3.401 \(\int \frac{1}{(a+b \sin ^3(c+d x))^2} \, dx\)

Optimal. Leaf size=16 \[ \text{Unintegrable}\left (\frac{1}{\left (a+b \sin ^3(c+d x)\right )^2},x\right ) \]

[Out]

Unintegrable[(a + b*Sin[c + d*x]^3)^(-2), x]

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Rubi [A]  time = 0.0118343, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{1}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

Int[(a + b*Sin[c + d*x]^3)^(-2),x]

[Out]

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

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx &=\int \frac{1}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx\\ \end{align*}

Mathematica [A]  time = 0.459565, size = 502, normalized size = 31.38 \[ \frac{-\frac{12 b \cos (c+d x) (a \cos (2 (c+d x))-3 a+2 b \sin (c+d x))}{(a-b) (a+b) (4 a+3 b \sin (c+d x)-b \sin (3 (c+d x)))}+\frac{i \text{RootSum}\left [8 \text{$\#$1}^3 a+i \text{$\#$1}^6 b-3 i \text{$\#$1}^4 b+3 i \text{$\#$1}^2 b-i b\& ,\frac{12 i \text{$\#$1}^2 a^2 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )-24 \text{$\#$1}^2 a^2 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )-2 \text{$\#$1}^3 a b \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+2 \text{$\#$1} a b \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )-4 i \text{$\#$1}^3 a b \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )-i \text{$\#$1}^4 b^2 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )-6 i \text{$\#$1}^2 b^2 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )-i b^2 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+2 \text{$\#$1}^4 b^2 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )+12 \text{$\#$1}^2 b^2 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )+4 i \text{$\#$1} a b \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )+2 b^2 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )}{-4 i \text{$\#$1}^2 a+\text{$\#$1}^5 b-2 \text{$\#$1}^3 b+\text{$\#$1} b}\& \right ]}{a^2-b^2}}{18 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I*RootSum[(-I)*b + (3*I)*b*#1^2 + 8*a*#1^3 - (3*I)*b*#1^4 + I*b*#1^6 & , (2*b^2*ArcTan[Sin[c + d*x]/(Cos[c +
d*x] - #1)] - I*b^2*Log[1 - 2*Cos[c + d*x]*#1 + #1^2] + (4*I)*a*b*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1
+ 2*a*b*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1 - 24*a^2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^2 + 12*b^2*
ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^2 + (12*I)*a^2*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2 - (6*I)*b^2*
Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2 - (4*I)*a*b*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^3 - 2*a*b*Log[1
- 2*Cos[c + d*x]*#1 + #1^2]*#1^3 + 2*b^2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^4 - I*b^2*Log[1 - 2*Cos[
c + d*x]*#1 + #1^2]*#1^4)/(b*#1 - (4*I)*a*#1^2 - 2*b*#1^3 + b*#1^5) & ])/(a^2 - b^2) - (12*b*Cos[c + d*x]*(-3*
a + a*Cos[2*(c + d*x)] + 2*b*Sin[c + d*x]))/((a - b)*(a + b)*(4*a + 3*b*Sin[c + d*x] - b*Sin[3*(c + d*x)])))/(
18*a*d)

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Maple [A]  time = 0.198, size = 658, normalized size = 41.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/d/(tan(1/2*d*x+1/2*c)^6*a+3*tan(1/2*d*x+1/2*c)^4*a+8*b*tan(1/2*d*x+1/2*c)^3+3*tan(1/2*d*x+1/2*c)^2*a+a)*b^
2/a/(a^2-b^2)*tan(1/2*d*x+1/2*c)^5-2/3/d/(tan(1/2*d*x+1/2*c)^6*a+3*tan(1/2*d*x+1/2*c)^4*a+8*b*tan(1/2*d*x+1/2*
c)^3+3*tan(1/2*d*x+1/2*c)^2*a+a)/(a^2-b^2)*b*tan(1/2*d*x+1/2*c)^4+8/3/d/(tan(1/2*d*x+1/2*c)^6*a+3*tan(1/2*d*x+
1/2*c)^4*a+8*b*tan(1/2*d*x+1/2*c)^3+3*tan(1/2*d*x+1/2*c)^2*a+a)*b^2/a/(a^2-b^2)*tan(1/2*d*x+1/2*c)^3+8/3/d/(ta
n(1/2*d*x+1/2*c)^6*a+3*tan(1/2*d*x+1/2*c)^4*a+8*b*tan(1/2*d*x+1/2*c)^3+3*tan(1/2*d*x+1/2*c)^2*a+a)/(a^2-b^2)*b
*tan(1/2*d*x+1/2*c)^2-2/3/d/(tan(1/2*d*x+1/2*c)^6*a+3*tan(1/2*d*x+1/2*c)^4*a+8*b*tan(1/2*d*x+1/2*c)^3+3*tan(1/
2*d*x+1/2*c)^2*a+a)*b^2/a/(a^2-b^2)*tan(1/2*d*x+1/2*c)+2/3/d/(tan(1/2*d*x+1/2*c)^6*a+3*tan(1/2*d*x+1/2*c)^4*a+
8*b*tan(1/2*d*x+1/2*c)^3+3*tan(1/2*d*x+1/2*c)^2*a+a)/(a^2-b^2)*b+1/9/d/a/(a^2-b^2)*sum(((3*a^2-2*b^2)*_R^4-2*_
R^3*a*b+6*_R^2*a^2-2*_R*a*b+3*a^2-2*b^2)/(_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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(d*x+c)^3)^2,x, algorithm="maxima")

[Out]

Timed out

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(d*x+c)^3)^2,x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \sin \left (d x + c\right )^{3} + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(d*x+c)^3)^2,x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c)^3 + a)^(-2), x)