Optimal. Leaf size=193 \[ -\frac{2 b \left (a^2+2 b^2\right ) \sqrt{\tan (c+d x)}}{a^2 d \left (a^2+b^2\right ) \sqrt{a+b \tan (c+d x)}}+\frac{\tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (-b+i a)^{3/2}}-\frac{2}{a d \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (b+i a)^{3/2}} \]
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Rubi [A] time = 0.644499, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {3569, 3649, 3616, 3615, 93, 203, 206} \[ -\frac{2 b \left (a^2+2 b^2\right ) \sqrt{\tan (c+d x)}}{a^2 d \left (a^2+b^2\right ) \sqrt{a+b \tan (c+d x)}}+\frac{\tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (-b+i a)^{3/2}}-\frac{2}{a d \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (b+i a)^{3/2}} \]
Antiderivative was successfully verified.
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Rule 3569
Rule 3649
Rule 3616
Rule 3615
Rule 93
Rule 203
Rule 206
Rubi steps
\begin{align*} \int \frac{1}{\tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx &=-\frac{2}{a d \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}-\frac{2 \int \frac{b+\frac{1}{2} a \tan (c+d x)+b \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx}{a}\\ &=-\frac{2}{a d \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}-\frac{2 b \left (a^2+2 b^2\right ) \sqrt{\tan (c+d x)}}{a^2 \left (a^2+b^2\right ) d \sqrt{a+b \tan (c+d x)}}-\frac{4 \int \frac{\frac{a^2 b}{4}+\frac{1}{4} a^3 \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{a^2 \left (a^2+b^2\right )}\\ &=-\frac{2}{a d \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}-\frac{2 b \left (a^2+2 b^2\right ) \sqrt{\tan (c+d x)}}{a^2 \left (a^2+b^2\right ) d \sqrt{a+b \tan (c+d x)}}+\frac{\int \frac{1-i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{2 (i a-b)}-\frac{\int \frac{1+i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{2 (i a+b)}\\ &=-\frac{2}{a d \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}-\frac{2 b \left (a^2+2 b^2\right ) \sqrt{\tan (c+d x)}}{a^2 \left (a^2+b^2\right ) d \sqrt{a+b \tan (c+d x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{(1+i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (i a-b) d}-\frac{\operatorname{Subst}\left (\int \frac{1}{(1-i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (i a+b) d}\\ &=-\frac{2}{a d \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}-\frac{2 b \left (a^2+2 b^2\right ) \sqrt{\tan (c+d x)}}{a^2 \left (a^2+b^2\right ) d \sqrt{a+b \tan (c+d x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{1-(-i a+b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{(i a-b) d}-\frac{\operatorname{Subst}\left (\int \frac{1}{1-(i a+b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{(i a+b) d}\\ &=\frac{\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{\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 d \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}-\frac{2 b \left (a^2+2 b^2\right ) \sqrt{\tan (c+d x)}}{a^2 \left (a^2+b^2\right ) d \sqrt{a+b \tan (c+d x)}}\\ \end{align*}
Mathematica [A] time = 2.94325, size = 201, normalized size = 1.04 \[ \frac{\frac{-\frac{2 \left (b \left (a^2+2 b^2\right ) \tan (c+d x)+a \left (a^2+b^2\right )\right )}{a^2 \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}+\frac{\sqrt [4]{-1} (a+i b) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{a-i b}}}{a^2+b^2}+\frac{\sqrt [4]{-1} \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{(-a-i b)^{3/2}}}{d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.727, size = 798721, normalized size = 4138.5 \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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \tan \left (d x + c\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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\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(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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