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

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

[Out]

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

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Rubi [A]  time = 0.0432045, 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 = {} \[ \int \frac{\sec ^2(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [A]  time = 1.59132, size = 845, normalized size = 33.8 \[ \frac{-\frac{i b \text{RootSum}\left [i b \text{$\#$1}^6-3 i b \text{$\#$1}^4+8 a \text{$\#$1}^3+3 i b \text{$\#$1}^2-i b\& ,\frac{2 b^3 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right ) \text{$\#$1}^4+16 a^2 b \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right ) \text{$\#$1}^4-i b^3 \log \left (\text{$\#$1}^2-2 \cos (c+d x) \text{$\#$1}+1\right ) \text{$\#$1}^4-8 i a^2 b \log \left (\text{$\#$1}^2-2 \cos (c+d x) \text{$\#$1}+1\right ) \text{$\#$1}^4-20 i a^3 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right ) \text{$\#$1}^3-16 i a b^2 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right ) \text{$\#$1}^3-10 a^3 \log \left (\text{$\#$1}^2-2 \cos (c+d x) \text{$\#$1}+1\right ) \text{$\#$1}^3-8 a b^2 \log \left (\text{$\#$1}^2-2 \cos (c+d x) \text{$\#$1}+1\right ) \text{$\#$1}^3+12 b^3 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right ) \text{$\#$1}^2-120 a^2 b \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right ) \text{$\#$1}^2-6 i b^3 \log \left (\text{$\#$1}^2-2 \cos (c+d x) \text{$\#$1}+1\right ) \text{$\#$1}^2+60 i a^2 b \log \left (\text{$\#$1}^2-2 \cos (c+d x) \text{$\#$1}+1\right ) \text{$\#$1}^2+20 i a^3 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right ) \text{$\#$1}+16 i a b^2 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right ) \text{$\#$1}+10 a^3 \log \left (\text{$\#$1}^2-2 \cos (c+d x) \text{$\#$1}+1\right ) \text{$\#$1}+8 a b^2 \log \left (\text{$\#$1}^2-2 \cos (c+d x) \text{$\#$1}+1\right ) \text{$\#$1}+2 b^3 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )+16 a^2 b \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )-i b^3 \log \left (\text{$\#$1}^2-2 \cos (c+d x) \text{$\#$1}+1\right )-8 i a^2 b \log \left (\text{$\#$1}^2-2 \cos (c+d x) \text{$\#$1}+1\right )}{b \text{$\#$1}^5-2 b \text{$\#$1}^3-4 i a \text{$\#$1}^2+b \text{$\#$1}}\& \right ]}{a \left (a^2-b^2\right )^2}+\frac{18 \sin \left (\frac{1}{2} (c+d x)\right )}{(a+b)^2 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{18 \sin \left (\frac{1}{2} (c+d x)\right )}{(a-b)^2 \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{12 b \cos (c+d x) \left (-2 a^3-7 b^2 a+3 b^2 \cos (2 (c+d x)) a+2 b \left (2 a^2+b^2\right ) \sin (c+d x)\right )}{a (a-b)^2 (a+b)^2 (4 a+3 b \sin (c+d x)-b \sin (3 (c+d x)))}}{18 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-I)*b*RootSum[(-I)*b + (3*I)*b*#1^2 + 8*a*#1^3 - (3*I)*b*#1^4 + I*b*#1^6 & , (16*a^2*b*ArcTan[Sin[c + d*x]/
(Cos[c + d*x] - #1)] + 2*b^3*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)] - (8*I)*a^2*b*Log[1 - 2*Cos[c + d*x]*#1
+ #1^2] - I*b^3*Log[1 - 2*Cos[c + d*x]*#1 + #1^2] + (20*I)*a^3*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1 + (
16*I)*a*b^2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1 + 10*a^3*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1 + 8*a*b^
2*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1 - 120*a^2*b*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^2 + 12*b^3*ArcT
an[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^2 + (60*I)*a^2*b*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2 - (6*I)*b^3*Lo
g[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2 - (20*I)*a^3*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^3 - (16*I)*a*b^2
*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^3 - 10*a^3*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^3 - 8*a*b^2*Log[1
- 2*Cos[c + d*x]*#1 + #1^2]*#1^3 + 16*a^2*b*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^4 + 2*b^3*ArcTan[Sin[
c + d*x]/(Cos[c + d*x] - #1)]*#1^4 - (8*I)*a^2*b*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^4 - I*b^3*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*(a^2 - b^2)^2) + (18*Sin[(c + d*x)
/2])/((a + b)^2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (18*Sin[(c + d*x)/2])/((a - b)^2*(Cos[(c + d*x)/2] +
Sin[(c + d*x)/2])) + (12*b*Cos[c + d*x]*(-2*a^3 - 7*a*b^2 + 3*a*b^2*Cos[2*(c + d*x)] + 2*b*(2*a^2 + b^2)*Sin[c
+ d*x]))/(a*(a - b)^2*(a + b)^2*(4*a + 3*b*Sin[c + d*x] - b*Sin[3*(c + d*x)])))/(18*d)

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Maple [A]  time = 0.269, size = 1276, normalized size = 51. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/3/d*b^2/(a-b)^2/(a+b)^2/(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*b^4/(a-b)^2/(a+b)^2/(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*b/(a-b)^2/(a+b)^2/(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)*tan(1/2*d*x
+1/2*c)^4*a^2+8/3/d*b^3/(a-b)^2/(a+b)^2/(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)*tan(1/2*d*x+1/2*c)^4-8/3/d*b^2/(a-b)^2/(a+b)^2/(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)^3-16/3/d*b^4/(a-b)
^2/(a+b)^2/(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)^3-4/3/d*b/(a-b)^2/(a+b)^2/(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)*tan(1/2*d*x+1/2*c)^2*a^2-20/3/d*b^3/(a-b)^2/(a+b)^2/(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)*tan(1/2*d*x+1/2*c)^2+4/3
/d*b^2/(a-b)^2/(a+b)^2/(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)+2/3/d*b^4/(a-b)^2/(a+b)^2/(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)-2/3/d*b/(a-b)^2/(a+b)^2/(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^2-4/3/d*b^3/(a-b)
^2/(a+b)^2/(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)-1/9/d*b/(a-b)^2/(a+b)^2/a*sum((b*(11*a^2-2*b^2)*_R^4+2*a*(-5*a^2-4*b^2)*_R^3+54*_R^2*a^2*b+2*a*(-5*a^2-4*b^
2)*_R+11*a^2*b-2*b^3)/(_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))-1/d/(a+b)^2/(tan(1/2*d*x+1/2*c)-1)-1/d/(a-b)^2/(tan(1/2*d*x+1/2*c)+1)

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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(sec(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(sec(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(sec(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{\sec \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(sec(d*x+c)^2/(a+b*sin(d*x+c)^3)^2,x, algorithm="giac")

[Out]

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