Optimal. Leaf size=251 \[ \frac{2 \left (A d^2-B c d+c^2 C\right ) \sqrt{a+b \tan (e+f x)}}{f \left (c^2+d^2\right ) (b c-a d) \sqrt{c+d \tan (e+f x)}}-\frac{(B+i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{a-i b} (c-i d)^{3/2}}+\frac{(i A-B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{a+i b} (c+i d)^{3/2}} \]
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Rubi [A] time = 1.00079, antiderivative size = 251, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{2 \left (A d^2-B c d+c^2 C\right ) \sqrt{a+b \tan (e+f x)}}{f \left (c^2+d^2\right ) (b c-a d) \sqrt{c+d \tan (e+f x)}}-\frac{(B+i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{a-i b} (c-i d)^{3/2}}+\frac{(i A-B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{a+i b} (c+i d)^{3/2}} \]
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)}{\sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx &=\frac{2 \left (c^2 C-B c d+A d^2\right ) \sqrt{a+b \tan (e+f x)}}{(b c-a d) \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{2 \int \frac{\frac{1}{2} (b c-a d) (A c-c C+B d)+\frac{1}{2} (b c-a d) (B c-(A-C) d) \tan (e+f x)}{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{(b c-a d) \left (c^2+d^2\right )}\\ &=\frac{2 \left (c^2 C-B c d+A d^2\right ) \sqrt{a+b \tan (e+f x)}}{(b c-a d) \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 (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 (c+i d)}\\ &=\frac{2 \left (c^2 C-B c d+A d^2\right ) \sqrt{a+b \tan (e+f x)}}{(b c-a d) \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 (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 (c+i d) f}\\ &=\frac{2 \left (c^2 C-B c d+A d^2\right ) \sqrt{a+b \tan (e+f x)}}{(b c-a d) \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 )}{(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 )}{(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 )}{\sqrt{a-i b} (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 )}{\sqrt{a+i b} (c+i d)^{3/2} f}+\frac{2 \left (c^2 C-B c d+A d^2\right ) \sqrt{a+b \tan (e+f x)}}{(b c-a d) \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}\\ \end{align*}
Mathematica [A] time = 3.11554, size = 275, normalized size = 1.1 \[ -\frac{\frac{2 \left (A d^2-B c d+c^2 C\right ) \sqrt{a+b \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}+(b c-a d) \left (\frac{(d+i c) (A+i B-C) \tan ^{-1}\left (\frac{\sqrt{-c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{a+i b} \sqrt{-c-i d}}+\frac{(c+i d) (i A+B-i C) \tan ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{-a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{-a+i b} \sqrt{c-i d}}\right )}{f \left (c^2+d^2\right ) (a d-b c)} \]
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}){\frac{1}{\sqrt{a+b\tan \left ( fx+e \right ) }}} \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] 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}{\sqrt{b \tan \left (f x + e\right ) + a}{\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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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 )}}{\sqrt{a + b \tan{\left (e + f x \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(-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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