Optimal. Leaf size=209 \[ -\frac{(a+i b)^2 (-B+i A) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{-b+i a}}-\frac{(b+i a)^{3/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a A \sqrt{a+b \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}+\frac{2 b^{3/2} B \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d} \]
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Rubi [A] time = 1.68511, antiderivative size = 209, 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} \[ -\frac{(a+i b)^2 (-B+i A) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{-b+i a}}-\frac{(b+i a)^{3/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a A \sqrt{a+b \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}+\frac{2 b^{3/2} B \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d} \]
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{(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac{3}{2}}(c+d x)} \, dx &=-\frac{2 a A \sqrt{a+b \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}+2 \int \frac{\frac{1}{2} a (2 A b+a B)-\frac{1}{2} \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)+\frac{1}{2} b^2 B \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{2 a A \sqrt{a+b \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}+\frac{2 \operatorname{Subst}\left (\int \frac{\frac{1}{2} a (2 A b+a B)+\frac{1}{2} \left (-a^2 A+A b^2+2 a b B\right ) x+\frac{1}{2} b^2 B x^2}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{2 a A \sqrt{a+b \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}+\frac{2 \operatorname{Subst}\left (\int \left (\frac{b^2 B}{2 \sqrt{x} \sqrt{a+b x}}+\frac{2 a A b+a^2 B-b^2 B-\left (a^2 A-A b^2-2 a b B\right ) x}{2 \sqrt{x} \sqrt{a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{2 a A \sqrt{a+b \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}+\frac{\operatorname{Subst}\left (\int \frac{2 a A b+a^2 B-b^2 B-\left (a^2 A-A b^2-2 a b B\right ) x}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}+\frac{\left (b^2 B\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{2 a A \sqrt{a+b \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}+\frac{\operatorname{Subst}\left (\int \left (\frac{a^2 A-A b^2-2 a b B+i \left (2 a A b+a^2 B-b^2 B\right )}{2 (i-x) \sqrt{x} \sqrt{a+b x}}+\frac{-a^2 A+A b^2+2 a b B+i \left (2 a A b+a^2 B-b^2 B\right )}{2 \sqrt{x} (i+x) \sqrt{a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{d}+\frac{\left (2 b^2 B\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=-\frac{2 a A \sqrt{a+b \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{\left ((a-i b)^2 (A-i B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} (i+x) \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{\left ((a+i b)^2 (A+i B)\right ) \operatorname{Subst}\left (\int \frac{1}{(i-x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{\left (2 b^2 B\right ) \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 )}{d}\\ &=\frac{2 b^{3/2} B \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a A \sqrt{a+b \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{\left ((a-i b)^2 (A-i B)\right ) \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 )}{d}+\frac{\left ((a+i b)^2 (A+i B)\right ) \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 )}{d}\\ &=-\frac{(a+i b)^2 (i A-B) \tan ^{-1}\left (\frac{\sqrt{i a-b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{i a-b} d}+\frac{2 b^{3/2} B \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{(i a+b)^{3/2} (i A+B) \tanh ^{-1}\left (\frac{\sqrt{i a+b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a A \sqrt{a+b \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}\\ \end{align*}
Mathematica [C] time = 39.1459, size = 121803, normalized size = 582.79 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.814, size = 2396071, normalized size = 11464.5 \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(-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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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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \tan{\left (c + d x \right )}\right ) \left (a + b \tan{\left (c + d x \right )}\right )^{\frac{3}{2}}}{\tan ^{\frac{3}{2}}{\left (c + d x \right )}}\, dx \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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