Optimal. Leaf size=271 \[ -\frac{2 (b c-a d)^{3/2} \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{b^{5/2} f \left (a^2+b^2\right )}-\frac{(c-i d)^{3/2} (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (a-i b)}-\frac{(c+i d)^{3/2} (A+i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (-b+i a)}+\frac{2 (-a C d+b B d+b c C) \sqrt{c+d \tan (e+f x)}}{b^2 f}+\frac{2 C (c+d \tan (e+f x))^{3/2}}{3 b f} \]
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Rubi [A] time = 1.81409, antiderivative size = 271, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 7, integrand size = 47, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.149, Rules used = {3647, 3653, 3539, 3537, 63, 208, 3634} \[ -\frac{2 (b c-a d)^{3/2} \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{b^{5/2} f \left (a^2+b^2\right )}-\frac{(c-i d)^{3/2} (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (a-i b)}-\frac{(c+i d)^{3/2} (A+i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (-b+i a)}+\frac{2 (-a C d+b B d+b c C) \sqrt{c+d \tan (e+f x)}}{b^2 f}+\frac{2 C (c+d \tan (e+f x))^{3/2}}{3 b f} \]
Antiderivative was successfully verified.
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Rule 3647
Rule 3653
Rule 3539
Rule 3537
Rule 63
Rule 208
Rule 3634
Rubi steps
\begin{align*} \int \frac{(c+d \tan (e+f x))^{3/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{a+b \tan (e+f x)} \, dx &=\frac{2 C (c+d \tan (e+f x))^{3/2}}{3 b f}+\frac{2 \int \frac{\sqrt{c+d \tan (e+f x)} \left (\frac{3}{2} (A b c-a C d)+\frac{3}{2} b (B c+(A-C) d) \tan (e+f x)+\frac{3}{2} (b c C+b B d-a C d) \tan ^2(e+f x)\right )}{a+b \tan (e+f x)} \, dx}{3 b}\\ &=\frac{2 (b c C+b B d-a C d) \sqrt{c+d \tan (e+f x)}}{b^2 f}+\frac{2 C (c+d \tan (e+f x))^{3/2}}{3 b f}+\frac{4 \int \frac{\frac{3}{4} \left (A b^2 c^2+a d (a C d-b (2 c C+B d))\right )+\frac{3}{4} b^2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) \tan (e+f x)+\frac{3}{4} \left (b^2 d (B c+(A-C) d)+(b c-a d) (b c C+b B d-a C d)\right ) \tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}} \, dx}{3 b^2}\\ &=\frac{2 (b c C+b B d-a C d) \sqrt{c+d \tan (e+f x)}}{b^2 f}+\frac{2 C (c+d \tan (e+f x))^{3/2}}{3 b f}+\frac{4 \int \frac{-\frac{3}{4} b^2 \left (a \left (c^2 C+2 B c d-C d^2-A \left (c^2-d^2\right )\right )-b \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )\right )+\frac{3}{4} b^2 \left (2 a A c d-2 a c C d-A b \left (c^2-d^2\right )+a B \left (c^2-d^2\right )+b \left (c^2 C+2 B c d-C d^2\right )\right ) \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{3 b^2 \left (a^2+b^2\right )}+\frac{\left (\left (A b^2-a (b B-a C)\right ) (b c-a d)^2\right ) \int \frac{1+\tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}} \, dx}{b^2 \left (a^2+b^2\right )}\\ &=\frac{2 (b c C+b B d-a C d) \sqrt{c+d \tan (e+f x)}}{b^2 f}+\frac{2 C (c+d \tan (e+f x))^{3/2}}{3 b f}+\frac{\left ((A-i B-C) (c-i d)^2\right ) \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (a-i b)}+\frac{\left ((A+i B-C) (c+i d)^2\right ) \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (a+i b)}+\frac{\left (\left (A b^2-a (b B-a C)\right ) (b c-a d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{b^2 \left (a^2+b^2\right ) f}\\ &=\frac{2 (b c C+b B d-a C d) \sqrt{c+d \tan (e+f x)}}{b^2 f}+\frac{2 C (c+d \tan (e+f x))^{3/2}}{3 b f}+\frac{\left ((i A+B-i C) (c-i d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 (a-i b) f}-\frac{\left (i (A+i B-C) (c+i d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 (a+i b) f}+\frac{\left (2 \left (A b^2-a (b B-a C)\right ) (b c-a d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b c}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{b^2 \left (a^2+b^2\right ) d f}\\ &=-\frac{2 \left (A b^2-a (b B-a C)\right ) (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{b^{5/2} \left (a^2+b^2\right ) f}+\frac{2 (b c C+b B d-a C d) \sqrt{c+d \tan (e+f x)}}{b^2 f}+\frac{2 C (c+d \tan (e+f x))^{3/2}}{3 b f}-\frac{\left ((A-i B-C) (c-i d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{-1-\frac{i c}{d}+\frac{i x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{(a-i b) d f}-\frac{\left ((A+i B-C) (c+i d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{-1+\frac{i c}{d}-\frac{i x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{(a+i b) d f}\\ &=-\frac{(i A+B-i C) (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{(a-i b) f}-\frac{(A+i B-C) (c+i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{(i a-b) f}-\frac{2 \left (A b^2-a (b B-a C)\right ) (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{b^{5/2} \left (a^2+b^2\right ) f}+\frac{2 (b c C+b B d-a C d) \sqrt{c+d \tan (e+f x)}}{b^2 f}+\frac{2 C (c+d \tan (e+f x))^{3/2}}{3 b f}\\ \end{align*}
Mathematica [A] time = 2.42084, size = 266, normalized size = 0.98 \[ \frac{-\frac{6 (b c-a d)^{3/2} \left (a (a C-b B)+A b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{b^{3/2} \left (a^2+b^2\right )}+\frac{3 i b \left ((a-i b) (c+i d)^{3/2} (A+i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )-(a+i b) (c-i d)^{3/2} (A-i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )\right )}{a^2+b^2}+\frac{6 (-a C d+b B d+b c C) \sqrt{c+d \tan (e+f x)}}{b}+2 C (c+d \tan (e+f x))^{3/2}}{3 b f} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.192, size = 6055, normalized size = 22.3 \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(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )}{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}{b \tan \left (f x + e\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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