Optimal. Leaf size=503 \[ -\frac {2 \sqrt {c+d \tan (e+f x)} \left (-\left (a^2 \left (2 c d (A-C)+B \left (c^2-d^2\right )\right )\right )+2 a b \left (-A \left (c^2-d^2\right )+2 B c d+c^2 C-C d^2\right )+b^2 \left (2 c d (A-C)+B \left (c^2-d^2\right )\right )\right )}{f}+\frac {2 (c+d \tan (e+f x))^{7/2} \left (36 a^2 C d^2-22 a b d (2 c C-9 B d)+b^2 \left (99 d^2 (A-C)-22 B c d+8 c^2 C\right )\right )}{693 d^3 f}+\frac {2 \left (a^2 B+2 a b (A-C)-b^2 B\right ) (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 (c+d \tan (e+f x))^{3/2} \left (a^2 (d (A-C)+B c)+2 a b (A c-B d-c C)-b^2 (d (A-C)+B c)\right )}{3 f}-\frac {(a-i b)^2 (c-i d)^{5/2} (i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {(a+i b)^2 (c+i d)^{5/2} (i A-B-i C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}-\frac {2 b \tan (e+f x) (-4 a C d-11 b B d+4 b c C) (c+d \tan (e+f x))^{7/2}}{99 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{7/2}}{11 d f} \]
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Rubi [A] time = 2.31, antiderivative size = 503, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.170, Rules used = {3647, 3637, 3630, 3528, 3539, 3537, 63, 208} \[ \frac {2 (c+d \tan (e+f x))^{7/2} \left (36 a^2 C d^2-22 a b d (2 c C-9 B d)+b^2 \left (99 d^2 (A-C)-22 B c d+8 c^2 C\right )\right )}{693 d^3 f}-\frac {2 \sqrt {c+d \tan (e+f x)} \left (a^2 \left (-\left (2 c d (A-C)+B \left (c^2-d^2\right )\right )\right )+2 a b \left (-A \left (c^2-d^2\right )+2 B c d+c^2 C-C d^2\right )+b^2 \left (2 c d (A-C)+B \left (c^2-d^2\right )\right )\right )}{f}+\frac {2 \left (a^2 B+2 a b (A-C)-b^2 B\right ) (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 (c+d \tan (e+f x))^{3/2} \left (a^2 (d (A-C)+B c)+2 a b (A c-B d-c C)-b^2 (d (A-C)+B c)\right )}{3 f}-\frac {(a-i b)^2 (c-i d)^{5/2} (i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {(a+i b)^2 (c+i d)^{5/2} (i A-B-i C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}-\frac {2 b \tan (e+f x) (-4 a C d-11 b B d+4 b c C) (c+d \tan (e+f x))^{7/2}}{99 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{7/2}}{11 d f} \]
Antiderivative was successfully verified.
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Rule 63
Rule 208
Rule 3528
Rule 3537
Rule 3539
Rule 3630
Rule 3637
Rule 3647
Rubi steps
\begin {align*} \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2} \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))^{7/2}}{11 d f}+\frac {2 \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \left (\frac {1}{2} (-4 b c C+a (11 A-7 C) d)+\frac {11}{2} (A b+a B-b C) d \tan (e+f x)-\frac {1}{2} (4 b c C-11 b B d-4 a C d) \tan ^2(e+f x)\right ) \, dx}{11 d}\\ &=-\frac {2 b (4 b c C-11 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{7/2}}{99 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{7/2}}{11 d f}-\frac {4 \int (c+d \tan (e+f x))^{5/2} \left (\frac {1}{4} \left (44 a b c C d-9 a^2 (11 A-7 C) d^2-4 b^2 \left (2 c^2 C-\frac {11 B c d}{2}\right )\right )-\frac {99}{4} \left (a^2 B-b^2 B+2 a b (A-C)\right ) d^2 \tan (e+f x)-\frac {1}{4} \left (36 a^2 C d^2-22 a b d (2 c C-9 B d)+b^2 \left (8 c^2 C-22 B c d+99 (A-C) d^2\right )\right ) \tan ^2(e+f x)\right ) \, dx}{99 d^2}\\ &=\frac {2 \left (36 a^2 C d^2-22 a b d (2 c C-9 B d)+b^2 \left (8 c^2 C-22 B c d+99 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{7/2}}{693 d^3 f}-\frac {2 b (4 b c C-11 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{7/2}}{99 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{7/2}}{11 d f}-\frac {4 \int (c+d \tan (e+f x))^{5/2} \left (\frac {99}{4} \left (2 a b B-a^2 (A-C)+b^2 (A-C)\right ) d^2-\frac {99}{4} \left (a^2 B-b^2 B+2 a b (A-C)\right ) d^2 \tan (e+f x)\right ) \, dx}{99 d^2}\\ &=\frac {2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 \left (36 a^2 C d^2-22 a b d (2 c C-9 B d)+b^2 \left (8 c^2 C-22 B c d+99 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{7/2}}{693 d^3 f}-\frac {2 b (4 b c C-11 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{7/2}}{99 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{7/2}}{11 d f}-\frac {4 \int (c+d \tan (e+f x))^{3/2} \left (-\frac {99}{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 {99}{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}{99 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 ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 \left (36 a^2 C d^2-22 a b d (2 c C-9 B d)+b^2 \left (8 c^2 C-22 B c d+99 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{7/2}}{693 d^3 f}-\frac {2 b (4 b c C-11 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{7/2}}{99 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{7/2}}{11 d f}-\frac {4 \int \sqrt {c+d \tan (e+f x)} \left (\frac {99}{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 {99}{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)\right ) \, dx}{99 d^2}\\ &=-\frac {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 ) \sqrt {c+d \tan (e+f x)}}{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 ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 \left (36 a^2 C d^2-22 a b d (2 c C-9 B d)+b^2 \left (8 c^2 C-22 B c d+99 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{7/2}}{693 d^3 f}-\frac {2 b (4 b c C-11 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{7/2}}{99 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{7/2}}{11 d f}-\frac {4 \int \frac {-\frac {99}{4} d^2 \left (a^2 \left (A c^3-c^3 C-3 B c^2 d-3 A c d^2+3 c C d^2+B d^3\right )+b^2 \left (c^3 C+3 B c^2 d-3 c C d^2-B d^3-A \left (c^3-3 c d^2\right )\right )-2 a b \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right )\right )+\frac {99}{4} d^2 \left (2 a b \left (c^3 C+3 B c^2 d-3 c C d^2-B d^3-A \left (c^3-3 c d^2\right )\right )-a^2 \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right )+b^2 \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{99 d^2}\\ &=-\frac {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 ) \sqrt {c+d \tan (e+f x)}}{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 ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 \left (36 a^2 C d^2-22 a b d (2 c C-9 B d)+b^2 \left (8 c^2 C-22 B c d+99 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{7/2}}{693 d^3 f}-\frac {2 b (4 b c C-11 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{7/2}}{99 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{7/2}}{11 d f}+\frac {1}{2} \left ((a-i b)^2 (A-i B-C) (c-i d)^3\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)^3\right ) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx\\ &=-\frac {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 ) \sqrt {c+d \tan (e+f x)}}{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 ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 \left (36 a^2 C d^2-22 a b d (2 c C-9 B d)+b^2 \left (8 c^2 C-22 B c d+99 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{7/2}}{693 d^3 f}-\frac {2 b (4 b c C-11 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{7/2}}{99 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{7/2}}{11 d f}+\frac {\left ((a-i b)^2 (i A+B-i C) (c-i d)^3\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)^3\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 \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 ) \sqrt {c+d \tan (e+f x)}}{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 ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 \left (36 a^2 C d^2-22 a b d (2 c C-9 B d)+b^2 \left (8 c^2 C-22 B c d+99 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{7/2}}{693 d^3 f}-\frac {2 b (4 b c C-11 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{7/2}}{99 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{7/2}}{11 d f}-\frac {\left ((a-i b)^2 (A-i B-C) (c-i d)^3\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)^3\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)^{5/2} \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)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}-\frac {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 ) \sqrt {c+d \tan (e+f x)}}{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 ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 \left (36 a^2 C d^2-22 a b d (2 c C-9 B d)+b^2 \left (8 c^2 C-22 B c d+99 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{7/2}}{693 d^3 f}-\frac {2 b (4 b c C-11 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{7/2}}{99 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{7/2}}{11 d f}\\ \end {align*}
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Mathematica [A] time = 6.45, size = 564, normalized size = 1.12 \[ \frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{7/2}}{11 d f}+\frac {2 \left (\frac {b \tan (e+f x) (4 a C d+11 b B d-4 b c C) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {2 \left (\frac {(c+d \tan (e+f x))^{7/2} \left (-36 a^2 C d^2+22 a b d (2 c C-9 B d)-\left (b^2 \left (99 d^2 (A-C)-22 B c d+8 c^2 C\right )\right )\right )}{14 d f}+\frac {i \left (\frac {99}{4} d^2 \left (-\left (a^2 (A-C)\right )+2 a b B+b^2 (A-C)\right )+\frac {99}{4} i d^2 \left (a^2 B+2 a b (A-C)-b^2 B\right )\right ) \left (\frac {2}{5} (c+d \tan (e+f x))^{5/2}+(c-i d) \left (\frac {2}{3} (c+d \tan (e+f x))^{3/2}+(c-i d) \left (2 \sqrt {c+d \tan (e+f x)}+\frac {2 (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{-c+i d}\right )\right )\right )}{2 f}-\frac {i \left (\frac {99}{4} d^2 \left (-\left (a^2 (A-C)\right )+2 a b B+b^2 (A-C)\right )-\frac {99}{4} i d^2 \left (a^2 B+2 a b (A-C)-b^2 B\right )\right ) \left (\frac {2}{5} (c+d \tan (e+f x))^{5/2}+(c+i d) \left (\frac {2}{3} (c+d \tan (e+f x))^{3/2}+(c+i d) \left (2 \sqrt {c+d \tan (e+f x)}+\frac {2 (c+i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{-c-i d}\right )\right )\right )}{2 f}\right )}{9 d}\right )}{11 d} \]
Antiderivative was successfully verified.
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fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.60, size = 11478, normalized size = 22.82 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F(-1)] time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan {\left (e + f x \right )}\right )^{2} \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \left (A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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