Optimal. Leaf size=372 \[ -\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{3/2}}{b f \left (a^2+b^2\right ) (a+b \tan (e+f x))}+\frac{d \left (3 a^2 C-a b B+A b^2+2 b^2 C\right ) \sqrt{c+d \tan (e+f x)}}{b^2 f \left (a^2+b^2\right )}+\frac{\sqrt{b c-a d} \left (a^2 b^2 (d (A-7 C)+2 B c)+a^3 b B d-3 a^4 C d-a b^3 (4 A c-5 B d-4 c C)-b^4 (3 A d+2 B c)\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{b^{5/2} f \left (a^2+b^2\right )^2}-\frac{(c-i d)^{3/2} (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (a-i b)^2}-\frac{(c+i d)^{3/2} (B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (a+i b)^2} \]
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Rubi [A] time = 2.54731, antiderivative size = 372, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 47, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.17, Rules used = {3645, 3647, 3653, 3539, 3537, 63, 208, 3634} \[ -\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{3/2}}{b f \left (a^2+b^2\right ) (a+b \tan (e+f x))}+\frac{d \left (3 a^2 C-a b B+A b^2+2 b^2 C\right ) \sqrt{c+d \tan (e+f x)}}{b^2 f \left (a^2+b^2\right )}+\frac{\sqrt{b c-a d} \left (a^2 b^2 (d (A-7 C)+2 B c)+a^3 b B d-3 a^4 C d-a b^3 (4 A c-5 B d-4 c C)-b^4 (3 A d+2 B c)\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{b^{5/2} f \left (a^2+b^2\right )^2}-\frac{(c-i d)^{3/2} (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (a-i b)^2}-\frac{(c+i d)^{3/2} (B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (a+i b)^2} \]
Antiderivative was successfully verified.
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Rule 3645
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))^2} \, dx &=-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{3/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac{\int \frac{\sqrt{c+d \tan (e+f x)} \left (\frac{1}{2} \left (2 (b B-a C) \left (b c-\frac{3 a d}{2}\right )+2 A b \left (a c+\frac{3 b d}{2}\right )\right )-b ((A-C) (b c-a d)-B (a c+b d)) \tan (e+f x)+\frac{1}{2} \left (A b^2-a b B+3 a^2 C+2 b^2 C\right ) d \tan ^2(e+f x)\right )}{a+b \tan (e+f x)} \, dx}{b \left (a^2+b^2\right )}\\ &=\frac{\left (A b^2-a b B+3 a^2 C+2 b^2 C\right ) d \sqrt{c+d \tan (e+f x)}}{b^2 \left (a^2+b^2\right ) f}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{3/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac{2 \int \frac{\frac{1}{4} \left (-a \left (A b^2-a b B+3 a^2 C+2 b^2 C\right ) d^2+b c ((b B-a C) (2 b c-3 a d)+A b (2 a c+3 b d))\right )+\frac{1}{2} 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)-\frac{1}{4} d \left (3 a^3 C d+A b^2 (b c-a d)-2 b^3 (2 c C+B d)-a^2 b (3 c C+B d)-a b^2 (B c-4 C d)\right ) \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{\left (A b^2-a b B+3 a^2 C+2 b^2 C\right ) d \sqrt{c+d \tan (e+f x)}}{b^2 \left (a^2+b^2\right ) f}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{3/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac{2 \int \frac{-\frac{1}{2} b^2 \left (a^2 \left (c^2 C+2 B c d-C d^2-A \left (c^2-d^2\right )\right )-b^2 \left (c^2 C+2 B c d-C d^2-A \left (c^2-d^2\right )\right )-2 a b \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )\right )+\frac{1}{2} b^2 \left (2 a b \left (c^2 C+2 B c d-C d^2-A \left (c^2-d^2\right )\right )+a^2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )-b^2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )\right ) \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{b^2 \left (a^2+b^2\right )^2}-\frac{\left ((b c-a d) \left (a^3 b B d-3 a^4 C d-b^4 (2 B c+3 A d)-a b^3 (4 A c-4 c C-5 B d)+a^2 b^2 (2 B c+(A-7 C) d)\right )\right ) \int \frac{1+\tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}} \, dx}{2 b^2 \left (a^2+b^2\right )^2}\\ &=\frac{\left (A b^2-a b B+3 a^2 C+2 b^2 C\right ) d \sqrt{c+d \tan (e+f x)}}{b^2 \left (a^2+b^2\right ) f}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{3/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\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)^2}+\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)^2}-\frac{\left ((b c-a d) \left (a^3 b B d-3 a^4 C d-b^4 (2 B c+3 A d)-a b^3 (4 A c-4 c C-5 B d)+a^2 b^2 (2 B c+(A-7 C) d)\right )\right ) \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 b^2 \left (a^2+b^2\right )^2 f}\\ &=\frac{\left (A b^2-a b B+3 a^2 C+2 b^2 C\right ) d \sqrt{c+d \tan (e+f x)}}{b^2 \left (a^2+b^2\right ) f}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{3/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\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)^2 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)^2 f}-\frac{\left ((b c-a d) \left (a^3 b B d-3 a^4 C d-b^4 (2 B c+3 A d)-a b^3 (4 A c-4 c C-5 B d)+a^2 b^2 (2 B c+(A-7 C) d)\right )\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 )^2 d f}\\ &=\frac{\sqrt{b c-a d} \left (a^3 b B d-3 a^4 C d-b^4 (2 B c+3 A d)-a b^3 (4 A c-4 c C-5 B d)+a^2 b^2 (2 B c+(A-7 C) d)\right ) \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 )^2 f}+\frac{\left (A b^2-a b B+3 a^2 C+2 b^2 C\right ) d \sqrt{c+d \tan (e+f x)}}{b^2 \left (a^2+b^2\right ) f}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{3/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}-\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)^2 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)^2 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)^2 f}-\frac{(B-i (A-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)^2 f}+\frac{\sqrt{b c-a d} \left (a^3 b B d-3 a^4 C d-b^4 (2 B c+3 A d)-a b^3 (4 A c-4 c C-5 B d)+a^2 b^2 (2 B c+(A-7 C) d)\right ) \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 )^2 f}+\frac{\left (A b^2-a b B+3 a^2 C+2 b^2 C\right ) d \sqrt{c+d \tan (e+f x)}}{b^2 \left (a^2+b^2\right ) f}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{3/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}\\ \end{align*}
Mathematica [B] time = 6.22458, size = 1732, normalized size = 4.66 \[ \text{result too large to display} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.235, size = 9865, normalized size = 26.5 \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(-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}}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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