d x)+i a \text {RootSum}\left [b-4 b \text {$\#$1}^2-16 a \text {$\#$1}^4+6 b \text {$\#$1}^4-4 b \text {$\#$1}^6+b \text {$\#$1}^8\&,\frac {-2 \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}+i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}+2 \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^3-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^3}{-b-8 a \text {$\#$1}^2+3 b \text {$\#$1}^2-3 b \text {$\#$1}^4+b \text {$\#$1}^6}\&\right ]}{2 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^5/(a - b*Sin[c + d*x]^4),x]

[Out]

(2*Cos[c + d*x] + I*a*RootSum[b - 4*b*#1^2 - 16*a*#1^4 + 6*b*#1^4 - 4*b*#1^6 + b*#1^8 & , (-2*ArcTan[Sin[c + d
*x]/(Cos[c + d*x] - #1)]*#1 + I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1 + 2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #
1)]*#1^3 - I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^3)/(-b - 8*a*#1^2 + 3*b*#1^2 - 3*b*#1^4 + b*#1^6) & ])/(2*b*
d)

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Maple [A]
time = 0.50, size = 99, normalized size = 0.72

method	result	size
derivativedivides	\(\frac {\frac {\cos \left (d x +c \right )}{b}+a \left (-\frac {\arctanh \left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}+b \right ) b}}-\frac {\arctan \left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{d}\)	\(99\)
default	\(\frac {\frac {\cos \left (d x +c \right )}{b}+a \left (-\frac {\arctanh \left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}+b \right ) b}}-\frac {\arctan \left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{d}\)	\(99\)
risch	\(\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 b d}+\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 b d}-\frac {i \left (\munderset {\textit {\_R} =\RootOf \left (\left (a \,b^{5} d^{4}-b^{6} d^{4}\right ) \textit {\_Z}^{4}-128 a \,d^{2} \textit {\_Z}^{2} b^{3}-4096 a^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\left (\left (-\frac {i b^{4} d^{3}}{256 a}+\frac {i b^{5} d^{3}}{256 a^{2}}\right ) \textit {\_R}^{3}+\left (\frac {i b d}{4}+\frac {i b^{2} d}{4 a}\right ) \textit {\_R} \right ) {\mathrm e}^{i \left (d x +c \right )}+1\right )\right )}{32}\)	\(154\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^5/(a-b*sin(d*x+c)^4),x,method=_RETURNVERBOSE)

[Out]

1/d*(1/b*cos(d*x+c)+a*(-1/2/(a*b)^(1/2)/(((a*b)^(1/2)+b)*b)^(1/2)*arctanh(b*cos(d*x+c)/(((a*b)^(1/2)+b)*b)^(1/
2))-1/2/(a*b)^(1/2)/(((a*b)^(1/2)-b)*b)^(1/2)*arctan(b*cos(d*x+c)/(((a*b)^(1/2)-b)*b)^(1/2))))

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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)^5/(a-b*sin(d*x+c)^4),x, algorithm="maxima")

[Out]

(b*d*integrate(8*(4*a*b*cos(3*d*x + 3*c)*sin(2*d*x + 2*c) + 2*(8*a^2 - 3*a*b)*cos(3*d*x + 3*c)*sin(4*d*x + 4*c
) - 2*(8*a^2 - 3*a*b)*cos(4*d*x + 4*c)*sin(3*d*x + 3*c) - (a*b*sin(5*d*x + 5*c) - a*b*sin(3*d*x + 3*c))*cos(8*
d*x + 8*c) + 4*(a*b*sin(5*d*x + 5*c) - a*b*sin(3*d*x + 3*c))*cos(6*d*x + 6*c) - 2*(2*a*b*sin(2*d*x + 2*c) + (8
*a^2 - 3*a*b)*sin(4*d*x + 4*c))*cos(5*d*x + 5*c) + (a*b*cos(5*d*x + 5*c) - a*b*cos(3*d*x + 3*c))*sin(8*d*x + 8
*c) - 4*(a*b*cos(5*d*x + 5*c) - a*b*cos(3*d*x + 3*c))*sin(6*d*x + 6*c) + (4*a*b*cos(2*d*x + 2*c) - a*b + 2*(8*
a^2 - 3*a*b)*cos(4*d*x + 4*c))*sin(5*d*x + 5*c) - (4*a*b*cos(2*d*x + 2*c) - a*b)*sin(3*d*x + 3*c))/(b^3*cos(8*
d*x + 8*c)^2 + 16*b^3*cos(6*d*x + 6*c)^2 + 16*b^3*cos(2*d*x + 2*c)^2 + b^3*sin(8*d*x + 8*c)^2 + 16*b^3*sin(6*d
*x + 6*c)^2 + 16*b^3*sin(2*d*x + 2*c)^2 - 8*b^3*cos(2*d*x + 2*c) + b^3 + 4*(64*a^2*b - 48*a*b^2 + 9*b^3)*cos(4
*d*x + 4*c)^2 + 4*(64*a^2*b - 48*a*b^2 + 9*b^3)*sin(4*d*x + 4*c)^2 + 16*(8*a*b^2 - 3*b^3)*sin(4*d*x + 4*c)*sin
(2*d*x + 2*c) - 2*(4*b^3*cos(6*d*x + 6*c) + 4*b^3*cos(2*d*x + 2*c) - b^3 + 2*(8*a*b^2 - 3*b^3)*cos(4*d*x + 4*c
))*cos(8*d*x + 8*c) + 8*(4*b^3*cos(2*d*x + 2*c) - b^3 + 2*(8*a*b^2 - 3*b^3)*cos(4*d*x + 4*c))*cos(6*d*x + 6*c)
- 4*(8*a*b^2 - 3*b^3 - 4*(8*a*b^2 - 3*b^3)*cos(2*d*x + 2*c))*cos(4*d*x + 4*c) - 4*(2*b^3*sin(6*d*x + 6*c) + 2
*b^3*sin(2*d*x + 2*c) + (8*a*b^2 - 3*b^3)*sin(4*d*x + 4*c))*sin(8*d*x + 8*c) + 16*(2*b^3*sin(2*d*x + 2*c) + (8
*a*b^2 - 3*b^3)*sin(4*d*x + 4*c))*sin(6*d*x + 6*c)), x) + cos(d*x + c))/(b*d)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 815 vs. \(2 (98) = 196\).
time = 0.47, size = 815, normalized size = 5.91 \begin {gather*} -\frac {b d \sqrt {-\frac {{\left (a b^{2} - b^{3}\right )} d^{2} \sqrt {\frac {a^{3}}{{\left (a^{2} b^{5} - 2 \, a b^{6} + b^{7}\right )} d^{4}}} + a}{{\left (a b^{2} - b^{3}\right )} d^{2}}} \log \left (a^{2} \cos \left (d x + c\right ) - {\left ({\left (a b^{4} - b^{5}\right )} d^{3} \sqrt {\frac {a^{3}}{{\left (a^{2} b^{5} - 2 \, a b^{6} + b^{7}\right )} d^{4}}} - a^{2} b d\right )} \sqrt {-\frac {{\left (a b^{2} - b^{3}\right )} d^{2} \sqrt {\frac {a^{3}}{{\left (a^{2} b^{5} - 2 \, a b^{6} + b^{7}\right )} d^{4}}} + a}{{\left (a b^{2} - b^{3}\right )} d^{2}}}\right ) - b d \sqrt {\frac {{\left (a b^{2} - b^{3}\right )} d^{2} \sqrt {\frac {a^{3}}{{\left (a^{2} b^{5} - 2 \, a b^{6} + b^{7}\right )} d^{4}}} - a}{{\left (a b^{2} - b^{3}\right )} d^{2}}} \log \left (a^{2} \cos \left (d x + c\right ) - {\left ({\left (a b^{4} - b^{5}\right )} d^{3} \sqrt {\frac {a^{3}}{{\left (a^{2} b^{5} - 2 \, a b^{6} + b^{7}\right )} d^{4}}} + a^{2} b d\right )} \sqrt {\frac {{\left (a b^{2} - b^{3}\right )} d^{2} \sqrt {\frac {a^{3}}{{\left (a^{2} b^{5} - 2 \, a b^{6} + b^{7}\right )} d^{4}}} - a}{{\left (a b^{2} - b^{3}\right )} d^{2}}}\right ) - b d \sqrt {-\frac {{\left (a b^{2} - b^{3}\right )} d^{2} \sqrt {\frac {a^{3}}{{\left (a^{2} b^{5} - 2 \, a b^{6} + b^{7}\right )} d^{4}}} + a}{{\left (a b^{2} - b^{3}\right )} d^{2}}} \log \left (-a^{2} \cos \left (d x + c\right ) - {\left ({\left (a b^{4} - b^{5}\right )} d^{3} \sqrt {\frac {a^{3}}{{\left (a^{2} b^{5} - 2 \, a b^{6} + b^{7}\right )} d^{4}}} - a^{2} b d\right )} \sqrt {-\frac {{\left (a b^{2} - b^{3}\right )} d^{2} \sqrt {\frac {a^{3}}{{\left (a^{2} b^{5} - 2 \, a b^{6} + b^{7}\right )} d^{4}}} + a}{{\left (a b^{2} - b^{3}\right )} d^{2}}}\right ) + b d \sqrt {\frac {{\left (a b^{2} - b^{3}\right )} d^{2} \sqrt {\frac {a^{3}}{{\left (a^{2} b^{5} - 2 \, a b^{6} + b^{7}\right )} d^{4}}} - a}{{\left (a b^{2} - b^{3}\right )} d^{2}}} \log \left (-a^{2} \cos \left (d x + c\right ) - {\left ({\left (a b^{4} - b^{5}\right )} d^{3} \sqrt {\frac {a^{3}}{{\left (a^{2} b^{5} - 2 \, a b^{6} + b^{7}\right )} d^{4}}} + a^{2} b d\right )} \sqrt {\frac {{\left (a b^{2} - b^{3}\right )} d^{2} \sqrt {\frac {a^{3}}{{\left (a^{2} b^{5} - 2 \, a b^{6} + b^{7}\right )} d^{4}}} - a}{{\left (a b^{2} - b^{3}\right )} d^{2}}}\right ) - 4 \, \cos \left (d x + c\right )}{4 \, b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^5/(a-b*sin(d*x+c)^4),x, algorithm="fricas")

[Out]

-1/4*(b*d*sqrt(-((a*b^2 - b^3)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) + a)/((a*b^2 - b^3)*d^2))*log(a^2
*cos(d*x + c) - ((a*b^4 - b^5)*d^3*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) - a^2*b*d)*sqrt(-((a*b^2 - b^3)*d
^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) + a)/((a*b^2 - b^3)*d^2))) - b*d*sqrt(((a*b^2 - b^3)*d^2*sqrt(a^3
/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) - a)/((a*b^2 - b^3)*d^2))*log(a^2*cos(d*x + c) - ((a*b^4 - b^5)*d^3*sqrt(a^3
/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) + a^2*b*d)*sqrt(((a*b^2 - b^3)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4))
- a)/((a*b^2 - b^3)*d^2))) - b*d*sqrt(-((a*b^2 - b^3)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) + a)/((a*
b^2 - b^3)*d^2))*log(-a^2*cos(d*x + c) - ((a*b^4 - b^5)*d^3*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) - a^2*b*
d)*sqrt(-((a*b^2 - b^3)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) + a)/((a*b^2 - b^3)*d^2))) + b*d*sqrt(((
a*b^2 - b^3)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) - a)/((a*b^2 - b^3)*d^2))*log(-a^2*cos(d*x + c) - (
(a*b^4 - b^5)*d^3*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) + a^2*b*d)*sqrt(((a*b^2 - b^3)*d^2*sqrt(a^3/((a^2*
b^5 - 2*a*b^6 + b^7)*d^4)) - a)/((a*b^2 - b^3)*d^2))) - 4*cos(d*x + c))/(b*d)

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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(sin(d*x+c)**5/(a-b*sin(d*x+c)**4),x)

[Out]

Timed out

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^5/(a-b*sin(d*x+c)^4),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Warning, need to choose a branch for the
root of a polynomial with parameters. This might be wrong.The choice was done assuming [sageVARa,sageVARb]=[-
36,13]Warning, need t

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Mupad [B]
time = 14.26, size = 1001, normalized size = 7.25 \begin {gather*} \frac {\cos \left (c+d\,x\right )}{b\,d}-\frac {2\,\mathrm {atanh}\left (\frac {8\,a^2\,b^7\,\cos \left (c+d\,x\right )\,\sqrt {-\frac {\sqrt {a^3\,b^5}}{16\,\left (a\,b^5-b^6\right )}-\frac {a\,b^3}{16\,\left (a\,b^5-b^6\right )}}}{\frac {2\,a^3\,b^{11}}{a\,b^5-b^6}-\frac {2\,a^4\,b^{10}}{a\,b^5-b^6}+\frac {2\,a^2\,b^8\,\sqrt {a^3\,b^5}}{a\,b^5-b^6}-\frac {2\,a^3\,b^7\,\sqrt {a^3\,b^5}}{a\,b^5-b^6}}-\frac {8\,a^2\,b\,\cos \left (c+d\,x\right )\,\sqrt {-\frac {\sqrt {a^3\,b^5}}{16\,\left (a\,b^5-b^6\right )}-\frac {a\,b^3}{16\,\left (a\,b^5-b^6\right )}}}{\frac {2\,a^3\,b^5}{a\,b^5-b^6}+\frac {2\,a^2\,b^2\,\sqrt {a^3\,b^5}}{a\,b^5-b^6}}+\frac {8\,a\,b^4\,\cos \left (c+d\,x\right )\,\sqrt {-\frac {\sqrt {a^3\,b^5}}{16\,\left (a\,b^5-b^6\right )}-\frac {a\,b^3}{16\,\left (a\,b^5-b^6\right )}}\,\sqrt {a^3\,b^5}}{\frac {2\,a^3\,b^{11}}{a\,b^5-b^6}-\frac {2\,a^4\,b^{10}}{a\,b^5-b^6}+\frac {2\,a^2\,b^8\,\sqrt {a^3\,b^5}}{a\,b^5-b^6}-\frac {2\,a^3\,b^7\,\sqrt {a^3\,b^5}}{a\,b^5-b^6}}\right )\,\sqrt {-\frac {\sqrt {a^3\,b^5}+a\,b^3}{16\,\left (a\,b^5-b^6\right )}}}{d}+\frac {2\,\mathrm {atanh}\left (\frac {8\,a^2\,b\,\cos \left (c+d\,x\right )\,\sqrt {\frac {\sqrt {a^3\,b^5}}{16\,\left (a\,b^5-b^6\right )}-\frac {a\,b^3}{16\,\left (a\,b^5-b^6\right )}}}{\frac {2\,a^3\,b^5}{a\,b^5-b^6}-\frac {2\,a^2\,b^2\,\sqrt {a^3\,b^5}}{a\,b^5-b^6}}-\frac {8\,a^2\,b^7\,\cos \left (c+d\,x\right )\,\sqrt {\frac {\sqrt {a^3\,b^5}}{16\,\left (a\,b^5-b^6\right )}-\frac {a\,b^3}{16\,\left (a\,b^5-b^6\right )}}}{\frac {2\,a^3\,b^{11}}{a\,b^5-b^6}-\frac {2\,a^4\,b^{10}}{a\,b^5-b^6}-\frac {2\,a^2\,b^8\,\sqrt {a^3\,b^5}}{a\,b^5-b^6}+\frac {2\,a^3\,b^7\,\sqrt {a^3\,b^5}}{a\,b^5-b^6}}+\frac {8\,a\,b^4\,\cos \left (c+d\,x\right )\,\sqrt {\frac {\sqrt {a^3\,b^5}}{16\,\left (a\,b^5-b^6\right )}-\frac {a\,b^3}{16\,\left (a\,b^5-b^6\right )}}\,\sqrt {a^3\,b^5}}{\frac {2\,a^3\,b^{11}}{a\,b^5-b^6}-\frac {2\,a^4\,b^{10}}{a\,b^5-b^6}-\frac {2\,a^2\,b^8\,\sqrt {a^3\,b^5}}{a\,b^5-b^6}+\frac {2\,a^3\,b^7\,\sqrt {a^3\,b^5}}{a\,b^5-b^6}}\right )\,\sqrt {\frac {\sqrt {a^3\,b^5}-a\,b^3}{16\,\left (a\,b^5-b^6\right )}}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^5/(a - b*sin(c + d*x)^4),x)

[Out]

cos(c + d*x)/(b*d) - (2*atanh((8*a^2*b^7*cos(c + d*x)*(- (a^3*b^5)^(1/2)/(16*(a*b^5 - b^6)) - (a*b^3)/(16*(a*b
^5 - b^6)))^(1/2))/((2*a^3*b^11)/(a*b^5 - b^6) - (2*a^4*b^10)/(a*b^5 - b^6) + (2*a^2*b^8*(a^3*b^5)^(1/2))/(a*b
^5 - b^6) - (2*a^3*b^7*(a^3*b^5)^(1/2))/(a*b^5 - b^6)) - (8*a^2*b*cos(c + d*x)*(- (a^3*b^5)^(1/2)/(16*(a*b^5 -
b^6)) - (a*b^3)/(16*(a*b^5 - b^6)))^(1/2))/((2*a^3*b^5)/(a*b^5 - b^6) + (2*a^2*b^2*(a^3*b^5)^(1/2))/(a*b^5 -
b^6)) + (8*a*b^4*cos(c + d*x)*(- (a^3*b^5)^(1/2)/(16*(a*b^5 - b^6)) - (a*b^3)/(16*(a*b^5 - b^6)))^(1/2)*(a^3*b
^5)^(1/2))/((2*a^3*b^11)/(a*b^5 - b^6) - (2*a^4*b^10)/(a*b^5 - b^6) + (2*a^2*b^8*(a^3*b^5)^(1/2))/(a*b^5 - b^6
) - (2*a^3*b^7*(a^3*b^5)^(1/2))/(a*b^5 - b^6)))*(-((a^3*b^5)^(1/2) + a*b^3)/(16*(a*b^5 - b^6)))^(1/2))/d + (2*
atanh((8*a^2*b*cos(c + d*x)*((a^3*b^5)^(1/2)/(16*(a*b^5 - b^6)) - (a*b^3)/(16*(a*b^5 - b^6)))^(1/2))/((2*a^3*b
^5)/(a*b^5 - b^6) - (2*a^2*b^2*(a^3*b^5)^(1/2))/(a*b^5 - b^6)) - (8*a^2*b^7*cos(c + d*x)*((a^3*b^5)^(1/2)/(16*
(a*b^5 - b^6)) - (a*b^3)/(16*(a*b^5 - b^6)))^(1/2))/((2*a^3*b^11)/(a*b^5 - b^6) - (2*a^4*b^10)/(a*b^5 - b^6) -
(2*a^2*b^8*(a^3*b^5)^(1/2))/(a*b^5 - b^6) + (2*a^3*b^7*(a^3*b^5)^(1/2))/(a*b^5 - b^6)) + (8*a*b^4*cos(c + d*x
)*((a^3*b^5)^(1/2)/(16*(a*b^5 - b^6)) - (a*b^3)/(16*(a*b^5 - b^6)))^(1/2)*(a^3*b^5)^(1/2))/((2*a^3*b^11)/(a*b^
5 - b^6) - (2*a^4*b^10)/(a*b^5 - b^6) - (2*a^2*b^8*(a^3*b^5)^(1/2))/(a*b^5 - b^6) + (2*a^3*b^7*(a^3*b^5)^(1/2)
)/(a*b^5 - b^6)))*(((a^3*b^5)^(1/2) - a*b^3)/(16*(a*b^5 - b^6)))^(1/2))/d

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