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

Optimal. Leaf size=299 \[ \frac{2 (-1)^{2/3} b^{2/3} \tan ^{-1}\left (\frac{\sqrt [3]{-1} \sqrt [3]{b}-\sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 a^{2/3} d \left (a^{2/3}-(-1)^{2/3} b^{2/3}\right )^{3/2}}-\frac{2 b^{2/3} \tan ^{-1}\left (\frac{\sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt{a^{2/3}-b^{2/3}}}\right )}{3 a^{2/3} d \left (a^{2/3}-b^{2/3}\right )^{3/2}}+\frac{2 \sqrt [3]{-1} b^{2/3} \tan ^{-1}\left (\frac{\sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt{a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 a^{2/3} d \left (a^{2/3}+\sqrt [3]{-1} b^{2/3}\right )^{3/2}}+\frac{\sec (c+d x) (b-a \sin (c+d x))}{d \left (b^2-a^2\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [F]  time = 0.0453848, 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 = {} \[ \int \frac{\sec ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [C]  time = 0.253172, size = 432, normalized size = 1.44 \[ \frac{-i b \cos (c+d x) \text{RootSum}\left [8 \text{$\#$1}^3 a+i \text{$\#$1}^6 b-3 i \text{$\#$1}^4 b+3 i \text{$\#$1}^2 b-i b\& ,\frac{-2 \text{$\#$1}^3 a \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+2 \text{$\#$1} a \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )-4 i \text{$\#$1}^3 a \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )-i \text{$\#$1}^4 b \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+6 i \text{$\#$1}^2 b \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )-i b \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+2 \text{$\#$1}^4 b \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )-12 \text{$\#$1}^2 b \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )+4 i \text{$\#$1} a \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )+2 b \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )}{-4 i \text{$\#$1}^2 a+\text{$\#$1}^5 b-2 \text{$\#$1}^3 b+\text{$\#$1} b}\& \right ]+6 a \sin (c+d x)+6 b \cos (c+d x)-6 b}{6 d (a-b) (a+b) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.213, size = 164, normalized size = 0.6 \begin{align*} -{\frac{b}{3\,d \left ( a-b \right ) \left ( a+b \right ) }\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}b-2\,{{\it \_R}}^{3}a+6\,{{\it \_R}}^{2}b-2\,{\it \_R}\,a+b}{{{\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 ) }}-2\,{\frac{1}{d \left ( 2\,a-2\,b \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }}-2\,{\frac{1}{d \left ( 2\,a+2\,b \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }} \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),x)

[Out]

-1/3/d*b/(a-b)/(a+b)*sum((_R^4*b-2*_R^3*a+6*_R^2*b-2*_R*a+b)/(_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))-2/d/(2*a-2*b)/(tan(1/2*d*x+1/2*c)+1)-2/d/(2*a+2*b)/(t
an(1/2*d*x+1/2*c)-1)

________________________________________________________________________________________

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),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

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),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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),x)

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

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