Optimal. Leaf size=175 \[ \frac{a \tan ^{m+1}(c+d x) \sqrt{a+b \tan (c+d x)} F_1\left (m+1;-\frac{3}{2},1;m+2;-\frac{b \tan (c+d x)}{a},-i \tan (c+d x)\right )}{2 d (m+1) \sqrt{\frac{b \tan (c+d x)}{a}+1}}+\frac{a \tan ^{m+1}(c+d x) \sqrt{a+b \tan (c+d x)} F_1\left (m+1;-\frac{3}{2},1;m+2;-\frac{b \tan (c+d x)}{a},i \tan (c+d x)\right )}{2 d (m+1) \sqrt{\frac{b \tan (c+d x)}{a}+1}} \]
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Rubi [A] time = 0.200525, antiderivative size = 175, 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} \[ \frac{a \tan ^{m+1}(c+d x) \sqrt{a+b \tan (c+d x)} F_1\left (m+1;-\frac{3}{2},1;m+2;-\frac{b \tan (c+d x)}{a},-i \tan (c+d x)\right )}{2 d (m+1) \sqrt{\frac{b \tan (c+d x)}{a}+1}}+\frac{a \tan ^{m+1}(c+d x) \sqrt{a+b \tan (c+d x)} F_1\left (m+1;-\frac{3}{2},1;m+2;-\frac{b \tan (c+d x)}{a},i \tan (c+d x)\right )}{2 d (m+1) \sqrt{\frac{b \tan (c+d x)}{a}+1}} \]
Antiderivative was successfully verified.
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Rule 3575
Rule 912
Rule 135
Rule 133
Rubi steps
\begin{align*} \int \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}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{i x^m (a+b x)^{3/2}}{2 (i-x)}+\frac{i x^m (a+b x)^{3/2}}{2 (i+x)}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{i \operatorname{Subst}\left (\int \frac{x^m (a+b x)^{3/2}}{i-x} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{i \operatorname{Subst}\left (\int \frac{x^m (a+b x)^{3/2}}{i+x} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac{\left (i a \sqrt{a+b \tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{x^m \left (1+\frac{b x}{a}\right )^{3/2}}{i-x} \, dx,x,\tan (c+d x)\right )}{2 d \sqrt{1+\frac{b \tan (c+d x)}{a}}}+\frac{\left (i a \sqrt{a+b \tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{x^m \left (1+\frac{b x}{a}\right )^{3/2}}{i+x} \, dx,x,\tan (c+d x)\right )}{2 d \sqrt{1+\frac{b \tan (c+d x)}{a}}}\\ &=\frac{a 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{a+b \tan (c+d x)}}{2 d (1+m) \sqrt{1+\frac{b \tan (c+d x)}{a}}}+\frac{a 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{a+b \tan (c+d x)}}{2 d (1+m) \sqrt{1+\frac{b \tan (c+d x)}{a}}}\\ \end{align*}
Mathematica [F] time = 8.72538, size = 0, normalized size = 0. \[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^{3/2} \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.333, 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{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \tan \left (d x + c\right )^{m}\,{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 ({\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \tan \left (d x + c\right )^{m}, 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 \left (a + b \tan{\left (c + d x \right )}\right )^{\frac{3}{2}} \tan ^{m}{\left (c + d x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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