Optimal. Leaf size=229 \[ \frac{2 b^3 \left (3 a^2-b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{5/2}}+\frac{2 b^3 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{5/2}}+\frac{b^4 \cos (c+d x)}{a d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac{\cos (c+d x)}{2 d (a+b)^2 (1-\sin (c+d x))}+\frac{\cos (c+d x)}{2 d (a-b)^2 (\sin (c+d x)+1)} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.310499, antiderivative size = 229, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {2897, 3770, 2648, 2664, 12, 2660, 618, 204} \[ \frac{2 b^3 \left (3 a^2-b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{5/2}}+\frac{2 b^3 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{5/2}}+\frac{b^4 \cos (c+d x)}{a d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac{\cos (c+d x)}{2 d (a+b)^2 (1-\sin (c+d x))}+\frac{\cos (c+d x)}{2 d (a-b)^2 (\sin (c+d x)+1)} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 2897
Rule 3770
Rule 2648
Rule 2664
Rule 12
Rule 2660
Rule 618
Rule 204
Rubi steps
\begin{align*} \int \frac{\csc (c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\int \left (\frac{\csc (c+d x)}{a^2}-\frac{1}{2 (a+b)^2 (-1+\sin (c+d x))}-\frac{1}{2 (a-b)^2 (1+\sin (c+d x))}-\frac{b^3}{a \left (-a^2+b^2\right ) (a+b \sin (c+d x))^2}+\frac{3 a^2 b^3-b^5}{a^2 \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}\right ) \, dx\\ &=\frac{\int \csc (c+d x) \, dx}{a^2}-\frac{\int \frac{1}{1+\sin (c+d x)} \, dx}{2 (a-b)^2}-\frac{\int \frac{1}{-1+\sin (c+d x)} \, dx}{2 (a+b)^2}+\frac{b^3 \int \frac{1}{(a+b \sin (c+d x))^2} \, dx}{a \left (a^2-b^2\right )}+\frac{\left (b^3 \left (3 a^2-b^2\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )^2}\\ &=-\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac{\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}+\frac{\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}+\frac{b^4 \cos (c+d x)}{a \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac{b^3 \int \frac{a}{a+b \sin (c+d x)} \, dx}{a \left (a^2-b^2\right )^2}+\frac{\left (2 b^3 \left (3 a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^2 \left (a^2-b^2\right )^2 d}\\ &=-\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac{\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}+\frac{\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}+\frac{b^4 \cos (c+d x)}{a \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac{b^3 \int \frac{1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^2}-\frac{\left (4 b^3 \left (3 a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^2 \left (a^2-b^2\right )^2 d}\\ &=\frac{2 b^3 \left (3 a^2-b^2\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{5/2} d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac{\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}+\frac{\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}+\frac{b^4 \cos (c+d x)}{a \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac{\left (2 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^2 d}\\ &=\frac{2 b^3 \left (3 a^2-b^2\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{5/2} d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac{\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}+\frac{\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}+\frac{b^4 \cos (c+d x)}{a \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac{\left (4 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^2 d}\\ &=\frac{2 b^3 \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2} d}+\frac{2 b^3 \left (3 a^2-b^2\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{5/2} d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac{\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}+\frac{\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}+\frac{b^4 \cos (c+d x)}{a \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}\\ \end{align*}
Mathematica [A] time = 2.21898, size = 203, normalized size = 0.89 \[ \frac{-\frac{2 \left (b^5-4 a^2 b^3\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{5/2}}+\frac{\frac{a b^4 \cos (c+d x)}{(a-b)^2 (a+b)^2 (a+b \sin (c+d x))}+\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{a^2}+\sin \left (\frac{1}{2} (c+d x)\right ) \left (\frac{1}{(a+b)^2 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{1}{(a-b)^2 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}\right )}{d} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [A] time = 0.15, size = 304, normalized size = 1.3 \begin{align*} -{\frac{1}{d \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+2\,{\frac{{b}^{5}\tan \left ( 1/2\,dx+c/2 \right ) }{d \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}{a}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) }}+2\,{\frac{{b}^{4}}{d \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}a \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) }}+8\,{\frac{{b}^{3}}{d \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-2\,{\frac{{b}^{5}}{d \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}{a}^{2}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+{\frac{1}{d \left ( a-b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{1}{d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [B] time = 7.80245, size = 2133, normalized size = 9.31 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[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]
[Out]
________________________________________________________________________________________
Giac [A] time = 1.2666, size = 424, normalized size = 1.85 \begin{align*} \frac{\frac{2 \,{\left (4 \, a^{2} b^{3} - b^{5}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \sqrt{a^{2} - b^{2}}} + \frac{2 \,{\left (2 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 2 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a^{5} - a^{3} b^{2} - a b^{4}\right )}}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a\right )}} + \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{2}}}{d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]