Optimal. Leaf size=156 \[ \frac{B \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}-\frac{B \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}-\frac{2 B}{3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{B \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{B \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d} \]
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Rubi [A] time = 0.0989388, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {21, 3474, 3476, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac{B \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}-\frac{B \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}-\frac{2 B}{3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{B \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{B \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d} \]
Antiderivative was successfully verified.
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Rule 21
Rule 3474
Rule 3476
Rule 329
Rule 211
Rule 1165
Rule 628
Rule 1162
Rule 617
Rule 204
Rubi steps
\begin{align*} \int \frac{a B+b B \tan (c+d x)}{\tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx &=B \int \frac{1}{\tan ^{\frac{5}{2}}(c+d x)} \, dx\\ &=-\frac{2 B}{3 d \tan ^{\frac{3}{2}}(c+d x)}-B \int \frac{1}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{2 B}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{B \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{2 B}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{(2 B) \operatorname{Subst}\left (\int \frac{1}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=-\frac{2 B}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{B \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}-\frac{B \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=-\frac{2 B}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{B \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 d}-\frac{B \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 d}+\frac{B \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \sqrt{2} d}+\frac{B \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \sqrt{2} d}\\ &=\frac{B \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} d}-\frac{B \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} d}-\frac{2 B}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{B \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{B \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}\\ &=\frac{B \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}-\frac{B \tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{B \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} d}-\frac{B \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} d}-\frac{2 B}{3 d \tan ^{\frac{3}{2}}(c+d x)}\\ \end{align*}
Mathematica [C] time = 0.0279196, size = 36, normalized size = 0.23 \[ -\frac{2 B \text{Hypergeometric2F1}\left (-\frac{3}{4},1,\frac{1}{4},-\tan ^2(c+d x)\right )}{3 d \tan ^{\frac{3}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.027, size = 118, normalized size = 0.8 \begin{align*} -{\frac{2\,B}{3\,d} \left ( \tan \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}-{\frac{B\sqrt{2}}{2\,d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{B\sqrt{2}}{2\,d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{B\sqrt{2}}{4\,d}\ln \left ({ \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) \left ( 1-\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) ^{-1}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.79317, size = 167, normalized size = 1.07 \begin{align*} -\frac{6 \, \sqrt{2} B \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right ) + 6 \, \sqrt{2} B \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right ) + 3 \, \sqrt{2} B \log \left (\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - 3 \, \sqrt{2} B \log \left (-\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \frac{8 \, B}{\tan \left (d x + c\right )^{\frac{3}{2}}}}{12 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.13258, size = 1526, normalized size = 9.78 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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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.
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Giac [A] time = 1.76265, size = 178, normalized size = 1.14 \begin{align*} -\frac{1}{12} \, B{\left (\frac{6 \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right )}{d} + \frac{6 \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right )}{d} + \frac{3 \, \sqrt{2} \log \left (\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{d} - \frac{3 \, \sqrt{2} \log \left (-\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{d} + \frac{8}{d \tan \left (d x + c\right )^{\frac{3}{2}}}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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