Optimal. Leaf size=204 \[ -\frac{(-b+i a)^{3/2} (A+i B) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{\sqrt{b} (3 a B+2 A b) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{(b+i a)^{3/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{b B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{d} \]
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Rubi [A] time = 1.73133, antiderivative size = 204, 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} \[ -\frac{(-b+i a)^{3/2} (A+i B) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{\sqrt{b} (3 a B+2 A b) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{(b+i a)^{3/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{b B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{d} \]
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{(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\sqrt{\tan (c+d x)}} \, dx &=\frac{b B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}+\int \frac{\frac{1}{2} a (2 a A-b B)+\left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)+\frac{1}{2} b (2 A b+3 a B) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx\\ &=\frac{b B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}+\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} a (2 a A-b B)+\left (2 a A b+a^2 B-b^2 B\right ) x+\frac{1}{2} b (2 A b+3 a B) x^2}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{b B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}+\frac{\operatorname{Subst}\left (\int \left (\frac{b (2 A b+3 a B)}{2 \sqrt{x} \sqrt{a+b x}}+\frac{a^2 A-A b^2-2 a b B+\left (2 a A b+a^2 B-b^2 B\right ) x}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{b B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}+\frac{\operatorname{Subst}\left (\int \frac{a^2 A-A b^2-2 a b B+\left (2 a A b+a^2 B-b^2 B\right ) x}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}+\frac{(b (2 A b+3 a B)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac{b B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}+\frac{\operatorname{Subst}\left (\int \left (\frac{-2 a A b-a^2 B+b^2 B+i \left (a^2 A-A b^2-2 a b B\right )}{2 (i-x) \sqrt{x} \sqrt{a+b x}}+\frac{2 a A b+a^2 B-b^2 B+i \left (a^2 A-A b^2-2 a b B\right )}{2 \sqrt{x} (i+x) \sqrt{a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{d}+\frac{(b (2 A b+3 a B)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=\frac{b B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}+\frac{\left ((a+i b)^2 (i A-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 ((a-i b)^2 (i A+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{(b (2 A b+3 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 )}{d}\\ &=\frac{\sqrt{b} (2 A b+3 a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{b B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}+\frac{\left ((a+i b)^2 (i A-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 (i A+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{(i a-b)^{3/2} (A+i B) \tan ^{-1}\left (\frac{\sqrt{i a-b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{\sqrt{b} (2 A b+3 a 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} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{i a+b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{b B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}\\ \end{align*}
Mathematica [A] time = 0.962479, size = 243, normalized size = 1.19 \[ \frac{-(-1)^{3/4} (-a-i b)^{3/2} (A+i B) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )-\sqrt [4]{-1} (a-i b)^{3/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )+\frac{\sqrt{a} \sqrt{b} (3 a B+2 A b) \sqrt{\frac{b \tan (c+d x)}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{\sqrt{a+b \tan (c+d x)}}+b B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.826, size = 2396041, normalized size = 11745.3 \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 )}{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}{\sqrt{\tan \left (d x + c\right )}}\,{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] 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}}}{\sqrt{\tan{\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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