Optimal. Leaf size=244 \[ \frac{2 a (A b-a B) \sqrt{\tan (c+d x)}}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac{2 \left (2 a^2 A b+a^3 B+7 a b^2 B-4 A b^3\right ) \sqrt{\tan (c+d x)}}{3 b d \left (a^2+b^2\right )^2 \sqrt{a+b \tan (c+d x)}}+\frac{(A+i B) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}+\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}} \]
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Rubi [A] time = 0.989244, antiderivative size = 244, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3605, 3649, 3616, 3615, 93, 203, 206} \[ \frac{2 a (A b-a B) \sqrt{\tan (c+d x)}}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac{2 \left (2 a^2 A b+a^3 B+7 a b^2 B-4 A b^3\right ) \sqrt{\tan (c+d x)}}{3 b d \left (a^2+b^2\right )^2 \sqrt{a+b \tan (c+d x)}}+\frac{(A+i B) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}+\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 3605
Rule 3649
Rule 3616
Rule 3615
Rule 93
Rule 203
Rule 206
Rubi steps
\begin{align*} \int \frac{\tan ^{\frac{3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx &=\frac{2 a (A b-a B) \sqrt{\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 \int \frac{-\frac{1}{2} a (A b-a B)+\frac{3}{2} b (A b-a B) \tan (c+d x)+\frac{1}{2} \left (2 a A b+a^2 B+3 b^2 B\right ) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx}{3 b \left (a^2+b^2\right )}\\ &=\frac{2 a (A b-a B) \sqrt{\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 \left (2 a^2 A b-4 A b^3+a^3 B+7 a b^2 B\right ) \sqrt{\tan (c+d x)}}{3 b \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}+\frac{4 \int \frac{-\frac{3}{4} a b \left (a^2 A-A b^2+2 a b B\right )+\frac{3}{4} a b \left (2 a A b-a^2 B+b^2 B\right ) \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{3 a b \left (a^2+b^2\right )^2}\\ &=\frac{2 a (A b-a B) \sqrt{\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 \left (2 a^2 A b-4 A b^3+a^3 B+7 a b^2 B\right ) \sqrt{\tan (c+d x)}}{3 b \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}-\frac{(A-i B) \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{(A+i B) \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 (A b-a B) \sqrt{\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 \left (2 a^2 A b-4 A b^3+a^3 B+7 a b^2 B\right ) \sqrt{\tan (c+d x)}}{3 b \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}-\frac{(A-i B) \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{(A+i B) \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 (A b-a B) \sqrt{\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 \left (2 a^2 A b-4 A b^3+a^3 B+7 a b^2 B\right ) \sqrt{\tan (c+d x)}}{3 b \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}-\frac{(A-i B) \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{(A+i B) \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{(A+i B) \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{(A-i B) \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 (A b-a B) \sqrt{\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 \left (2 a^2 A b-4 A b^3+a^3 B+7 a b^2 B\right ) \sqrt{\tan (c+d x)}}{3 b \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}\\ \end{align*}
Mathematica [A] time = 2.84999, size = 308, normalized size = 1.26 \[ \frac{\frac{\left (a^2 B+2 a A b+3 b^2 B\right ) \sqrt{\tan (c+d x)}}{\left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac{\frac{2 \left (2 a^2 A b+a^3 B+7 a b^2 B-4 A b^3\right ) \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}+3 \sqrt [4]{-1} b \left (\frac{i (a-i b)^2 (A+i B) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{-a-i b}}+\frac{(a+i b)^2 (B+i A) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{a-i b}}\right )}{\left (a^2+b^2\right )^2}-\frac{3 B \sqrt{\tan (c+d x)}}{(a+b \tan (c+d x))^{3/2}}}{3 b d} \]
Antiderivative was successfully verified.
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Maple [B] time = 2.301, size = 2976654, normalized size = 12199.4 \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}}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{5}{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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