Optimal. Leaf size=396 \[ \frac{2 (c+d \tan (e+f x))^{5/2} \left (28 a^2 C d^2-18 a b d (2 c C-7 B d)+b^2 \left (63 d^2 (A-C)-18 B c d+8 c^2 C\right )\right )}{315 d^3 f}+\frac{2 \left (a^2 B+2 a b (A-C)-b^2 B\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{2 \sqrt{c+d \tan (e+f x)} \left (a^2 (d (A-C)+B c)+2 a b (A c-B d-c C)-b^2 (d (A-C)+B c)\right )}{f}-\frac{(a-i b)^2 (c-i d)^{3/2} (B+i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f}+\frac{(a+i b)^2 (c+i d)^{3/2} (i A-B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f}-\frac{2 b \tan (e+f x) (-4 a C d-9 b B d+4 b c C) (c+d \tan (e+f x))^{5/2}}{63 d^2 f}+\frac{2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}}{9 d f} \]
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Rubi [A] time = 1.72651, antiderivative size = 396, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 47, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.17, Rules used = {3647, 3637, 3630, 3528, 3539, 3537, 63, 208} \[ \frac{2 (c+d \tan (e+f x))^{5/2} \left (28 a^2 C d^2-18 a b d (2 c C-7 B d)+b^2 \left (63 d^2 (A-C)-18 B c d+8 c^2 C\right )\right )}{315 d^3 f}+\frac{2 \left (a^2 B+2 a b (A-C)-b^2 B\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{2 \sqrt{c+d \tan (e+f x)} \left (a^2 (d (A-C)+B c)+2 a b (A c-B d-c C)-b^2 (d (A-C)+B c)\right )}{f}-\frac{(a-i b)^2 (c-i d)^{3/2} (B+i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f}+\frac{(a+i b)^2 (c+i d)^{3/2} (i A-B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f}-\frac{2 b \tan (e+f x) (-4 a C d-9 b B d+4 b c C) (c+d \tan (e+f x))^{5/2}}{63 d^2 f}+\frac{2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}}{9 d f} \]
Antiderivative was successfully verified.
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Rule 3647
Rule 3637
Rule 3630
Rule 3528
Rule 3539
Rule 3537
Rule 63
Rule 208
Rubi steps
\begin{align*} \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx &=\frac{2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}}{9 d f}+\frac{2 \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \left (\frac{1}{2} (-4 b c C+a (9 A-5 C) d)+\frac{9}{2} (A b+a B-b C) d \tan (e+f x)-\frac{1}{2} (4 b c C-9 b B d-4 a C d) \tan ^2(e+f x)\right ) \, dx}{9 d}\\ &=-\frac{2 b (4 b c C-9 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{63 d^2 f}+\frac{2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}}{9 d f}-\frac{4 \int (c+d \tan (e+f x))^{3/2} \left (\frac{1}{4} \left (36 a b c C d-7 a^2 (9 A-5 C) d^2-4 b^2 \left (2 c^2 C-\frac{9 B c d}{2}\right )\right )-\frac{63}{4} \left (a^2 B-b^2 B+2 a b (A-C)\right ) d^2 \tan (e+f x)-\frac{1}{4} \left (28 a^2 C d^2-18 a b d (2 c C-7 B d)+b^2 \left (8 c^2 C-18 B c d+63 (A-C) d^2\right )\right ) \tan ^2(e+f x)\right ) \, dx}{63 d^2}\\ &=\frac{2 \left (28 a^2 C d^2-18 a b d (2 c C-7 B d)+b^2 \left (8 c^2 C-18 B c d+63 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{5/2}}{315 d^3 f}-\frac{2 b (4 b c C-9 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{63 d^2 f}+\frac{2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}}{9 d f}-\frac{4 \int (c+d \tan (e+f x))^{3/2} \left (\frac{63}{4} \left (2 a b B-a^2 (A-C)+b^2 (A-C)\right ) d^2-\frac{63}{4} \left (a^2 B-b^2 B+2 a b (A-C)\right ) d^2 \tan (e+f x)\right ) \, dx}{63 d^2}\\ &=\frac{2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{2 \left (28 a^2 C d^2-18 a b d (2 c C-7 B d)+b^2 \left (8 c^2 C-18 B c d+63 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{5/2}}{315 d^3 f}-\frac{2 b (4 b c C-9 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{63 d^2 f}+\frac{2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}}{9 d f}-\frac{4 \int \sqrt{c+d \tan (e+f x)} \left (-\frac{63}{4} d^2 \left (a^2 (A c-c C-B d)-b^2 (A c-c C-B d)-2 a b (B c+(A-C) d)\right )-\frac{63}{4} d^2 \left (2 a b (A c-c C-B d)+a^2 (B c+(A-C) d)-b^2 (B c+(A-C) d)\right ) \tan (e+f x)\right ) \, dx}{63 d^2}\\ &=\frac{2 \left (2 a b (A c-c C-B d)+a^2 (B c+(A-C) d)-b^2 (B c+(A-C) d)\right ) \sqrt{c+d \tan (e+f x)}}{f}+\frac{2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{2 \left (28 a^2 C d^2-18 a b d (2 c C-7 B d)+b^2 \left (8 c^2 C-18 B c d+63 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{5/2}}{315 d^3 f}-\frac{2 b (4 b c C-9 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{63 d^2 f}+\frac{2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}}{9 d f}-\frac{4 \int \frac{\frac{63}{4} 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{63}{4} 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}{63 d^2}\\ &=\frac{2 \left (2 a b (A c-c C-B d)+a^2 (B c+(A-C) d)-b^2 (B c+(A-C) d)\right ) \sqrt{c+d \tan (e+f x)}}{f}+\frac{2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{2 \left (28 a^2 C d^2-18 a b d (2 c C-7 B d)+b^2 \left (8 c^2 C-18 B c d+63 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{5/2}}{315 d^3 f}-\frac{2 b (4 b c C-9 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{63 d^2 f}+\frac{2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}}{9 d f}+\frac{1}{2} \left ((a-i b)^2 (A-i B-C) (c-i d)^2\right ) \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx+\frac{1}{2} \left ((a+i b)^2 (A+i B-C) (c+i d)^2\right ) \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx\\ &=\frac{2 \left (2 a b (A c-c C-B d)+a^2 (B c+(A-C) d)-b^2 (B c+(A-C) d)\right ) \sqrt{c+d \tan (e+f x)}}{f}+\frac{2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{2 \left (28 a^2 C d^2-18 a b d (2 c C-7 B d)+b^2 \left (8 c^2 C-18 B c d+63 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{5/2}}{315 d^3 f}-\frac{2 b (4 b c C-9 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{63 d^2 f}+\frac{2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}}{9 d f}+\frac{\left ((a-i b)^2 (i A+B-i C) (c-i d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 f}-\frac{\left (i (a+i b)^2 (A+i B-C) (c+i d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 f}\\ &=\frac{2 \left (2 a b (A c-c C-B d)+a^2 (B c+(A-C) d)-b^2 (B c+(A-C) d)\right ) \sqrt{c+d \tan (e+f x)}}{f}+\frac{2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{2 \left (28 a^2 C d^2-18 a b d (2 c C-7 B d)+b^2 \left (8 c^2 C-18 B c d+63 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{5/2}}{315 d^3 f}-\frac{2 b (4 b c C-9 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{63 d^2 f}+\frac{2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}}{9 d f}-\frac{\left ((a-i b)^2 (A-i B-C) (c-i d)^2\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 )}{d f}-\frac{\left ((a+i b)^2 (A+i B-C) (c+i d)^2\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 )}{d f}\\ &=-\frac{(a-i b)^2 (i A+B-i C) (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f}-\frac{(a+i b)^2 (B-i (A-C)) (c+i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f}+\frac{2 \left (2 a b (A c-c C-B d)+a^2 (B c+(A-C) d)-b^2 (B c+(A-C) d)\right ) \sqrt{c+d \tan (e+f x)}}{f}+\frac{2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{2 \left (28 a^2 C d^2-18 a b d (2 c C-7 B d)+b^2 \left (8 c^2 C-18 B c d+63 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{5/2}}{315 d^3 f}-\frac{2 b (4 b c C-9 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{63 d^2 f}+\frac{2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}}{9 d f}\\ \end{align*}
Mathematica [A] time = 6.1604, size = 350, normalized size = 0.88 \[ \frac{2 \left ((c+d \tan (e+f x))^{5/2} \left (28 a^2 C d^2+18 a b d (7 B d-2 c C)+b^2 \left (63 d^2 (A-C)-18 B c d+8 c^2 C\right )\right )+\frac{105}{2} d^3 (a-i b)^2 (i A+B-i C) \left (\sqrt{c+d \tan (e+f x)} (4 c+d \tan (e+f x)-3 i d)-3 (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )\right )+\frac{105}{2} d^3 (a+i b)^2 (-i A+B+i C) \left (\sqrt{c+d \tan (e+f x)} (4 c+d \tan (e+f x)+3 i d)-3 (c+i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )\right )+5 b d \tan (e+f x) (4 a C d+9 b B d-4 b c C) (c+d \tan (e+f x))^{5/2}+35 C d^2 (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}\right )}{315 d^3 f} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.18, size = 8031, normalized size = 20.3 \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 \left (a + b \tan{\left (e + f x \right )}\right )^{2} \left (c + d \tan{\left (e + f x \right )}\right )^{\frac{3}{2}} \left (A + B \tan{\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )\, 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{\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{3}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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