Optimal. Leaf size=265 \[ -\frac{2 b \left (3 a^2+4 b^2\right ) \sqrt{\tan (c+d x)}}{3 a^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac{2 b \left (17 a^2 b^2+3 a^4+8 b^4\right ) \sqrt{\tan (c+d x)}}{3 a^3 d \left (a^2+b^2\right )^2 \sqrt{a+b \tan (c+d x)}}+\frac{i \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}-\frac{2}{a d \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}} \]
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Rubi [A] time = 0.959065, antiderivative size = 265, normalized size of antiderivative = 1., number of steps used = 10, 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} \[ -\frac{2 b \left (3 a^2+4 b^2\right ) \sqrt{\tan (c+d x)}}{3 a^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac{2 b \left (17 a^2 b^2+3 a^4+8 b^4\right ) \sqrt{\tan (c+d x)}}{3 a^3 d \left (a^2+b^2\right )^2 \sqrt{a+b \tan (c+d x)}}+\frac{i \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}-\frac{2}{a d \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}} \]
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))^{5/2}} \, dx &=-\frac{2}{a d \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac{2 \int \frac{2 b+\frac{1}{2} a \tan (c+d x)+2 b \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}} \, dx}{a}\\ &=-\frac{2}{a d \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac{2 b \left (3 a^2+4 b^2\right ) \sqrt{\tan (c+d x)}}{3 a^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac{4 \int \frac{\frac{1}{4} b \left (9 a^2+8 b^2\right )+\frac{3}{4} a^3 \tan (c+d x)+\frac{1}{2} b \left (3 a^2+4 b^2\right ) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx}{3 a^2 \left (a^2+b^2\right )}\\ &=-\frac{2}{a d \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac{2 b \left (3 a^2+4 b^2\right ) \sqrt{\tan (c+d x)}}{3 a^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac{2 b \left (3 a^4+17 a^2 b^2+8 b^4\right ) \sqrt{\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}-\frac{8 \int \frac{\frac{3 a^4 b}{4}+\frac{3}{8} a^3 \left (a^2-b^2\right ) \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{3 a^3 \left (a^2+b^2\right )^2}\\ &=-\frac{2}{a d \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac{2 b \left (3 a^2+4 b^2\right ) \sqrt{\tan (c+d x)}}{3 a^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac{2 b \left (3 a^4+17 a^2 b^2+8 b^4\right ) \sqrt{\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}+\frac{i \int \frac{1+i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{2 (a-i b)^2}-\frac{i \int \frac{1-i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{2 (a+i b)^2}\\ &=-\frac{2}{a d \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac{2 b \left (3 a^2+4 b^2\right ) \sqrt{\tan (c+d x)}}{3 a^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac{2 b \left (3 a^4+17 a^2 b^2+8 b^4\right ) \sqrt{\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}+\frac{i \operatorname{Subst}\left (\int \frac{1}{(1-i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a-i b)^2 d}-\frac{i \operatorname{Subst}\left (\int \frac{1}{(1+i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a+i b)^2 d}\\ &=-\frac{2}{a d \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac{2 b \left (3 a^2+4 b^2\right ) \sqrt{\tan (c+d x)}}{3 a^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac{2 b \left (3 a^4+17 a^2 b^2+8 b^4\right ) \sqrt{\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}+\frac{i \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 )}{(a-i b)^2 d}-\frac{i \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 )}{(a+i b)^2 d}\\ &=\frac{i \tan ^{-1}\left (\frac{\sqrt{i a-b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{(i a-b)^{5/2} d}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{i a+b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}-\frac{2}{a d \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac{2 b \left (3 a^2+4 b^2\right ) \sqrt{\tan (c+d x)}}{3 a^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac{2 b \left (3 a^4+17 a^2 b^2+8 b^4\right ) \sqrt{\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}\\ \end{align*}
Mathematica [A] time = 2.66033, size = 294, normalized size = 1.11 \[ -\frac{\frac{2 b \left (3 a^2+4 b^2\right ) \sqrt{\tan (c+d x)}}{\left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (17 a^2 b^2+3 a^4+8 b^4\right ) \sqrt{\tan (c+d x)}}{a \left (a^2+b^2\right )^2 \sqrt{a+b \tan (c+d x)}}+\frac{3 \sqrt [4]{-1} a^2 \left (\frac{(a-i b)^2 \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{-a-i b}}-\frac{(a+i b)^2 \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{a-i b}}\right )}{\left (a^2+b^2\right )^2}+\frac{6 a}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}}{3 a^2 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.955, size = 1489246, normalized size = 5619.8 \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{5}{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(-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(-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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