Optimal. Leaf size=219 \[ \frac{2 a (A b-a B) \sqrt{\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt{a+b \tan (c+d x)}}-\frac{(-B+i A) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (-b+i a)^{3/2}}-\frac{(B+i A) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (b+i a)^{3/2}}+\frac{2 B \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{b^{3/2} d} \]
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Rubi [A] time = 1.76962, antiderivative size = 219, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.257, Rules used = {3605, 3655, 6725, 63, 217, 206, 93, 205, 208} \[ \frac{2 a (A b-a B) \sqrt{\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt{a+b \tan (c+d x)}}-\frac{(-B+i A) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (-b+i a)^{3/2}}-\frac{(B+i A) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (b+i a)^{3/2}}+\frac{2 B \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{b^{3/2} d} \]
Antiderivative was successfully verified.
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Rule 3605
Rule 3655
Rule 6725
Rule 63
Rule 217
Rule 206
Rule 93
Rule 205
Rule 208
Rubi steps
\begin{align*} \int \frac{\tan ^{\frac{3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx &=\frac{2 a (A b-a B) \sqrt{\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt{a+b \tan (c+d x)}}+\frac{2 \int \frac{-\frac{1}{2} a (A b-a B)+\frac{1}{2} b (A b-a B) \tan (c+d x)+\frac{1}{2} \left (a^2+b^2\right ) B \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{b \left (a^2+b^2\right )}\\ &=\frac{2 a (A b-a B) \sqrt{\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt{a+b \tan (c+d x)}}+\frac{2 \operatorname{Subst}\left (\int \frac{-\frac{1}{2} a (A b-a B)+\frac{1}{2} b (A b-a B) x+\frac{1}{2} \left (a^2+b^2\right ) B x^2}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{b \left (a^2+b^2\right ) d}\\ &=\frac{2 a (A b-a B) \sqrt{\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt{a+b \tan (c+d x)}}+\frac{2 \operatorname{Subst}\left (\int \left (\frac{\left (a^2+b^2\right ) B}{2 \sqrt{x} \sqrt{a+b x}}-\frac{b (a A+b B)-b (A b-a B) x}{2 \sqrt{x} \sqrt{a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{b \left (a^2+b^2\right ) d}\\ &=\frac{2 a (A b-a B) \sqrt{\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt{a+b \tan (c+d x)}}-\frac{\operatorname{Subst}\left (\int \frac{b (a A+b B)-b (A b-a B) x}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{b \left (a^2+b^2\right ) d}+\frac{B \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{b d}\\ &=\frac{2 a (A b-a B) \sqrt{\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt{a+b \tan (c+d x)}}-\frac{\operatorname{Subst}\left (\int \left (\frac{b (A b-a B)+i b (a A+b B)}{2 (i-x) \sqrt{x} \sqrt{a+b x}}+\frac{-b (A b-a B)+i b (a A+b B)}{2 \sqrt{x} (i+x) \sqrt{a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{b \left (a^2+b^2\right ) d}+\frac{(2 B) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{b d}\\ &=\frac{2 a (A b-a B) \sqrt{\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt{a+b \tan (c+d x)}}-\frac{((i a+b) (A+i B)) \operatorname{Subst}\left (\int \frac{1}{(i-x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 \left (a^2+b^2\right ) d}+\frac{(2 B) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{b d}-\frac{(i A+B) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} (i+x) \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a-i b) d}\\ &=\frac{2 B \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{b^{3/2} d}+\frac{2 a (A b-a B) \sqrt{\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt{a+b \tan (c+d x)}}-\frac{((i a+b) (A+i B)) \operatorname{Subst}\left (\int \frac{1}{i-(a+i b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\left (a^2+b^2\right ) d}-\frac{(i A+B) \operatorname{Subst}\left (\int \frac{1}{i-(-a+i b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{(a-i b) d}\\ &=-\frac{(i A-B) \tan ^{-1}\left (\frac{\sqrt{i a-b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{(i a-b)^{3/2} d}+\frac{2 B \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{b^{3/2} d}-\frac{(i A+B) \tanh ^{-1}\left (\frac{\sqrt{i a+b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{(i a+b)^{3/2} d}+\frac{2 a (A b-a B) \sqrt{\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt{a+b \tan (c+d x)}}\\ \end{align*}
Mathematica [C] time = 39.9383, size = 177751, normalized size = 811.65 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 1.622, size = 1561442, normalized size = 7129.9 \begin{align*} \text{output too large to display} \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{{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{\frac{3}{2}}}{{\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(-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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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(-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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