Optimal. Leaf size=532 \[ -\frac{\left (-2 a^3 b^3 \left (12 c d (A-C)+B \left (4 c^2-9 d^2\right )\right )+a^2 b^4 \left (24 A c^2-26 A d^2-48 B c d-24 c^2 C+35 C d^2\right )+a^4 b^2 d (3 d (A+2 C)+4 B c)+a^5 b B d^2+3 a^6 C d^2+a b^5 \left (40 c d (A-C)+3 B \left (8 c^2-5 d^2\right )\right )-b^6 \left (8 A c^2-3 A d^2-12 B c d-8 c^2 C\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{4 b^{5/2} f \left (a^2+b^2\right )^3 \sqrt{b c-a d}}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{3/2}}{2 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^2}-\frac{\sqrt{c+d \tan (e+f x)} \left (-a^2 b^2 (5 A d+4 B c-11 C d)+a^3 b B d+3 a^4 C d+a b^3 (8 A c-7 B d-8 c C)+b^4 (3 A d+4 B c)\right )}{4 b^2 f \left (a^2+b^2\right )^2 (a+b \tan (e+f x))}-\frac{(c-i d)^{3/2} (A-i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (b+i a)^3}+\frac{(c+i d)^{3/2} (A+i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (-b+i a)^3} \]
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Rubi [A] time = 4.08696, antiderivative size = 532, 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 = {3645, 3653, 3539, 3537, 63, 208, 3634} \[ -\frac{\left (-2 a^3 b^3 \left (12 c d (A-C)+B \left (4 c^2-9 d^2\right )\right )+a^2 b^4 \left (24 A c^2-26 A d^2-48 B c d-24 c^2 C+35 C d^2\right )+a^4 b^2 d (3 d (A+2 C)+4 B c)+a^5 b B d^2+3 a^6 C d^2+a b^5 \left (40 c d (A-C)+3 B \left (8 c^2-5 d^2\right )\right )-b^6 \left (8 A c^2-3 A d^2-12 B c d-8 c^2 C\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{4 b^{5/2} f \left (a^2+b^2\right )^3 \sqrt{b c-a d}}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{3/2}}{2 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^2}-\frac{\sqrt{c+d \tan (e+f x)} \left (-a^2 b^2 (5 A d+4 B c-11 C d)+a^3 b B d+3 a^4 C d+a b^3 (8 A c-7 B d-8 c C)+b^4 (3 A d+4 B c)\right )}{4 b^2 f \left (a^2+b^2\right )^2 (a+b \tan (e+f x))}-\frac{(c-i d)^{3/2} (A-i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (b+i a)^3}+\frac{(c+i d)^{3/2} (A+i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (-b+i a)^3} \]
Antiderivative was successfully verified.
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Rule 3645
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))^3} \, dx &=-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{3/2}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}+\frac{\int \frac{\sqrt{c+d \tan (e+f x)} \left (\frac{1}{2} \left (2 (b B-a C) \left (2 b c-\frac{3 a d}{2}\right )+2 A b \left (2 a c+\frac{3 b d}{2}\right )\right )-2 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-4 b^2 C\right ) d \tan ^2(e+f x)\right )}{(a+b \tan (e+f x))^2} \, dx}{2 b \left (a^2+b^2\right )}\\ &=-\frac{\left (a^3 b B d+3 a^4 C d+b^4 (4 B c+3 A d)+a b^3 (8 A c-8 c C-7 B d)-a^2 b^2 (4 B c+5 A d-11 C d)\right ) \sqrt{c+d \tan (e+f x)}}{4 b^2 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{3/2}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}+\frac{\int \frac{\frac{1}{4} \left (b (2 a c+b d) \left (3 a^2 C d+b^2 (4 B c+3 A d)+a b (4 A c-4 c C-3 B d)\right )-(2 b c-a d) \left (a^2 b B d+3 a^3 C d+A b^2 (4 b c-5 a d)-4 b^3 (c C+B d)-4 a b^2 (B c-2 C d)\right )\right )+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)+\frac{1}{4} d \left (a^3 b B d+3 a^4 C d-a b^3 (8 A c-8 c C-9 B d)-b^4 (4 B c+5 A d-8 C d)+a^2 b^2 (4 B c+3 (A+C) d)\right ) \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^3 b B d+3 a^4 C d+b^4 (4 B c+3 A d)+a b^3 (8 A c-8 c C-7 B d)-a^2 b^2 (4 B c+5 A d-11 C d)\right ) \sqrt{c+d \tan (e+f x)}}{4 b^2 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{3/2}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}+\frac{\int \frac{-2 b^2 \left (a^3 \left (c^2 C+2 B c d-C d^2-A \left (c^2-d^2\right )\right )-3 a b^2 \left (c^2 C+2 B c d-C d^2-A \left (c^2-d^2\right )\right )-3 a^2 b \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )+b^3 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )\right )+2 b^2 \left (3 a^2 b \left (c^2 C+2 B c d-C d^2-A \left (c^2-d^2\right )\right )-b^3 \left (c^2 C+2 B c d-C d^2-A \left (c^2-d^2\right )\right )+a^3 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )-3 a 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}{2 b^2 \left (a^2+b^2\right )^3}+\frac{\left (a^5 b B d^2+3 a^6 C d^2+a^4 b^2 d (4 B c+3 (A+2 C) d)-b^6 \left (8 A c^2-8 c^2 C-12 B c d-3 A d^2\right )+a^2 b^4 \left (24 A c^2-24 c^2 C-48 B c d-26 A d^2+35 C d^2\right )-2 a^3 b^3 \left (12 c (A-C) d+B \left (4 c^2-9 d^2\right )\right )+a b^5 \left (40 c (A-C) d+3 B \left (8 c^2-5 d^2\right )\right )\right ) \int \frac{1+\tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}} \, dx}{8 b^2 \left (a^2+b^2\right )^3}\\ &=-\frac{\left (a^3 b B d+3 a^4 C d+b^4 (4 B c+3 A d)+a b^3 (8 A c-8 c C-7 B d)-a^2 b^2 (4 B c+5 A d-11 C d)\right ) \sqrt{c+d \tan (e+f x)}}{4 b^2 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{3/2}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^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)^3}+\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)^3}+\frac{\left (a^5 b B d^2+3 a^6 C d^2+a^4 b^2 d (4 B c+3 (A+2 C) d)-b^6 \left (8 A c^2-8 c^2 C-12 B c d-3 A d^2\right )+a^2 b^4 \left (24 A c^2-24 c^2 C-48 B c d-26 A d^2+35 C d^2\right )-2 a^3 b^3 \left (12 c (A-C) d+B \left (4 c^2-9 d^2\right )\right )+a b^5 \left (40 c (A-C) d+3 B \left (8 c^2-5 d^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{8 b^2 \left (a^2+b^2\right )^3 f}\\ &=-\frac{\left (a^3 b B d+3 a^4 C d+b^4 (4 B c+3 A d)+a b^3 (8 A c-8 c C-7 B d)-a^2 b^2 (4 B c+5 A d-11 C d)\right ) \sqrt{c+d \tan (e+f x)}}{4 b^2 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{3/2}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}+\frac{\left ((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 (i a+b)^3 f}-\frac{\left ((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 (i a-b)^3 f}+\frac{\left (a^5 b B d^2+3 a^6 C d^2+a^4 b^2 d (4 B c+3 (A+2 C) d)-b^6 \left (8 A c^2-8 c^2 C-12 B c d-3 A d^2\right )+a^2 b^4 \left (24 A c^2-24 c^2 C-48 B c d-26 A d^2+35 C d^2\right )-2 a^3 b^3 \left (12 c (A-C) d+B \left (4 c^2-9 d^2\right )\right )+a b^5 \left (40 c (A-C) d+3 B \left (8 c^2-5 d^2\right )\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 )}{4 b^2 \left (a^2+b^2\right )^3 d f}\\ &=-\frac{\left (a^5 b B d^2+3 a^6 C d^2+a^4 b^2 d (4 B c+3 (A+2 C) d)-b^6 \left (8 A c^2-8 c^2 C-12 B c d-3 A d^2\right )+a^2 b^4 \left (24 A c^2-24 c^2 C-48 B c d-26 A d^2+35 C d^2\right )-2 a^3 b^3 \left (12 c (A-C) d+B \left (4 c^2-9 d^2\right )\right )+a b^5 \left (40 c (A-C) d+3 B \left (8 c^2-5 d^2\right )\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{4 b^{5/2} \left (a^2+b^2\right )^3 \sqrt{b c-a d} f}-\frac{\left (a^3 b B d+3 a^4 C d+b^4 (4 B c+3 A d)+a b^3 (8 A c-8 c C-7 B d)-a^2 b^2 (4 B c+5 A d-11 C d)\right ) \sqrt{c+d \tan (e+f x)}}{4 b^2 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{3/2}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}-\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)^3 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)^3 d 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)^3 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)^3 f}-\frac{\left (a^5 b B d^2+3 a^6 C d^2+a^4 b^2 d (4 B c+3 (A+2 C) d)-b^6 \left (8 A c^2-8 c^2 C-12 B c d-3 A d^2\right )+a^2 b^4 \left (24 A c^2-24 c^2 C-48 B c d-26 A d^2+35 C d^2\right )-2 a^3 b^3 \left (12 c (A-C) d+B \left (4 c^2-9 d^2\right )\right )+a b^5 \left (40 c (A-C) d+3 B \left (8 c^2-5 d^2\right )\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{4 b^{5/2} \left (a^2+b^2\right )^3 \sqrt{b c-a d} f}-\frac{\left (a^3 b B d+3 a^4 C d+b^4 (4 B c+3 A d)+a b^3 (8 A c-8 c C-7 B d)-a^2 b^2 (4 B c+5 A d-11 C d)\right ) \sqrt{c+d \tan (e+f x)}}{4 b^2 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{3/2}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}\\ \end{align*}
Mathematica [B] time = 6.54976, size = 7678, normalized size = 14.43 \[ \text{Result too large to show} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.251, size = 14441, normalized size = 27.1 \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 )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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