Optimal. Leaf size=598 \[ -\frac{2 d \sqrt{a+b \tan (e+f x)} \left (-a^2 b^2 \left (11 A c^2 d+17 A d^3+3 B c^3-3 B c d^2+5 c^2 C d-C d^3\right )+a^4 (-d) \left (d^2 (3 A+5 C)-3 B c d+8 c^2 C\right )+8 a^3 b B d \left (c^2+d^2\right )+2 a b^3 \left (c^2+d^2\right ) (3 A c+B d-3 c C)-b^4 \left (d \left (5 A c^2+8 A d^2+3 c^2 C\right )-3 B \left (c^3+2 c d^2\right )\right )\right )}{3 f \left (a^2+b^2\right )^2 \left (c^2+d^2\right ) (b c-a d)^3 \sqrt{c+d \tan (e+f x)}}-\frac{2 \left (A b^2-a (b B-a C)\right )}{3 f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}-\frac{2 \left (-a^2 b^2 (2 d (5 A-C)+3 B c)+7 a^3 b B d-4 a^4 C d+a b^3 (6 A c+B d-6 c C)+b^4 (3 B c-4 A d)\right )}{3 f \left (a^2+b^2\right )^2 (b c-a d)^2 \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}-\frac{(i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a-i b)^{5/2} (c-i d)^{3/2}}-\frac{(B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a+i b)^{5/2} (c+i d)^{3/2}} \]
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Rubi [A] time = 3.43465, antiderivative size = 598, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 49, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.102, Rules used = {3649, 3616, 3615, 93, 208} \[ -\frac{2 d \sqrt{a+b \tan (e+f x)} \left (-a^2 b^2 \left (11 A c^2 d+17 A d^3+3 B c^3-3 B c d^2+5 c^2 C d-C d^3\right )+a^4 (-d) \left (d^2 (3 A+5 C)-3 B c d+8 c^2 C\right )+8 a^3 b B d \left (c^2+d^2\right )+2 a b^3 \left (c^2+d^2\right ) (3 A c+B d-3 c C)-b^4 \left (d \left (5 A c^2+8 A d^2+3 c^2 C\right )-3 B \left (c^3+2 c d^2\right )\right )\right )}{3 f \left (a^2+b^2\right )^2 \left (c^2+d^2\right ) (b c-a d)^3 \sqrt{c+d \tan (e+f x)}}-\frac{2 \left (A b^2-a (b B-a C)\right )}{3 f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}-\frac{2 \left (-a^2 b^2 (2 d (5 A-C)+3 B c)+7 a^3 b B d-4 a^4 C d+a b^3 (6 A c+B d-6 c C)+b^4 (3 B c-4 A d)\right )}{3 f \left (a^2+b^2\right )^2 (b c-a d)^2 \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}-\frac{(i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a-i b)^{5/2} (c-i d)^{3/2}}-\frac{(B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a+i b)^{5/2} (c+i d)^{3/2}} \]
Antiderivative was successfully verified.
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Rule 3649
Rule 3616
Rule 3615
Rule 93
Rule 208
Rubi steps
\begin{align*} \int \frac{A+B \tan (e+f x)+C \tan ^2(e+f x)}{(a+b \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2}} \, dx &=-\frac{2 \left (A b^2-a (b B-a C)\right )}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}-\frac{2 \int \frac{\frac{1}{2} \left (4 A b^2 d-3 a A (b c-a d)-(b B-a C) (3 b c+a d)\right )+\frac{3}{2} (A b-a B-b C) (b c-a d) \tan (e+f x)+2 \left (A b^2-a (b B-a C)\right ) d \tan ^2(e+f x)}{(a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx}{3 \left (a^2+b^2\right ) (b c-a d)}\\ &=-\frac{2 \left (A b^2-a (b B-a C)\right )}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}-\frac{2 \left (7 a^3 b B d-4 a^4 C d+b^4 (3 B c-4 A d)+a b^3 (6 A c-6 c C+B d)-a^2 b^2 (3 B c+2 (5 A-C) d)\right )}{3 \left (a^2+b^2\right )^2 (b c-a d)^2 f \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}+\frac{4 \int \frac{\frac{1}{4} \left (-\left (a b c-a^2 d-2 b^2 d\right ) \left (a^2 (3 A+C) d-b^2 (3 B c-4 A d)-a b (3 A c-3 c C+B d)\right )+(b c+a d) \left (3 b^3 c C-7 a^2 b B d+4 a^3 C d-A b^2 (3 b c-7 a d)+3 a b^2 (B c-C d)\right )\right )+\frac{3}{4} \left (a^2 B-b^2 B-2 a b (A-C)\right ) (b c-a d)^2 \tan (e+f x)-\frac{1}{2} d \left (7 a^3 b B d-4 a^4 C d+b^4 (3 B c-4 A d)+a b^3 (6 A c-6 c C+B d)-a^2 b^2 (3 B c+10 A d-2 C d)\right ) \tan ^2(e+f x)}{\sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx}{3 \left (a^2+b^2\right )^2 (b c-a d)^2}\\ &=-\frac{2 \left (A b^2-a (b B-a C)\right )}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}-\frac{2 \left (7 a^3 b B d-4 a^4 C d+b^4 (3 B c-4 A d)+a b^3 (6 A c-6 c C+B d)-a^2 b^2 (3 B c+2 (5 A-C) d)\right )}{3 \left (a^2+b^2\right )^2 (b c-a d)^2 f \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}-\frac{2 d \left (8 a^3 b B d \left (c^2+d^2\right )+2 a b^3 (3 A c-3 c C+B d) \left (c^2+d^2\right )-a^4 d \left (8 c^2 C-3 B c d+(3 A+5 C) d^2\right )-a^2 b^2 \left (3 B c^3+11 A c^2 d+5 c^2 C d-3 B c d^2+17 A d^3-C d^3\right )-b^4 \left (d \left (5 A c^2+3 c^2 C+8 A d^2\right )-3 B \left (c^3+2 c d^2\right )\right )\right ) \sqrt{a+b \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d)^3 \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{8 \int \frac{\frac{3}{8} (b c-a d)^3 \left (a^2 (A c-c C+B d)-b^2 (A c-c C+B d)+2 a b (B c-(A-C) d)\right )-\frac{3}{8} (b c-a d)^3 \left (2 a b (A c-c C+B d)-a^2 (B c-(A-C) d)+b^2 (B c-(A-C) d)\right ) \tan (e+f x)}{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{3 \left (a^2+b^2\right )^2 (b c-a d)^3 \left (c^2+d^2\right )}\\ &=-\frac{2 \left (A b^2-a (b B-a C)\right )}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}-\frac{2 \left (7 a^3 b B d-4 a^4 C d+b^4 (3 B c-4 A d)+a b^3 (6 A c-6 c C+B d)-a^2 b^2 (3 B c+2 (5 A-C) d)\right )}{3 \left (a^2+b^2\right )^2 (b c-a d)^2 f \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}-\frac{2 d \left (8 a^3 b B d \left (c^2+d^2\right )+2 a b^3 (3 A c-3 c C+B d) \left (c^2+d^2\right )-a^4 d \left (8 c^2 C-3 B c d+(3 A+5 C) d^2\right )-a^2 b^2 \left (3 B c^3+11 A c^2 d+5 c^2 C d-3 B c d^2+17 A d^3-C d^3\right )-b^4 \left (d \left (5 A c^2+3 c^2 C+8 A d^2\right )-3 B \left (c^3+2 c d^2\right )\right )\right ) \sqrt{a+b \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d)^3 \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{(A-i B-C) \int \frac{1+i \tan (e+f x)}{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{2 (a-i b)^2 (c-i d)}+\frac{(A+i B-C) \int \frac{1-i \tan (e+f x)}{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{2 (a+i b)^2 (c+i d)}\\ &=-\frac{2 \left (A b^2-a (b B-a C)\right )}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}-\frac{2 \left (7 a^3 b B d-4 a^4 C d+b^4 (3 B c-4 A d)+a b^3 (6 A c-6 c C+B d)-a^2 b^2 (3 B c+2 (5 A-C) d)\right )}{3 \left (a^2+b^2\right )^2 (b c-a d)^2 f \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}-\frac{2 d \left (8 a^3 b B d \left (c^2+d^2\right )+2 a b^3 (3 A c-3 c C+B d) \left (c^2+d^2\right )-a^4 d \left (8 c^2 C-3 B c d+(3 A+5 C) d^2\right )-a^2 b^2 \left (3 B c^3+11 A c^2 d+5 c^2 C d-3 B c d^2+17 A d^3-C d^3\right )-b^4 \left (d \left (5 A c^2+3 c^2 C+8 A d^2\right )-3 B \left (c^3+2 c d^2\right )\right )\right ) \sqrt{a+b \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d)^3 \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{(A-i B-C) \operatorname{Subst}\left (\int \frac{1}{(1-i x) \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 (a-i b)^2 (c-i d) f}+\frac{(A+i B-C) \operatorname{Subst}\left (\int \frac{1}{(1+i x) \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 (a+i b)^2 (c+i d) f}\\ &=-\frac{2 \left (A b^2-a (b B-a C)\right )}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}-\frac{2 \left (7 a^3 b B d-4 a^4 C d+b^4 (3 B c-4 A d)+a b^3 (6 A c-6 c C+B d)-a^2 b^2 (3 B c+2 (5 A-C) d)\right )}{3 \left (a^2+b^2\right )^2 (b c-a d)^2 f \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}-\frac{2 d \left (8 a^3 b B d \left (c^2+d^2\right )+2 a b^3 (3 A c-3 c C+B d) \left (c^2+d^2\right )-a^4 d \left (8 c^2 C-3 B c d+(3 A+5 C) d^2\right )-a^2 b^2 \left (3 B c^3+11 A c^2 d+5 c^2 C d-3 B c d^2+17 A d^3-C d^3\right )-b^4 \left (d \left (5 A c^2+3 c^2 C+8 A d^2\right )-3 B \left (c^3+2 c d^2\right )\right )\right ) \sqrt{a+b \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d)^3 \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{(A-i B-C) \operatorname{Subst}\left (\int \frac{1}{i a+b-(i c+d) x^2} \, dx,x,\frac{\sqrt{a+b \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}\right )}{(a-i b)^2 (c-i d) f}+\frac{(A+i B-C) \operatorname{Subst}\left (\int \frac{1}{-i a+b-(-i c+d) x^2} \, dx,x,\frac{\sqrt{a+b \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}\right )}{(a+i b)^2 (c+i d) f}\\ &=-\frac{(i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{(a-i b)^{5/2} (c-i d)^{3/2} f}-\frac{(B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{(a+i b)^{5/2} (c+i d)^{3/2} f}-\frac{2 \left (A b^2-a (b B-a C)\right )}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}-\frac{2 \left (7 a^3 b B d-4 a^4 C d+b^4 (3 B c-4 A d)+a b^3 (6 A c-6 c C+B d)-a^2 b^2 (3 B c+2 (5 A-C) d)\right )}{3 \left (a^2+b^2\right )^2 (b c-a d)^2 f \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}-\frac{2 d \left (8 a^3 b B d \left (c^2+d^2\right )+2 a b^3 (3 A c-3 c C+B d) \left (c^2+d^2\right )-a^4 d \left (8 c^2 C-3 B c d+(3 A+5 C) d^2\right )-a^2 b^2 \left (3 B c^3+11 A c^2 d+5 c^2 C d-3 B c d^2+17 A d^3-C d^3\right )-b^4 \left (d \left (5 A c^2+3 c^2 C+8 A d^2\right )-3 B \left (c^3+2 c d^2\right )\right )\right ) \sqrt{a+b \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d)^3 \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}\\ \end{align*}
Mathematica [A] time = 6.84906, size = 902, normalized size = 1.51 \[ -\frac{2 \left (A b^2-a (b B-a C)\right )}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}-\frac{2 \left (-\frac{2 \left (\frac{1}{2} b^2 \left (4 A d b^2-3 a A (b c-a d)-(b B-a C) (3 b c+a d)\right )-a \left (\frac{3}{2} b (A b-C b-a B) (b c-a d)-2 a \left (A b^2-a (b B-a C)\right ) d\right )\right )}{\left (a^2+b^2\right ) (b c-a d) f \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}-\frac{2 \left (-\frac{3 \left (\frac{(A+i B-C) (i c+d) \tan ^{-1}\left (\frac{\sqrt{-c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right ) (a-i b)^2}{\sqrt{a+i b} \sqrt{-c-i d}}+\frac{(a+i b)^2 (i A+B-i C) (c+i d) \tan ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{i b-a} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{i b-a} \sqrt{c-i d}}\right ) (b c-a d)^3}{4 (a d-b c) \left (c^2+d^2\right ) f}-\frac{2 \left (d^2 \left (\left (-\frac{b c}{2}-\frac{a d}{2}\right ) \left (\frac{3}{2} b (A b-C b-a B) (b c-a d)-2 a \left (A b^2-a (b B-a C)\right ) d\right )+\frac{1}{2} \left (b^2 d-\frac{1}{2} a (b c-a d)\right ) \left (4 A d b^2-3 a A (b c-a d)-(b B-a C) (3 b c+a d)\right )\right )-c \left (\frac{1}{2} d (b c-a d) \left (-2 b \left (A b^2-a (b B-a C)\right ) d-\frac{3}{2} a (A b-C b-a B) (b c-a d)+\frac{1}{2} b \left (4 A d b^2-3 a A (b c-a d)-(b B-a C) (3 b c+a d)\right )\right )-c d \left (\frac{1}{2} b^2 \left (4 A d b^2-3 a A (b c-a d)-(b B-a C) (3 b c+a d)\right )-a \left (\frac{3}{2} b (A b-C b-a B) (b c-a d)-2 a \left (A b^2-a (b B-a C)\right ) d\right )\right )\right )\right ) \sqrt{a+b \tan (e+f x)}}{(a d-b c) \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}\right )}{\left (a^2+b^2\right ) (b c-a d)}\right )}{3 \left (a^2+b^2\right ) (b c-a d)} \]
Antiderivative was successfully verified.
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Maple [F] time = 180., size = 0, normalized size = 0. \begin{align*} \int{(A+B\tan \left ( fx+e \right ) +C \left ( \tan \left ( fx+e \right ) \right ) ^{2}) \left ( a+b\tan \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}} \left ( c+d\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [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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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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