Optimal. Leaf size=358 \[ \frac{2 (b c-a d) \left (3 a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )+b \left (-2 c^2 d^2 (A-5 C)+4 A d^4-B c^3 d-7 B c d^3+4 c^4 C\right )\right )}{3 d^3 f \left (c^2+d^2\right )^2 \sqrt{c+d \tan (e+f x)}}-\frac{2 \left (A d^2-B c d+c^2 C\right ) (a+b \tan (e+f x))^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac{(a-i b)^2 (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{5/2}}-\frac{(a+i b)^2 (B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{5/2}}+\frac{2 b^2 \left (d^2 (A+3 C)-B c d+4 c^2 C\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 f \left (c^2+d^2\right )} \]
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Rubi [A] time = 1.55107, antiderivative size = 358, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 47, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.149, Rules used = {3645, 3635, 3630, 3539, 3537, 63, 208} \[ \frac{2 (b c-a d) \left (3 a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )+b \left (-2 c^2 d^2 (A-5 C)+4 A d^4-B c^3 d-7 B c d^3+4 c^4 C\right )\right )}{3 d^3 f \left (c^2+d^2\right )^2 \sqrt{c+d \tan (e+f x)}}-\frac{2 \left (A d^2-B c d+c^2 C\right ) (a+b \tan (e+f x))^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac{(a-i b)^2 (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{5/2}}-\frac{(a+i b)^2 (B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{5/2}}+\frac{2 b^2 \left (d^2 (A+3 C)-B c d+4 c^2 C\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 f \left (c^2+d^2\right )} \]
Antiderivative was successfully verified.
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Rule 3645
Rule 3635
Rule 3630
Rule 3539
Rule 3537
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{(a+b \tan (e+f x))^2 \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))^2}{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)) \left (\frac{1}{2} \left (2 A d \left (\frac{3 a c}{2}+2 b d\right )+2 \left (2 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{1}{2} b \left (4 c^2 C-B c d+(A+3 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))^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 (b c-a d) \left (b \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 )+3 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{3 d^3 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{2 \int \frac{\frac{1}{2} \left (b^2 \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 )-3 a^2 d^2 \left (c^2 C-2 B c d-C d^2-A \left (c^2-d^2\right )\right )+6 a b d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )-\frac{3}{2} 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{1}{2} b^2 \left (c^2+d^2\right ) \left (4 c^2 C-B c d+(A+3 C) d^2\right ) \tan ^2(e+f x)}{\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))^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 (b c-a d) \left (b \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 )+3 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{3 d^3 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{2 b^2 \left (4 c^2 C-B c d+(A+3 C) d^2\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}+\frac{2 \int \frac{-\frac{3}{2} d^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{3}{2} 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)}{\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))^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 (b c-a d) \left (b \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 )+3 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{3 d^3 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{2 b^2 \left (4 c^2 C-B c d+(A+3 C) d^2\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}+\frac{\left ((a-i b)^2 (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)^2 (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))^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 (b c-a d) \left (b \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 )+3 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{3 d^3 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{2 b^2 \left (4 c^2 C-B c d+(A+3 C) d^2\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}+\frac{\left ((a-i b)^2 (i A+B-i 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)^2 (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))^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 (b c-a d) \left (b \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 )+3 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{3 d^3 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{2 b^2 \left (4 c^2 C-B c d+(A+3 C) d^2\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}-\frac{\left ((a-i b)^2 (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)^2 (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)^2 (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{(a+i b)^2 (B-i (A-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))^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 (b c-a d) \left (b \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 )+3 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{3 d^3 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{2 b^2 \left (4 c^2 C-B c d+(A+3 C) d^2\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}\\ \end{align*}
Mathematica [C] time = 6.47687, size = 502, normalized size = 1.4 \[ \frac{2 C (a+b \tan (e+f x))^2}{d f (c+d \tan (e+f x))^{3/2}}+\frac{2 \left (-\frac{(4 a C d+b B d-4 b c C) (a+b \tan (e+f x))}{d f (c+d \tan (e+f x))^{3/2}}-\frac{-\frac{2 \left (8 a^2 C d^2+a b B d^2-16 a b c C d-A b^2 d^2-2 b^2 B c d+8 b^2 c^2 C+b^2 C d^2\right )}{3 d (c+d \tan (e+f x))^{3/2}}+\frac{2 \left (\frac{\left (\frac{3}{2} c d^3 \left (a^2 B+2 a b (A-C)-b^2 B\right )+\frac{3}{2} d^4 \left (a^2 (-(A-C))+2 a b B+b^2 (A-C)\right )\right ) \left (\frac{\text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\frac{c+d \tan (e+f x)}{c+i d}\right )}{3 (-d+i c) (c+d \tan (e+f x))^{3/2}}-\frac{\text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\frac{c+d \tan (e+f x)}{c-i d}\right )}{3 (d+i c) (c+d \tan (e+f x))^{3/2}}\right )}{d}-\frac{3}{2} d^2 \left (a^2 B+2 a b (A-C)-b^2 B\right ) \left (\frac{\text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{c+d \tan (e+f x)}{c+i d}\right )}{(-d+i c) \sqrt{c+d \tan (e+f x)}}-\frac{\text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{c+d \tan (e+f x)}{c-i d}\right )}{(d+i c) \sqrt{c+d \tan (e+f x)}}\right )\right )}{3 d}}{2 d f}\right )}{d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.247, size = 61833, normalized size = 172.7 \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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \tan{\left (e + f x \right )}\right )^{2} \left (A + B \tan{\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )}{\left (c + d \tan{\left (e + f x \right )}\right )^{\frac{5}{2}}}\, dx \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 )}^{2}}{{\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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