Optimal. Leaf size=195 \[ -\frac{2 a^2 \sqrt{\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt{a+b \tan (c+d x)}}+\frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{b^{3/2} d}+\frac{\tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (-b+i a)^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (b+i a)^{3/2}} \]
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Rubi [A] time = 1.23784, antiderivative size = 195, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {3565, 3655, 6725, 63, 217, 206, 93, 205, 208} \[ -\frac{2 a^2 \sqrt{\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt{a+b \tan (c+d x)}}+\frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{b^{3/2} d}+\frac{\tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (-b+i a)^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (b+i a)^{3/2}} \]
Antiderivative was successfully verified.
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Rule 3565
Rule 3655
Rule 6725
Rule 63
Rule 217
Rule 206
Rule 93
Rule 205
Rule 208
Rubi steps
\begin{align*} \int \frac{\tan ^{\frac{5}{2}}(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx &=-\frac{2 a^2 \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{a^2}{2}-\frac{1}{2} a b \tan (c+d x)+\frac{1}{2} \left (a^2+b^2\right ) \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^2 \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{a^2}{2}-\frac{a b x}{2}+\frac{1}{2} \left (a^2+b^2\right ) 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^2 \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{a^2+b^2}{2 \sqrt{x} \sqrt{a+b x}}-\frac{b^2+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^2 \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{1}{\sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{b d}-\frac{\operatorname{Subst}\left (\int \frac{b^2+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{2 a^2 \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{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{b d}-\frac{\operatorname{Subst}\left (\int \left (\frac{-a b+i b^2}{2 (i-x) \sqrt{x} \sqrt{a+b x}}+\frac{a b+i b^2}{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 a^2 \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{1}{\sqrt{x} (i+x) \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a-i b) d}+\frac{\operatorname{Subst}\left (\int \frac{1}{(i-x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a+i b) d}+\frac{2 \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 \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^2 \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{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{\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{\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 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{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^2 \sqrt{\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt{a+b \tan (c+d x)}}\\ \end{align*}
Mathematica [A] time = 3.12952, size = 230, normalized size = 1.18 \[ \frac{\frac{2 \sqrt{a} \left (a^2+b^2\right ) \sqrt{\frac{b \tan (c+d x)}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )-2 a^2 \sqrt{b} \sqrt{\tan (c+d x)}}{b^{3/2} \left (a^2+b^2\right ) \sqrt{a+b \tan (c+d x)}}+\frac{\sqrt [4]{-1} \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{(-a-i b)^{3/2}}+\frac{\sqrt [4]{-1} \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{(a-i b)^{3/2}}}{d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.658, size = 798950, normalized size = 4097.2 \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{\tan \left (d x + c\right )^{\frac{5}{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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