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

Optimal. Leaf size=25 \[ \text{Unintegrable}\left (\frac{\cos ^2(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2},x\right ) \]

[Out]

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

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Rubi [A]  time = 0.0425309, 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 = {} \[ \int \frac{\cos ^2(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.245492, size = 273, normalized size = 10.92 \[ \frac{\frac{12 \sin (2 (c+d x))}{4 a+3 b \sin (c+d x)-b \sin (3 (c+d x))}-i \text{RootSum}\left [8 \text{$\#$1}^3 a+i \text{$\#$1}^6 b-3 i \text{$\#$1}^4 b+3 i \text{$\#$1}^2 b-i b\& ,\frac{-i \text{$\#$1}^4 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )-6 i \text{$\#$1}^2 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )-i \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+2 \text{$\#$1}^4 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )+12 \text{$\#$1}^2 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )+2 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )}{-4 i \text{$\#$1}^2 a+\text{$\#$1}^5 b-2 \text{$\#$1}^3 b+\text{$\#$1} b}\& \right ]}{18 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2/(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*ArcTan[Sin[c + d*x]/(Cos[c + d
*x] - #1)] - I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2] + 12*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^2 - (6*I)*Lo
g[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2 + 2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^4 - I*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) & ] + (12*Sin[2*(c + d*x)])/(4*a + 3*b*Sin[c +
d*x] - b*Sin[3*(c + d*x)]))/(18*a*d)

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Maple [A]  time = 0.247, size = 236, normalized size = 9.4 \begin{align*} -{\frac{2}{3\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5} \left ( \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{6}a+3\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}a+8\,b \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+3\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+a \right ) ^{-1}}+{\frac{2}{3\,da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{6}a+3\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}a+8\,b \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+3\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+a \right ) ^{-1}}+{\frac{2}{9\,da}\sum _{{\it \_R}={\it RootOf} \left ( a{{\it \_Z}}^{6}+3\,a{{\it \_Z}}^{4}+8\,b{{\it \_Z}}^{3}+3\,a{{\it \_Z}}^{2}+a \right ) }{\frac{{{\it \_R}}^{4}+1}{{{\it \_R}}^{5}a+2\,{{\it \_R}}^{3}a+4\,{{\it \_R}}^{2}b+{\it \_R}\,a}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -{\it \_R} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2/(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)/a
*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*tan(1/2*d*x+1/2*c)+2/9/d/a*sum((_R^4+1)/(_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(cos(d*x+c)^2/(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(cos(d*x+c)^2/(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(cos(d*x+c)**2/(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{\cos \left (d x + c\right )^{2}}{{\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(cos(d*x+c)^2/(a+b*sin(d*x+c)^3)^2,x, algorithm="giac")

[Out]

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