Optimal. Leaf size=585 \[ \frac{2 b \sqrt{c+d \tan (e+f x)} \left (6 a^2 d^3 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )+3 a b d \left (-c^2 d^2 (A-17 C)+d^4 (5 A+3 C)-2 B c^3 d-8 B c d^3+8 c^4 C\right )-b^2 \left (2 c^3 d^2 (A+15 C)+8 c d^4 (A+C)-17 B c^2 d^3-8 B c^4 d-3 B d^5+16 c^5 C\right )\right )}{3 d^4 f \left (c^2+d^2\right )^2}+\frac{2 b^2 \tan (e+f x) \sqrt{c+d \tan (e+f x)} \left (3 a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )+b \left (c^2 d^2 (A+15 C)+d^4 (7 A+C)-4 B c^3 d-10 B c d^3+8 c^4 C\right )\right )}{3 d^3 f \left (c^2+d^2\right )^2}-\frac{2 (a+b \tan (e+f x))^2 \left (a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )+b \left (2 A d^4-B c^3 d-3 B c d^3+4 c^2 C d^2+2 c^4 C\right )\right )}{d^2 f \left (c^2+d^2\right )^2 \sqrt{c+d \tan (e+f x)}}-\frac{2 \left (A d^2-B c d+c^2 C\right ) (a+b \tan (e+f x))^3}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac{(-b+i a)^3 (A+i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{5/2}}-\frac{(a-i b)^3 (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{5/2}} \]
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Rubi [A] time = 2.96752, antiderivative size = 585, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 47, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.149, Rules used = {3645, 3637, 3630, 3539, 3537, 63, 208} \[ \frac{2 b \sqrt{c+d \tan (e+f x)} \left (6 a^2 d^3 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )+3 a b d \left (-c^2 d^2 (A-17 C)+d^4 (5 A+3 C)-2 B c^3 d-8 B c d^3+8 c^4 C\right )-b^2 \left (2 c^3 d^2 (A+15 C)+8 c d^4 (A+C)-17 B c^2 d^3-8 B c^4 d-3 B d^5+16 c^5 C\right )\right )}{3 d^4 f \left (c^2+d^2\right )^2}+\frac{2 b^2 \tan (e+f x) \sqrt{c+d \tan (e+f x)} \left (3 a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )+b \left (c^2 d^2 (A+15 C)+d^4 (7 A+C)-4 B c^3 d-10 B c d^3+8 c^4 C\right )\right )}{3 d^3 f \left (c^2+d^2\right )^2}-\frac{2 (a+b \tan (e+f x))^2 \left (a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )+b \left (2 A d^4-B c^3 d-3 B c d^3+4 c^2 C d^2+2 c^4 C\right )\right )}{d^2 f \left (c^2+d^2\right )^2 \sqrt{c+d \tan (e+f x)}}-\frac{2 \left (A d^2-B c d+c^2 C\right ) (a+b \tan (e+f x))^3}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac{(-b+i a)^3 (A+i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{5/2}}-\frac{(a-i b)^3 (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 3645
Rule 3637
Rule 3630
Rule 3539
Rule 3537
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{(a+b \tan (e+f x))^3 \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{5/2}} \, dx &=-\frac{2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^3}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 \int \frac{(a+b \tan (e+f x))^2 \left (\frac{3}{2} \left (A d (a c+2 b d)+\frac{2}{3} \left (3 b c-\frac{3 a d}{2}\right ) (c C-B d)\right )+\frac{3}{2} d ((A-C) (b c-a d)+B (a c+b d)) \tan (e+f x)+\frac{3}{2} b \left (2 c^2 C-B c d+(A+C) d^2\right ) \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx}{3 d \left (c^2+d^2\right )}\\ &=-\frac{2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^3}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{2 \left (b \left (2 c^4 C-B c^3 d+4 c^2 C d^2-3 B c d^3+2 A d^4\right )+a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) (a+b \tan (e+f x))^2}{d^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{4 \int \frac{(a+b \tan (e+f x)) \left (-\frac{3}{4} \left ((4 b c-a d) \left (a d^2 (B c-(A-C) d)-b (2 c C-B d) \left (c^2+d^2\right )\right )-d (a c+4 b d) \left (a d (A c-c C+B d)+2 b \left (c^2 C-B c d+A d^2\right )\right )\right )-\frac{3}{4} d^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{3}{4} b \left (b \left (8 c^4 C-4 B c^3 d+c^2 (A+15 C) d^2-10 B c d^3+(7 A+C) d^4\right )+3 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \tan ^2(e+f x)\right )}{\sqrt{c+d \tan (e+f x)}} \, dx}{3 d^2 \left (c^2+d^2\right )^2}\\ &=-\frac{2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^3}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{2 \left (b \left (2 c^4 C-B c^3 d+4 c^2 C d^2-3 B c d^3+2 A d^4\right )+a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) (a+b \tan (e+f x))^2}{d^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{2 b^2 \left (b \left (8 c^4 C-4 B c^3 d+c^2 (A+15 C) d^2-10 B c d^3+(7 A+C) d^4\right )+3 a d^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)}}{3 d^3 \left (c^2+d^2\right )^2 f}-\frac{8 \int \frac{-\frac{3}{8} \left (6 a b^2 d \left (4 c^4 C-B c^3 d-2 c^2 (A-5 C) d^2-7 B c d^3+4 A d^4\right )-2 b^3 c \left (8 c^4 C-4 B c^3 d+c^2 (A+15 C) d^2-10 B c d^3+(7 A+C) d^4\right )-3 a^3 d^3 \left (c^2 C-2 B c d-C d^2-A \left (c^2-d^2\right )\right )+15 a^2 b d^3 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )+\frac{9}{8} d^3 \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)-\frac{3}{8} b \left (3 a b d \left (8 c^4 C-2 B c^3 d-c^2 (A-17 C) d^2-8 B c d^3+(5 A+3 C) d^4\right )-b^2 \left (16 c^5 C-8 B c^4 d+2 c^3 (A+15 C) d^2-17 B c^2 d^3+8 c (A+C) d^4-3 B d^5\right )+6 a^2 d^3 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \tan ^2(e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{9 d^3 \left (c^2+d^2\right )^2}\\ &=-\frac{2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^3}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{2 \left (b \left (2 c^4 C-B c^3 d+4 c^2 C d^2-3 B c d^3+2 A d^4\right )+a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) (a+b \tan (e+f x))^2}{d^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{2 b \left (3 a b d \left (8 c^4 C-2 B c^3 d-c^2 (A-17 C) d^2-8 B c d^3+(5 A+3 C) d^4\right )-b^2 \left (16 c^5 C-8 B c^4 d+2 c^3 (A+15 C) d^2-17 B c^2 d^3+8 c (A+C) d^4-3 B d^5\right )+6 a^2 d^3 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \sqrt{c+d \tan (e+f x)}}{3 d^4 \left (c^2+d^2\right )^2 f}+\frac{2 b^2 \left (b \left (8 c^4 C-4 B c^3 d+c^2 (A+15 C) d^2-10 B c d^3+(7 A+C) d^4\right )+3 a d^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)}}{3 d^3 \left (c^2+d^2\right )^2 f}-\frac{8 \int \frac{\frac{9}{8} d^3 \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 )+\frac{9}{8} d^3 \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}{9 d^3 \left (c^2+d^2\right )^2}\\ &=-\frac{2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^3}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{2 \left (b \left (2 c^4 C-B c^3 d+4 c^2 C d^2-3 B c d^3+2 A d^4\right )+a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) (a+b \tan (e+f x))^2}{d^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{2 b \left (3 a b d \left (8 c^4 C-2 B c^3 d-c^2 (A-17 C) d^2-8 B c d^3+(5 A+3 C) d^4\right )-b^2 \left (16 c^5 C-8 B c^4 d+2 c^3 (A+15 C) d^2-17 B c^2 d^3+8 c (A+C) d^4-3 B d^5\right )+6 a^2 d^3 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \sqrt{c+d \tan (e+f x)}}{3 d^4 \left (c^2+d^2\right )^2 f}+\frac{2 b^2 \left (b \left (8 c^4 C-4 B c^3 d+c^2 (A+15 C) d^2-10 B c d^3+(7 A+C) d^4\right )+3 a d^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)}}{3 d^3 \left (c^2+d^2\right )^2 f}+\frac{\left ((a-i b)^3 (A-i B-C)\right ) \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (c-i d)^2}+\frac{\left ((a+i b)^3 (A+i B-C)\right ) \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (c+i d)^2}\\ &=-\frac{2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^3}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{2 \left (b \left (2 c^4 C-B c^3 d+4 c^2 C d^2-3 B c d^3+2 A d^4\right )+a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) (a+b \tan (e+f x))^2}{d^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{2 b \left (3 a b d \left (8 c^4 C-2 B c^3 d-c^2 (A-17 C) d^2-8 B c d^3+(5 A+3 C) d^4\right )-b^2 \left (16 c^5 C-8 B c^4 d+2 c^3 (A+15 C) d^2-17 B c^2 d^3+8 c (A+C) d^4-3 B d^5\right )+6 a^2 d^3 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \sqrt{c+d \tan (e+f x)}}{3 d^4 \left (c^2+d^2\right )^2 f}+\frac{2 b^2 \left (b \left (8 c^4 C-4 B c^3 d+c^2 (A+15 C) d^2-10 B c d^3+(7 A+C) d^4\right )+3 a d^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)}}{3 d^3 \left (c^2+d^2\right )^2 f}+\frac{\left (i (a-i b)^3 (A-i B-C)\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 (c-i d)^2 f}-\frac{\left (i (a+i b)^3 (A+i B-C)\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 (c+i d)^2 f}\\ &=-\frac{2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^3}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{2 \left (b \left (2 c^4 C-B c^3 d+4 c^2 C d^2-3 B c d^3+2 A d^4\right )+a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) (a+b \tan (e+f x))^2}{d^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{2 b \left (3 a b d \left (8 c^4 C-2 B c^3 d-c^2 (A-17 C) d^2-8 B c d^3+(5 A+3 C) d^4\right )-b^2 \left (16 c^5 C-8 B c^4 d+2 c^3 (A+15 C) d^2-17 B c^2 d^3+8 c (A+C) d^4-3 B d^5\right )+6 a^2 d^3 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \sqrt{c+d \tan (e+f x)}}{3 d^4 \left (c^2+d^2\right )^2 f}+\frac{2 b^2 \left (b \left (8 c^4 C-4 B c^3 d+c^2 (A+15 C) d^2-10 B c d^3+(7 A+C) d^4\right )+3 a d^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)}}{3 d^3 \left (c^2+d^2\right )^2 f}-\frac{\left ((a-i b)^3 (A-i B-C)\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 )}{(c-i d)^2 d f}-\frac{\left ((a+i b)^3 (A+i B-C)\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 )}{(c+i d)^2 d f}\\ &=-\frac{(a-i b)^3 (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{(c-i d)^{5/2} f}-\frac{(i a-b)^3 (A+i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{(c+i d)^{5/2} f}-\frac{2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^3}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{2 \left (b \left (2 c^4 C-B c^3 d+4 c^2 C d^2-3 B c d^3+2 A d^4\right )+a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) (a+b \tan (e+f x))^2}{d^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{2 b \left (3 a b d \left (8 c^4 C-2 B c^3 d-c^2 (A-17 C) d^2-8 B c d^3+(5 A+3 C) d^4\right )-b^2 \left (16 c^5 C-8 B c^4 d+2 c^3 (A+15 C) d^2-17 B c^2 d^3+8 c (A+C) d^4-3 B d^5\right )+6 a^2 d^3 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \sqrt{c+d \tan (e+f x)}}{3 d^4 \left (c^2+d^2\right )^2 f}+\frac{2 b^2 \left (b \left (8 c^4 C-4 B c^3 d+c^2 (A+15 C) d^2-10 B c d^3+(7 A+C) d^4\right )+3 a d^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)}}{3 d^3 \left (c^2+d^2\right )^2 f}\\ \end{align*}
Mathematica [C] time = 6.83154, size = 670, normalized size = 1.15 \[ \frac{2 C (a+b \tan (e+f x))^3}{3 d f (c+d \tan (e+f x))^{3/2}}+\frac{2 \left (-\frac{3 (-2 a C d-b B d+2 b c C) (a+b \tan (e+f x))^2}{d f (c+d \tan (e+f x))^{3/2}}+\frac{2 \left (-\frac{3 (a+b \tan (e+f x)) \left (b d^2 (a B+A b-b C)+4 (b c-a d) (-2 a C d-b B d+2 b c C)\right )}{2 d f (c+d \tan (e+f x))^{3/2}}-\frac{3 \left (-\frac{2 \left (9 a^2 b B d^3-48 a^2 b c C d^2+16 a^3 C d^3-18 a b^2 B c d^2+48 a b^2 c^2 C d-2 A b^3 c d^2+8 b^3 B c^2 d+b^3 B d^3-16 b^3 c^3 C+2 b^3 c C d^2\right )}{3 d (c+d \tan (e+f x))^{3/2}}+\frac{2 \left (\frac{\left (\frac{3}{2} c d^4 \left (3 a^2 b (A-C)+a^3 B-3 a b^2 B-b^3 (A-C)\right )+\frac{3}{2} d^5 \left (a^3 (-(A-C))+3 a^2 b B+3 a b^2 (A-C)-b^3 B\right )\right ) \left (\frac{\text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\frac{c+d \tan (e+f x)}{c+i d}\right )}{3 (-d+i c) (c+d \tan (e+f x))^{3/2}}-\frac{\text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\frac{c+d \tan (e+f x)}{c-i d}\right )}{3 (d+i c) (c+d \tan (e+f x))^{3/2}}\right )}{d}-\frac{3}{2} d^3 \left (3 a^2 b (A-C)+a^3 B-3 a b^2 B-b^3 (A-C)\right ) \left (\frac{\text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{c+d \tan (e+f x)}{c+i d}\right )}{(-d+i c) \sqrt{c+d \tan (e+f x)}}-\frac{\text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{c+d \tan (e+f x)}{c-i d}\right )}{(d+i c) \sqrt{c+d \tan (e+f x)}}\right )\right )}{3 d}\right )}{4 d f}\right )}{d}\right )}{3 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.284, size = 85156, normalized size = 145.6 \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(-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] 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 (b \tan \left (f x + e\right ) + a\right )}^{3}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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