Optimal. Leaf size=209 \[ -\frac{2 \left (A d^2-B c d+c^2 C\right )}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac{2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right )^2 \sqrt{c+d \tan (e+f x)}}-\frac{(i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{5/2}}-\frac{(B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{5/2}} \]
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Rubi [A] time = 0.486027, antiderivative size = 209, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {3628, 3529, 3539, 3537, 63, 208} \[ -\frac{2 \left (A d^2-B c d+c^2 C\right )}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac{2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right )^2 \sqrt{c+d \tan (e+f x)}}-\frac{(i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{5/2}}-\frac{(B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 3628
Rule 3529
Rule 3539
Rule 3537
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{A+B \tan (e+f x)+C \tan ^2(e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx &=-\frac{2 \left (c^2 C-B c d+A d^2\right )}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{\int \frac{A c-c C+B d+(B c-(A-C) d) \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx}{c^2+d^2}\\ &=-\frac{2 \left (c^2 C-B c d+A d^2\right )}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )}{\left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{\int \frac{-c^2 C+2 B c d+C d^2+A \left (c^2-d^2\right )-\left (2 c (A-C) d-B \left (c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{\left (c^2+d^2\right )^2}\\ &=-\frac{2 \left (c^2 C-B c d+A d^2\right )}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )}{\left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{(A-i B-C) \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (c-i d)^2}+\frac{(A+i B-C) \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 )}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )}{\left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{(i A+B-i C) \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{(i (A+i B-C)) \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 )}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )}{\left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}-\frac{(A+i B-C) \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-C) \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 )}{d (i c+d)^2 f}\\ &=-\frac{(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{(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 )}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )}{\left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}\\ \end{align*}
Mathematica [C] time = 0.853108, size = 223, normalized size = 1.07 \[ -\frac{(d (C-A)+B c) \left (i (c+i d) \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\frac{c+d \tan (e+f x)}{c-i d}\right )-(d+i c) \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\frac{c+d \tan (e+f x)}{c+i d}\right )\right )-3 B (c+d \tan (e+f x)) \left (i (c+i d) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{c+d \tan (e+f x)}{c-i d}\right )-(d+i c) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{c+d \tan (e+f x)}{c+i d}\right )\right )+2 C \left (c^2+d^2\right )}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.149, size = 20647, normalized size = 98.8 \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{A + B \tan{\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \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{C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A}{{\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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