Optimal. Leaf size=201 \[ \frac{2 (b c-a d) \left (A d^2-B c d+c^2 C\right )}{d^2 f \left (c^2+d^2\right ) \sqrt{c+d \tan (e+f x)}}-\frac{(b+i a) (A-i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{3/2}}+\frac{(-b+i a) (A+i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{3/2}}+\frac{2 b C \sqrt{c+d \tan (e+f x)}}{d^2 f} \]
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Rubi [A] time = 0.554369, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {3635, 3630, 3539, 3537, 63, 208} \[ \frac{2 (b c-a d) \left (A d^2-B c d+c^2 C\right )}{d^2 f \left (c^2+d^2\right ) \sqrt{c+d \tan (e+f x)}}-\frac{(b+i a) (A-i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{3/2}}+\frac{(-b+i a) (A+i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{3/2}}+\frac{2 b C \sqrt{c+d \tan (e+f x)}}{d^2 f} \]
Antiderivative was successfully verified.
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Rule 3635
Rule 3630
Rule 3539
Rule 3537
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{(a+b \tan (e+f x)) \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx &=\frac{2 (b c-a d) \left (c^2 C-B c d+A d^2\right )}{d^2 \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{\int \frac{a d (A c-c C+B d)+b \left (c^2 C-B c d+A d^2\right )+d (A b c+a B c-b c C-a A d+b B d+a C d) \tan (e+f x)+b C \left (c^2+d^2\right ) \tan ^2(e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{d \left (c^2+d^2\right )}\\ &=\frac{2 (b c-a d) \left (c^2 C-B c d+A d^2\right )}{d^2 \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{2 b C \sqrt{c+d \tan (e+f x)}}{d^2 f}+\frac{\int \frac{d (a (A c-c C+B d)-b (B c-(A-C) d))+d (A b c+a B c-b c C-a A d+b B d+a C d) \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{d \left (c^2+d^2\right )}\\ &=\frac{2 (b c-a d) \left (c^2 C-B c d+A d^2\right )}{d^2 \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{2 b C \sqrt{c+d \tan (e+f x)}}{d^2 f}+\frac{((a-i b) (A-i B-C)) \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (c-i d)}+\frac{((a+i b) (A+i B-C)) \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (c+i d)}\\ &=\frac{2 (b c-a d) \left (c^2 C-B c d+A d^2\right )}{d^2 \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{2 b C \sqrt{c+d \tan (e+f x)}}{d^2 f}+\frac{(i (a-i b) (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) f}-\frac{((i a-b) (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) f}\\ &=\frac{2 (b c-a d) \left (c^2 C-B c d+A d^2\right )}{d^2 \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{2 b C \sqrt{c+d \tan (e+f x)}}{d^2 f}-\frac{((a-i b) (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) d f}-\frac{((a+i b) (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) d f}\\ &=-\frac{(i a+b) (A-i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{(c-i d)^{3/2} f}+\frac{(i a-b) (A+i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{(c+i d)^{3/2} f}+\frac{2 (b c-a d) \left (c^2 C-B c d+A d^2\right )}{d^2 \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{2 b C \sqrt{c+d \tan (e+f x)}}{d^2 f}\\ \end{align*}
Mathematica [C] time = 2.39151, size = 290, normalized size = 1.44 \[ \frac{\frac{(-a A d+a B c+a C d+A b c+b B d-b c C) \left ((d-i c) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{c+d \tan (e+f x)}{c-i d}\right )+(d+i c) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{c+d \tan (e+f x)}{c+i d}\right )\right )}{\left (c^2+d^2\right ) \sqrt{c+d \tan (e+f x)}}+(a B+A b-b C) \left (\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{\sqrt{c+i d}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{\sqrt{c-i d}}\right )-\frac{2 (2 a C d+b B d-2 b c C)}{d \sqrt{c+d \tan (e+f x)}}+\frac{2 C (a+b \tan (e+f x))}{\sqrt{c+d \tan (e+f x)}}}{d f} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.164, size = 23472, normalized size = 116.8 \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{\left (a + b \tan{\left (e + f x \right )}\right ) \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{3}{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 )}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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