Optimal. Leaf size=384 \[ \frac{\left (3 a^2 C d^2-2 a b d (2 B d+3 c C)+b^2 \left (8 d^2 (A-C)+12 B c d+3 c^2 C\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{4 b^{5/2} \sqrt{d} f}-\frac{(c-i d)^{3/2} (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{a-i b}}+\frac{(c+i d)^{3/2} (i A-B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{a+i b}}+\frac{(-3 a C d+4 b B d+3 b c C) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 b^2 f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 b f} \]
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Rubi [A] time = 4.31065, antiderivative size = 384, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 49, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.163, Rules used = {3647, 3655, 6725, 63, 217, 206, 93, 208} \[ \frac{\left (3 a^2 C d^2-2 a b d (2 B d+3 c C)+b^2 \left (8 d^2 (A-C)+12 B c d+3 c^2 C\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{4 b^{5/2} \sqrt{d} f}-\frac{(c-i d)^{3/2} (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{a-i b}}+\frac{(c+i d)^{3/2} (i A-B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f \sqrt{a+i b}}+\frac{(-3 a C d+4 b B d+3 b c C) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 b^2 f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 b f} \]
Antiderivative was successfully verified.
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Rule 3647
Rule 3655
Rule 6725
Rule 63
Rule 217
Rule 206
Rule 93
Rule 208
Rubi steps
\begin{align*} \int \frac{(c+d \tan (e+f x))^{3/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt{a+b \tan (e+f x)}} \, dx &=\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 b f}+\frac{\int \frac{\sqrt{c+d \tan (e+f x)} \left (\frac{1}{2} (4 A b c-C (b c+3 a d))+2 b (B c+(A-C) d) \tan (e+f x)+\frac{1}{2} (3 b c C+4 b B d-3 a C d) \tan ^2(e+f x)\right )}{\sqrt{a+b \tan (e+f x)}} \, dx}{2 b}\\ &=\frac{(3 b c C+4 b B d-3 a C d) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 b^2 f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 b f}+\frac{\int \frac{\frac{1}{4} \left (8 A b^2 c^2+3 a^2 C d^2-2 a b d (3 c C+2 B d)-b^2 c (5 c C+4 B d)\right )+2 b^2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) \tan (e+f x)+\frac{1}{4} \left (8 b^2 d (B c+(A-C) d)+(b c-a d) (3 b c C+4 b B d-3 a C d)\right ) \tan ^2(e+f x)}{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{2 b^2}\\ &=\frac{(3 b c C+4 b B d-3 a C d) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 b^2 f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 b f}+\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{4} \left (8 A b^2 c^2+3 a^2 C d^2-2 a b d (3 c C+2 B d)-b^2 c (5 c C+4 B d)\right )+2 b^2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) x+\frac{1}{4} \left (8 b^2 d (B c+(A-C) d)+(b c-a d) (3 b c C+4 b B d-3 a C d)\right ) x^2}{\sqrt{a+b x} \sqrt{c+d x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 b^2 f}\\ &=\frac{(3 b c C+4 b B d-3 a C d) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 b^2 f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 b f}+\frac{\operatorname{Subst}\left (\int \left (\frac{3 a^2 C d^2-2 a b d (3 c C+2 B d)+b^2 \left (3 c^2 C+12 B c d+8 (A-C) d^2\right )}{4 \sqrt{a+b x} \sqrt{c+d x}}+\frac{2 \left (-b^2 \left (c^2 C+2 B c d-C d^2-A \left (c^2-d^2\right )\right )+b^2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) x\right )}{\sqrt{a+b x} \sqrt{c+d x} \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{2 b^2 f}\\ &=\frac{(3 b c C+4 b B d-3 a C d) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 b^2 f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 b f}+\frac{\operatorname{Subst}\left (\int \frac{-b^2 \left (c^2 C+2 B c d-C d^2-A \left (c^2-d^2\right )\right )+b^2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) x}{\sqrt{a+b x} \sqrt{c+d x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{b^2 f}+\frac{\left (3 a^2 C d^2-2 a b d (3 c C+2 B d)+b^2 \left (3 c^2 C+12 B c d+8 (A-C) d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{8 b^2 f}\\ &=\frac{(3 b c C+4 b B d-3 a C d) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 b^2 f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 b f}+\frac{\operatorname{Subst}\left (\int \left (\frac{-i b^2 \left (c^2 C+2 B c d-C d^2-A \left (c^2-d^2\right )\right )-b^2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )}{2 (i-x) \sqrt{a+b x} \sqrt{c+d x}}+\frac{-i b^2 \left (c^2 C+2 B c d-C d^2-A \left (c^2-d^2\right )\right )+b^2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )}{2 (i+x) \sqrt{a+b x} \sqrt{c+d x}}\right ) \, dx,x,\tan (e+f x)\right )}{b^2 f}+\frac{\left (3 a^2 C d^2-2 a b d (3 c C+2 B d)+b^2 \left (3 c^2 C+12 B c d+8 (A-C) d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{a d}{b}+\frac{d x^2}{b}}} \, dx,x,\sqrt{a+b \tan (e+f x)}\right )}{4 b^3 f}\\ &=\frac{(3 b c C+4 b B d-3 a C d) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 b^2 f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 b f}+\frac{\left ((i A+B-i C) (c-i d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{(i+x) \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac{\left (3 a^2 C d^2-2 a b d (3 c C+2 B d)+b^2 \left (3 c^2 C+12 B c d+8 (A-C) d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{d x^2}{b}} \, dx,x,\frac{\sqrt{a+b \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}\right )}{4 b^3 f}-\frac{\left (i b^2 \left (c^2 C+2 B c d-C d^2-A \left (c^2-d^2\right )\right )+b^2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{(i-x) \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 b^2 f}\\ &=\frac{\left (3 a^2 C d^2-2 a b d (3 c C+2 B d)+b^2 \left (3 c^2 C+12 B c d+8 (A-C) d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{4 b^{5/2} \sqrt{d} f}+\frac{(3 b c C+4 b B d-3 a C d) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 b^2 f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 b f}+\frac{\left ((i A+B-i C) (c-i d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{-a+i b-(-c+i d) x^2} \, dx,x,\frac{\sqrt{a+b \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}\right )}{f}-\frac{\left (i b^2 \left (c^2 C+2 B c d-C d^2-A \left (c^2-d^2\right )\right )+b^2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+i b-(c+i d) x^2} \, dx,x,\frac{\sqrt{a+b \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}\right )}{b^2 f}\\ &=-\frac{(i A+B-i C) (c-i d)^{3/2} \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} f}+\frac{(i A-B-i C) (c+i d)^{3/2} \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} f}+\frac{\left (3 a^2 C d^2-2 a b d (3 c C+2 B d)+b^2 \left (3 c^2 C+12 B c d+8 (A-C) d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{4 b^{5/2} \sqrt{d} f}+\frac{(3 b c C+4 b B d-3 a C d) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 b^2 f}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 b f}\\ \end{align*}
Mathematica [A] time = 7.57049, size = 613, normalized size = 1.6 \[ \frac{\frac{\frac{\sqrt{b} \sqrt{c-\frac{a d}{b}} \left (3 a^2 C d^2-2 a b d (2 B d+3 c C)+b^2 \left (8 d^2 (A-C)+12 B c d+3 c^2 C\right )\right ) \sqrt{\frac{b c+b d \tan (e+f x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c-\frac{a d}{b}}}\right )}{2 \sqrt{d} \sqrt{c+d \tan (e+f x)}}-\frac{2 b^2 \left (\sqrt{-b^2} \left (-A \left (c^2-d^2\right )+2 B c d+c^2 C-C d^2\right )-b \left (2 c d (A-C)+B \left (c^2-d^2\right )\right )\right ) \tan ^{-1}\left (\frac{\sqrt{\frac{b d}{\sqrt{-b^2}}+c} \sqrt{a+b \tan (e+f x)}}{\sqrt{\sqrt{-b^2}-a} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{\sqrt{-b^2}-a} \sqrt{\frac{b d}{\sqrt{-b^2}}+c}}-\frac{2 b^2 \left (\sqrt{-b^2} \left (-A \left (c^2-d^2\right )+2 B c d+c^2 C-C d^2\right )+b \left (2 c d (A-C)+B \left (c^2-d^2\right )\right )\right ) \tan ^{-1}\left (\frac{\sqrt{-\frac{\sqrt{-b^2} d+b c}{b}} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+\sqrt{-b^2}} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{a+\sqrt{-b^2}} \sqrt{-\frac{\sqrt{-b^2} d+b c}{b}}}}{b^2 f}+\frac{(-3 a C d+4 b B d+3 b c C) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{2 b f}}{2 b}+\frac{C \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 b f} \]
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}) \left ( c+d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}{\frac{1}{\sqrt{a+b\tan \left ( fx+e \right ) }}}}\, dx \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(-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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