Optimal. Leaf size=189 \[ \frac{a (A+i B) \tan ^{m+1}(c+d x) \sqrt{a+b \tan (c+d x)} F_1\left (m+1;-\frac{3}{2},1;m+2;-\frac{b \tan (c+d x)}{a},-i \tan (c+d x)\right )}{2 d (m+1) \sqrt{\frac{b \tan (c+d x)}{a}+1}}+\frac{a (A-i B) \tan ^{m+1}(c+d x) \sqrt{a+b \tan (c+d x)} F_1\left (m+1;-\frac{3}{2},1;m+2;-\frac{b \tan (c+d x)}{a},i \tan (c+d x)\right )}{2 d (m+1) \sqrt{\frac{b \tan (c+d x)}{a}+1}} \]
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Rubi [A] time = 0.454937, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {3603, 3602, 135, 133} \[ \frac{a (A+i B) \tan ^{m+1}(c+d x) \sqrt{a+b \tan (c+d x)} F_1\left (m+1;-\frac{3}{2},1;m+2;-\frac{b \tan (c+d x)}{a},-i \tan (c+d x)\right )}{2 d (m+1) \sqrt{\frac{b \tan (c+d x)}{a}+1}}+\frac{a (A-i B) \tan ^{m+1}(c+d x) \sqrt{a+b \tan (c+d x)} F_1\left (m+1;-\frac{3}{2},1;m+2;-\frac{b \tan (c+d x)}{a},i \tan (c+d x)\right )}{2 d (m+1) \sqrt{\frac{b \tan (c+d x)}{a}+1}} \]
Antiderivative was successfully verified.
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Rule 3603
Rule 3602
Rule 135
Rule 133
Rubi steps
\begin{align*} \int \tan ^m(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx &=\frac{1}{2} (A-i B) \int (1+i \tan (c+d x)) \tan ^m(c+d x) (a+b \tan (c+d x))^{3/2} \, dx+\frac{1}{2} (A+i B) \int (1-i \tan (c+d x)) \tan ^m(c+d x) (a+b \tan (c+d x))^{3/2} \, dx\\ &=\frac{(A-i B) \operatorname{Subst}\left (\int \frac{x^m (a+b x)^{3/2}}{1-i x} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{(A+i B) \operatorname{Subst}\left (\int \frac{x^m (a+b x)^{3/2}}{1+i x} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac{\left (a (A-i B) \sqrt{a+b \tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{x^m \left (1+\frac{b x}{a}\right )^{3/2}}{1-i x} \, dx,x,\tan (c+d x)\right )}{2 d \sqrt{1+\frac{b \tan (c+d x)}{a}}}+\frac{\left (a (A+i B) \sqrt{a+b \tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{x^m \left (1+\frac{b x}{a}\right )^{3/2}}{1+i x} \, dx,x,\tan (c+d x)\right )}{2 d \sqrt{1+\frac{b \tan (c+d x)}{a}}}\\ &=\frac{a (A+i B) 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 (A-i B) 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 = 14.9155, size = 0, normalized size = 0. \[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.493, 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}}} \left ( A+B\tan \left ( dx+c \right ) \right ) \, 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 )}{\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 b \tan \left (d x + c\right )^{2} + A a +{\left (B a + A b\right )} \tan \left (d x + c\right )\right )} \sqrt{b \tan \left (d x + c\right ) + a} \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(-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 [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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