Optimal. Leaf size=179 \[ \frac{\tan ^{m+1}(c+d x) \sqrt{\frac{b \tan (c+d x)}{a}+1} F_1\left (m+1;\frac{3}{2},1;m+2;-\frac{b \tan (c+d x)}{a},-i \tan (c+d x)\right )}{2 a d (m+1) \sqrt{a+b \tan (c+d x)}}+\frac{\tan ^{m+1}(c+d x) \sqrt{\frac{b \tan (c+d x)}{a}+1} F_1\left (m+1;\frac{3}{2},1;m+2;-\frac{b \tan (c+d x)}{a},i \tan (c+d x)\right )}{2 a d (m+1) \sqrt{a+b \tan (c+d x)}} \]
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Rubi [A] time = 0.201852, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3575, 912, 135, 133} \[ \frac{\tan ^{m+1}(c+d x) \sqrt{\frac{b \tan (c+d x)}{a}+1} F_1\left (m+1;\frac{3}{2},1;m+2;-\frac{b \tan (c+d x)}{a},-i \tan (c+d x)\right )}{2 a d (m+1) \sqrt{a+b \tan (c+d x)}}+\frac{\tan ^{m+1}(c+d x) \sqrt{\frac{b \tan (c+d x)}{a}+1} F_1\left (m+1;\frac{3}{2},1;m+2;-\frac{b \tan (c+d x)}{a},i \tan (c+d x)\right )}{2 a d (m+1) \sqrt{a+b \tan (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 3575
Rule 912
Rule 135
Rule 133
Rubi steps
\begin{align*} \int \frac{\tan ^m(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^m}{(a+b x)^{3/2} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{i x^m}{2 (i-x) (a+b x)^{3/2}}+\frac{i x^m}{2 (i+x) (a+b x)^{3/2}}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{i \operatorname{Subst}\left (\int \frac{x^m}{(i-x) (a+b x)^{3/2}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{i \operatorname{Subst}\left (\int \frac{x^m}{(i+x) (a+b x)^{3/2}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac{\left (i \sqrt{1+\frac{b \tan (c+d x)}{a}}\right ) \operatorname{Subst}\left (\int \frac{x^m}{(i-x) \left (1+\frac{b x}{a}\right )^{3/2}} \, dx,x,\tan (c+d x)\right )}{2 a d \sqrt{a+b \tan (c+d x)}}+\frac{\left (i \sqrt{1+\frac{b \tan (c+d x)}{a}}\right ) \operatorname{Subst}\left (\int \frac{x^m}{(i+x) \left (1+\frac{b x}{a}\right )^{3/2}} \, dx,x,\tan (c+d x)\right )}{2 a d \sqrt{a+b \tan (c+d x)}}\\ &=\frac{F_1\left (1+m;\frac{3}{2},1;2+m;-\frac{b \tan (c+d x)}{a},-i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt{1+\frac{b \tan (c+d x)}{a}}}{2 a d (1+m) \sqrt{a+b \tan (c+d x)}}+\frac{F_1\left (1+m;\frac{3}{2},1;2+m;-\frac{b \tan (c+d x)}{a},i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt{1+\frac{b \tan (c+d x)}{a}}}{2 a d (1+m) \sqrt{a+b \tan (c+d x)}}\\ \end{align*}
Mathematica [F] time = 18.5182, size = 0, normalized size = 0. \[ \int \frac{\tan ^m(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.283, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \tan \left ( dx+c \right ) \right ) ^{m} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (d x + c\right )^{m}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{m}}{b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{m}{\left (c + d x \right )}}{\left (a + b \tan{\left (c + d x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (d x + c\right )^{m}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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