Optimal. Leaf size=597 \[ -\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{5 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{5/2}}-\frac {2 \sqrt {c+d \tan (e+f x)} \left (a^4 C d+4 a^3 b B d-a^2 b^2 (9 A d+5 B c-11 C d)+2 a b^3 (5 A c-3 B d-5 c C)+b^4 (A d+5 B c)\right )}{15 b f \left (a^2+b^2\right )^2 (b c-a d) (a+b \tan (e+f x))^{3/2}}+\frac {2 \sqrt {c+d \tan (e+f x)} \left (2 a^6 C d^2+8 a^5 b B d^2-a^4 b^2 d (33 A d+25 B c-39 C d)+a^3 b^3 \left (80 c d (A-C)+B \left (15 c^2-49 d^2\right )\right )-a^2 b^4 \left (45 A c^2-29 A d^2-90 B c d-45 c^2 C+23 C d^2\right )-a b^5 \left (40 c d (A-C)+B \left (45 c^2-3 d^2\right )\right )-b^6 \left (5 c (B d+3 c C)-A \left (15 c^2+2 d^2\right )\right )\right )}{15 b f \left (a^2+b^2\right )^3 (b c-a d)^2 \sqrt {a+b \tan (e+f x)}}-\frac {\sqrt {c-i d} (i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a-i b)^{7/2}}-\frac {\sqrt {c+i d} (B-i (A-C)) \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a+i b)^{7/2}} \]
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Rubi [A] time = 3.59, antiderivative size = 597, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {3645, 3649, 3616, 3615, 93, 208} \[ \frac {2 \sqrt {c+d \tan (e+f x)} \left (a^3 b^3 \left (80 c d (A-C)+B \left (15 c^2-49 d^2\right )\right )-a^2 b^4 \left (45 A c^2-29 A d^2-90 B c d-45 c^2 C+23 C d^2\right )-a^4 b^2 d (33 A d+25 B c-39 C d)+8 a^5 b B d^2+2 a^6 C d^2-a b^5 \left (40 c d (A-C)+B \left (45 c^2-3 d^2\right )\right )-b^6 \left (5 c (B d+3 c C)-A \left (15 c^2+2 d^2\right )\right )\right )}{15 b f \left (a^2+b^2\right )^3 (b c-a d)^2 \sqrt {a+b \tan (e+f x)}}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{5 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{5/2}}-\frac {2 \sqrt {c+d \tan (e+f x)} \left (-a^2 b^2 (9 A d+5 B c-11 C d)+4 a^3 b B d+a^4 C d+2 a b^3 (5 A c-3 B d-5 c C)+b^4 (A d+5 B c)\right )}{15 b f \left (a^2+b^2\right )^2 (b c-a d) (a+b \tan (e+f x))^{3/2}}-\frac {\sqrt {c-i d} (i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a-i b)^{7/2}}-\frac {\sqrt {c+i d} (B-i (A-C)) \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a+i b)^{7/2}} \]
Antiderivative was successfully verified.
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Rule 93
Rule 208
Rule 3615
Rule 3616
Rule 3645
Rule 3649
Rubi steps
\begin {align*} \int \frac {\sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(a+b \tan (e+f x))^{7/2}} \, dx &=-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{5 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}+\frac {2 \int \frac {\frac {1}{2} ((b B-a C) (5 b c-a d)+A b (5 a c+b d))-\frac {5}{2} b ((A-C) (b c-a d)-B (a c+b d)) \tan (e+f x)-\frac {1}{2} \left (4 A b^2-4 a b B-a^2 C-5 b^2 C\right ) d \tan ^2(e+f x)}{(a+b \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}} \, dx}{5 b \left (a^2+b^2\right )}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{5 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac {2 \left (4 a^3 b B d+a^4 C d+b^4 (5 B c+A d)+2 a b^3 (5 A c-5 c C-3 B d)-a^2 b^2 (5 B c+9 A d-11 C d)\right ) \sqrt {c+d \tan (e+f x)}}{15 b \left (a^2+b^2\right )^2 (b c-a d) f (a+b \tan (e+f x))^{3/2}}-\frac {4 \int \frac {\frac {1}{4} \left (2 \left (b^2 d-\frac {3}{2} a (b c-a d)\right ) ((b B-a C) (5 b c-a d)+A b (5 a c+b d))+(3 b c-a d) \left (4 a^2 b B d+a^3 C d+A b^2 (5 b c-9 a d)-5 b^3 (c C+B d)-5 a b^2 (B c-2 C d)\right )\right )+\frac {15}{4} b (b c-a d) \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)+\frac {1}{2} d \left (4 a^3 b B d+a^4 C d+b^4 (5 B c+A d)+2 a b^3 (5 A c-5 c C-3 B d)-a^2 b^2 (5 B c+9 A d-11 C d)\right ) \tan ^2(e+f x)}{(a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx}{15 b \left (a^2+b^2\right )^2 (b c-a d)}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{5 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac {2 \left (4 a^3 b B d+a^4 C d+b^4 (5 B c+A d)+2 a b^3 (5 A c-5 c C-3 B d)-a^2 b^2 (5 B c+9 A d-11 C d)\right ) \sqrt {c+d \tan (e+f x)}}{15 b \left (a^2+b^2\right )^2 (b c-a d) f (a+b \tan (e+f x))^{3/2}}+\frac {2 \left (8 a^5 b B d^2+2 a^6 C d^2-a^4 b^2 d (25 B c+33 A d-39 C d)-a^2 b^4 \left (45 A c^2-45 c^2 C-90 B c d-29 A d^2+23 C d^2\right )+a^3 b^3 \left (80 c (A-C) d+B \left (15 c^2-49 d^2\right )\right )-a b^5 \left (40 c (A-C) d+B \left (45 c^2-3 d^2\right )\right )-b^6 \left (5 c (3 c C+B d)-A \left (15 c^2+2 d^2\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 b \left (a^2+b^2\right )^3 (b c-a d)^2 f \sqrt {a+b \tan (e+f x)}}+\frac {8 \int \frac {\frac {15}{8} b (b c-a d)^2 \left (a^3 (A c-c C-B d)-3 a b^2 (A c-c C-B d)+3 a^2 b (B c+(A-C) d)-b^3 (B c+(A-C) d)\right )-\frac {15}{8} b (b c-a d)^2 \left (3 a^2 b (A c-c C-B d)-b^3 (A c-c C-B d)-a^3 (B c+(A-C) d)+3 a b^2 (B c+(A-C) d)\right ) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{15 b \left (a^2+b^2\right )^3 (b c-a d)^2}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{5 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac {2 \left (4 a^3 b B d+a^4 C d+b^4 (5 B c+A d)+2 a b^3 (5 A c-5 c C-3 B d)-a^2 b^2 (5 B c+9 A d-11 C d)\right ) \sqrt {c+d \tan (e+f x)}}{15 b \left (a^2+b^2\right )^2 (b c-a d) f (a+b \tan (e+f x))^{3/2}}+\frac {2 \left (8 a^5 b B d^2+2 a^6 C d^2-a^4 b^2 d (25 B c+33 A d-39 C d)-a^2 b^4 \left (45 A c^2-45 c^2 C-90 B c d-29 A d^2+23 C d^2\right )+a^3 b^3 \left (80 c (A-C) d+B \left (15 c^2-49 d^2\right )\right )-a b^5 \left (40 c (A-C) d+B \left (45 c^2-3 d^2\right )\right )-b^6 \left (5 c (3 c C+B d)-A \left (15 c^2+2 d^2\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 b \left (a^2+b^2\right )^3 (b c-a d)^2 f \sqrt {a+b \tan (e+f x)}}+\frac {((A-i B-C) (c-i d)) \int \frac {1+i \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{2 (a-i b)^3}+\frac {((A+i B-C) (c+i d)) \int \frac {1-i \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{2 (a+i b)^3}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{5 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac {2 \left (4 a^3 b B d+a^4 C d+b^4 (5 B c+A d)+2 a b^3 (5 A c-5 c C-3 B d)-a^2 b^2 (5 B c+9 A d-11 C d)\right ) \sqrt {c+d \tan (e+f x)}}{15 b \left (a^2+b^2\right )^2 (b c-a d) f (a+b \tan (e+f x))^{3/2}}+\frac {2 \left (8 a^5 b B d^2+2 a^6 C d^2-a^4 b^2 d (25 B c+33 A d-39 C d)-a^2 b^4 \left (45 A c^2-45 c^2 C-90 B c d-29 A d^2+23 C d^2\right )+a^3 b^3 \left (80 c (A-C) d+B \left (15 c^2-49 d^2\right )\right )-a b^5 \left (40 c (A-C) d+B \left (45 c^2-3 d^2\right )\right )-b^6 \left (5 c (3 c C+B d)-A \left (15 c^2+2 d^2\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 b \left (a^2+b^2\right )^3 (b c-a d)^2 f \sqrt {a+b \tan (e+f x)}}+\frac {((A-i B-C) (c-i d)) \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 (a-i b)^3 f}+\frac {((A+i B-C) (c+i d)) \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 (a+i b)^3 f}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{5 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac {2 \left (4 a^3 b B d+a^4 C d+b^4 (5 B c+A d)+2 a b^3 (5 A c-5 c C-3 B d)-a^2 b^2 (5 B c+9 A d-11 C d)\right ) \sqrt {c+d \tan (e+f x)}}{15 b \left (a^2+b^2\right )^2 (b c-a d) f (a+b \tan (e+f x))^{3/2}}+\frac {2 \left (8 a^5 b B d^2+2 a^6 C d^2-a^4 b^2 d (25 B c+33 A d-39 C d)-a^2 b^4 \left (45 A c^2-45 c^2 C-90 B c d-29 A d^2+23 C d^2\right )+a^3 b^3 \left (80 c (A-C) d+B \left (15 c^2-49 d^2\right )\right )-a b^5 \left (40 c (A-C) d+B \left (45 c^2-3 d^2\right )\right )-b^6 \left (5 c (3 c C+B d)-A \left (15 c^2+2 d^2\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 b \left (a^2+b^2\right )^3 (b c-a d)^2 f \sqrt {a+b \tan (e+f x)}}+\frac {((A-i B-C) (c-i d)) \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 )}{(a-i b)^3 f}+\frac {((A+i B-C) (c+i d)) \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 )}{(a+i b)^3 f}\\ &=-\frac {(i A+B-i C) \sqrt {c-i d} \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 )}{(a-i b)^{7/2} f}-\frac {(B-i (A-C)) \sqrt {c+i d} \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 )}{(a+i b)^{7/2} f}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{5 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac {2 \left (4 a^3 b B d+a^4 C d+b^4 (5 B c+A d)+2 a b^3 (5 A c-5 c C-3 B d)-a^2 b^2 (5 B c+9 A d-11 C d)\right ) \sqrt {c+d \tan (e+f x)}}{15 b \left (a^2+b^2\right )^2 (b c-a d) f (a+b \tan (e+f x))^{3/2}}+\frac {2 \left (8 a^5 b B d^2+2 a^6 C d^2-a^4 b^2 d (25 B c+33 A d-39 C d)-a^2 b^4 \left (45 A c^2-45 c^2 C-90 B c d-29 A d^2+23 C d^2\right )+a^3 b^3 \left (80 c (A-C) d+B \left (15 c^2-49 d^2\right )\right )-a b^5 \left (40 c (A-C) d+B \left (45 c^2-3 d^2\right )\right )-b^6 \left (5 c (3 c C+B d)-A \left (15 c^2+2 d^2\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 b \left (a^2+b^2\right )^3 (b c-a d)^2 f \sqrt {a+b \tan (e+f x)}}\\ \end {align*}
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Mathematica [A] time = 7.35, size = 1109, normalized size = 1.86 \[ -\frac {\sqrt {c+d \tan (e+f x)} C}{2 b f (a+b \tan (e+f x))^{5/2}}-\frac {-\frac {2 \sqrt {c+d \tan (e+f x)} \left (\frac {1}{2} b^2 (-4 A b c+5 b C c-a C d)-a \left (-2 (B c+(A-C) d) b^2-\frac {1}{2} a (b c C-a d C-4 b B d)\right )\right )}{5 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{5/2}}-\frac {2 \left (-\frac {2 \sqrt {c+d \tan (e+f x)} \left (b^2 (b c-a d) \left (C d a^2+b (5 A c-5 C c-B d) a+b^2 (5 B c+A d)\right )-a \left (a \left (-C a^2-4 b B a+4 A b^2-5 b^2 C\right ) d (b c-a d)-5 b^2 (b c-a d) (A b c-a B c-b C c-a A d-b B d+a C d)\right )\right )}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2}}-\frac {2 \left (-\frac {15 b \left (\frac {(i a-b)^3 (A-i B-C) \sqrt {i d-c} \tanh ^{-1}\left (\frac {\sqrt {i d-c} \sqrt {a+b \tan (e+f x)}}{\sqrt {i b-a} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {i b-a}}-\frac {(i a+b)^3 (A+i B-C) \sqrt {c+i d} \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}}\right ) (b c-a d)^2}{2 \left (a^2+b^2\right ) f}-\frac {2 \left (b^2 \left ((b c-a d) \left (b^2 d-\frac {3}{2} a (b c-a d)\right ) \left (C d a^2+b (5 A c-5 C c-B d) a+b^2 (5 B c+A d)\right )+\left (\frac {a d}{2}-\frac {3 b c}{2}\right ) \left (a \left (-C a^2-4 b B a+4 A b^2-5 b^2 C\right ) d (b c-a d)-5 b^2 (b c-a d) (A b c-a B c-b C c-a A d-b B d+a C d)\right )\right )-a \left (\frac {3}{2} b (b c-a d) \left (b \left (-C a^2-4 b B a+4 A b^2-5 b^2 C\right ) d (b c-a d)+5 a b (A b c-a B c-b C c-a A d-b B d+a C d) (b c-a d)+b \left (C d a^2+b (5 A c-5 C c-B d) a+b^2 (5 B c+A d)\right ) (b c-a d)\right )-a d \left (b^2 (b c-a d) \left (C d a^2+b (5 A c-5 C c-B d) a+b^2 (5 B c+A d)\right )-a \left (a \left (-C a^2-4 b B a+4 A b^2-5 b^2 C\right ) d (b c-a d)-5 b^2 (b c-a d) (A b c-a B c-b C c-a A d-b B d+a C d)\right )\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{\left (a^2+b^2\right ) f \sqrt {a+b \tan (e+f x)} (b c-a d)}\right )}{3 \left (a^2+b^2\right ) (b c-a d)}\right )}{5 \left (a^2+b^2\right ) (b c-a d)}}{2 b} \]
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 [F(-1)] time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c +d \tan \left (f x +e \right )}\, \left (A +B \tan \left (f x +e \right )+C \left (\tan ^{2}\left (f x +e \right )\right )\right )}{\left (a +b \tan \left (f x +e \right )\right )^{\frac {7}{2}}}\, dx \]
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 \[ \int \frac {\sqrt {c + d \tan {\left (e + f x \right )}} \left (A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )}{\left (a + b \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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