Optimal. Leaf size=206 \[ \frac{(2 A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{b^{3/2} d}-\frac{(A+i B) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{-b+i a}}-\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{b+i a}}+\frac{B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{b d} \]
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Rubi [A] time = 1.36837, antiderivative size = 206, 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 = {3607, 3655, 6725, 63, 217, 206, 93, 205, 208} \[ \frac{(2 A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{b^{3/2} d}-\frac{(A+i B) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{-b+i a}}-\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{b+i a}}+\frac{B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{b d} \]
Antiderivative was successfully verified.
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Rule 3607
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))}{\sqrt{a+b \tan (c+d x)}} \, dx &=\frac{B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{b d}+\frac{\int \frac{-\frac{a B}{2}-b B \tan (c+d x)+\frac{1}{2} (2 A b-a B) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{b}\\ &=\frac{B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{b d}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{a B}{2}-b B x+\frac{1}{2} (2 A b-a B) x^2}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{b d}\\ &=\frac{B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{b d}+\frac{\operatorname{Subst}\left (\int \left (\frac{2 A b-a B}{2 \sqrt{x} \sqrt{a+b x}}-\frac{A b+b B x}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{b d}\\ &=\frac{B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{b d}-\frac{\operatorname{Subst}\left (\int \frac{A b+b B x}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{b d}+\frac{(2 A b-a B) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 b d}\\ &=\frac{B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{b d}-\frac{\operatorname{Subst}\left (\int \left (\frac{i A b-b B}{2 (i-x) \sqrt{x} \sqrt{a+b x}}+\frac{i A b+b B}{2 \sqrt{x} (i+x) \sqrt{a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{b d}+\frac{(2 A b-a B) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{b d}\\ &=\frac{B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{b d}-\frac{(i A-B) \operatorname{Subst}\left (\int \frac{1}{(i-x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 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 d}+\frac{(2 A b-a 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{(2 A b-a 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{B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{b 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 )}{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 )}{d}\\ &=-\frac{(A+i 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 A b-a 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{(A-i B) \tanh ^{-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{B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{b d}\\ \end{align*}
Mathematica [A] time = 2.34423, size = 245, normalized size = 1.19 \[ \frac{\frac{(-1)^{3/4} b (A+i B) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{-a-i b}}+\frac{\sqrt [4]{-1} b (B+i A) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{a-i b}}+\frac{\sqrt{a} (2 A b-a B) \sqrt{\frac{b \tan (c+d x)}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{\sqrt{b} \sqrt{a+b \tan (c+d x)}}+B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{b d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.885, size = 1888526, normalized size = 9167.6 \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}}}{\sqrt{b \tan \left (d x + c\right ) + a}}\,{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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