Optimal. Leaf size=473 \[ -\frac {\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b f \left (a^2+b^2\right ) (a+b \tan (e+f x))}+\frac {d \left (5 a^2 C-3 a b B+3 A b^2+2 b^2 C\right ) (c+d \tan (e+f x))^{3/2}}{3 b^2 f \left (a^2+b^2\right )}-\frac {d \sqrt {c+d \tan (e+f x)} \left (5 a^3 C d-a^2 b (3 B d+5 c C)-A b^2 (b c-a d)+a b^2 (B c+4 C d)-2 b^3 (B d+2 c C)\right )}{b^3 f \left (a^2+b^2\right )}+\frac {(b c-a d)^{3/2} \left (-5 a^4 C d+3 a^3 b B d+a^2 b^2 (2 B c-d (A+9 C))-a b^3 (4 A c-7 B d-4 c C)-b^4 (5 A d+2 B c)\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{b^{7/2} f \left (a^2+b^2\right )^2}-\frac {(c-i d)^{5/2} (i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f (a-i b)^2}-\frac {(c+i d)^{5/2} (B-i (A-C)) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f (a+i b)^2} \]
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Rubi [A] time = 3.90, antiderivative size = 473, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.170, Rules used = {3645, 3647, 3653, 3539, 3537, 63, 208, 3634} \[ -\frac {\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b f \left (a^2+b^2\right ) (a+b \tan (e+f x))}+\frac {d \left (5 a^2 C-3 a b B+3 A b^2+2 b^2 C\right ) (c+d \tan (e+f x))^{3/2}}{3 b^2 f \left (a^2+b^2\right )}-\frac {d \sqrt {c+d \tan (e+f x)} \left (-a^2 b (3 B d+5 c C)+5 a^3 C d-A b^2 (b c-a d)+a b^2 (B c+4 C d)-2 b^3 (B d+2 c C)\right )}{b^3 f \left (a^2+b^2\right )}+\frac {(b c-a d)^{3/2} \left (a^2 b^2 (2 B c-d (A+9 C))+3 a^3 b B d-5 a^4 C d-a b^3 (4 A c-7 B d-4 c C)-b^4 (5 A d+2 B c)\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{b^{7/2} f \left (a^2+b^2\right )^2}-\frac {(c-i d)^{5/2} (i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f (a-i b)^2}-\frac {(c+i d)^{5/2} (B-i (A-C)) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f (a+i b)^2} \]
Antiderivative was successfully verified.
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Rule 63
Rule 208
Rule 3537
Rule 3539
Rule 3634
Rule 3645
Rule 3647
Rule 3653
Rubi steps
\begin {align*} \int \frac {(c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(a+b \tan (e+f x))^2} \, dx &=-\frac {\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac {\int \frac {(c+d \tan (e+f x))^{3/2} \left (\frac {1}{2} \left (2 (b B-a C) \left (b c-\frac {5 a d}{2}\right )+2 A b \left (a c+\frac {5 b d}{2}\right )\right )-b ((A-C) (b c-a d)-B (a c+b d)) \tan (e+f x)+\frac {1}{2} \left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d \tan ^2(e+f x)\right )}{a+b \tan (e+f x)} \, dx}{b \left (a^2+b^2\right )}\\ &=\frac {\left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d (c+d \tan (e+f x))^{3/2}}{3 b^2 \left (a^2+b^2\right ) f}-\frac {\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac {2 \int \frac {\sqrt {c+d \tan (e+f x)} \left (-\frac {3}{4} \left (a \left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d^2-b c ((b B-a C) (2 b c-5 a d)+A b (2 a c+5 b d))\right )+\frac {3}{2} b^2 \left (2 a A c d-2 a c C d-A b \left (c^2-d^2\right )+a B \left (c^2-d^2\right )+b \left (c^2 C+2 B c d-C d^2\right )\right ) \tan (e+f x)-\frac {3}{4} d \left (5 a^3 C d-A b^2 (b c-a d)-2 b^3 (2 c C+B d)-a^2 b (5 c C+3 B d)+a b^2 (B c+4 C d)\right ) \tan ^2(e+f x)\right )}{a+b \tan (e+f x)} \, dx}{3 b^2 \left (a^2+b^2\right )}\\ &=-\frac {d \left (5 a^3 C d-A b^2 (b c-a d)-2 b^3 (2 c C+B d)-a^2 b (5 c C+3 B d)+a b^2 (B c+4 C d)\right ) \sqrt {c+d \tan (e+f x)}}{b^3 \left (a^2+b^2\right ) f}+\frac {\left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d (c+d \tan (e+f x))^{3/2}}{3 b^2 \left (a^2+b^2\right ) f}-\frac {\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac {4 \int \frac {\frac {3}{8} \left (5 a^4 C d^3+b^4 c^2 (2 B c+5 A d)-a^3 b d^2 (10 c C+3 B d)+a^2 b^2 d \left (5 c^2 C+4 B c d+(A+4 C) d^2\right )+a b^3 \left (2 A c^3-2 c^3 C-5 B c^2 d-4 A c d^2-6 c C d^2-2 B d^3\right )\right )+\frac {3}{4} b^3 \left (a A d \left (3 c^2-d^2\right )-A b \left (c^3-3 c d^2\right )+b \left (c^3 C+3 B c^2 d-3 c C d^2-B d^3\right )-a \left (C d \left (3 c^2-d^2\right )-B \left (c^3-3 c d^2\right )\right )\right ) \tan (e+f x)+\frac {3}{8} d \left (5 a^4 C d^2-a^3 b d (10 c C+3 B d)+a b^3 \left (B c^2-12 c C d-4 B d^2\right )+2 b^4 \left (3 c^2 C+3 B c d-C d^2\right )+a^2 b^2 \left (5 c^2 C+4 B c d+4 C d^2\right )+A b^2 \left (2 a b c d+a^2 d^2-b^2 \left (c^2-2 d^2\right )\right )\right ) \tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx}{3 b^3 \left (a^2+b^2\right )}\\ &=-\frac {d \left (5 a^3 C d-A b^2 (b c-a d)-2 b^3 (2 c C+B d)-a^2 b (5 c C+3 B d)+a b^2 (B c+4 C d)\right ) \sqrt {c+d \tan (e+f x)}}{b^3 \left (a^2+b^2\right ) f}+\frac {\left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d (c+d \tan (e+f x))^{3/2}}{3 b^2 \left (a^2+b^2\right ) f}-\frac {\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac {4 \int \frac {-\frac {3}{4} b^3 \left (b^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 )+a^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 {3}{4} b^3 \left (2 a b \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 )-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}{3 b^3 \left (a^2+b^2\right )^2}-\frac {\left ((b c-a d)^2 \left (3 a^3 b B d-5 a^4 C d-b^4 (2 B c+5 A d)-a b^3 (4 A c-4 c C-7 B d)+a^2 b^2 (2 B c-(A+9 C) d)\right )\right ) \int \frac {1+\tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx}{2 b^3 \left (a^2+b^2\right )^2}\\ &=-\frac {d \left (5 a^3 C d-A b^2 (b c-a d)-2 b^3 (2 c C+B d)-a^2 b (5 c C+3 B d)+a b^2 (B c+4 C d)\right ) \sqrt {c+d \tan (e+f x)}}{b^3 \left (a^2+b^2\right ) f}+\frac {\left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d (c+d \tan (e+f x))^{3/2}}{3 b^2 \left (a^2+b^2\right ) f}-\frac {\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac {\left ((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}{2 (a-i b)^2}+\frac {\left ((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}{2 (a+i b)^2}-\frac {\left ((b c-a d)^2 \left (3 a^3 b B d-5 a^4 C d-b^4 (2 B c+5 A d)-a b^3 (4 A c-4 c C-7 B d)+a^2 b^2 (2 B c-(A+9 C) d)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{2 b^3 \left (a^2+b^2\right )^2 f}\\ &=-\frac {d \left (5 a^3 C d-A b^2 (b c-a d)-2 b^3 (2 c C+B d)-a^2 b (5 c C+3 B d)+a b^2 (B c+4 C d)\right ) \sqrt {c+d \tan (e+f x)}}{b^3 \left (a^2+b^2\right ) f}+\frac {\left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d (c+d \tan (e+f x))^{3/2}}{3 b^2 \left (a^2+b^2\right ) f}-\frac {\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac {\left ((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 (a-i b)^2 f}-\frac {\left (i (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 (a+i b)^2 f}-\frac {\left ((b c-a d)^2 \left (3 a^3 b B d-5 a^4 C d-b^4 (2 B c+5 A d)-a b^3 (4 A c-4 c C-7 B d)+a^2 b^2 (2 B c-(A+9 C) d)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{b^3 \left (a^2+b^2\right )^2 d f}\\ &=\frac {(b c-a d)^{3/2} \left (3 a^3 b B d-5 a^4 C d-b^4 (2 B c+5 A d)-a b^3 (4 A c-4 c C-7 B d)+a^2 b^2 (2 B c-(A+9 C) d)\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{b^{7/2} \left (a^2+b^2\right )^2 f}-\frac {d \left (5 a^3 C d-A b^2 (b c-a d)-2 b^3 (2 c C+B d)-a^2 b (5 c C+3 B d)+a b^2 (B c+4 C d)\right ) \sqrt {c+d \tan (e+f x)}}{b^3 \left (a^2+b^2\right ) f}+\frac {\left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d (c+d \tan (e+f x))^{3/2}}{3 b^2 \left (a^2+b^2\right ) f}-\frac {\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}-\frac {\left ((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 )}{(a-i b)^2 d f}-\frac {\left ((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 )}{(a+i b)^2 d f}\\ &=-\frac {(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 )}{(a-i b)^2 f}-\frac {(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 )}{(a+i b)^2 f}+\frac {(b c-a d)^{3/2} \left (3 a^3 b B d-5 a^4 C d-b^4 (2 B c+5 A d)-a b^3 (4 A c-4 c C-7 B d)+a^2 b^2 (2 B c-(A+9 C) d)\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{b^{7/2} \left (a^2+b^2\right )^2 f}-\frac {d \left (5 a^3 C d-A b^2 (b c-a d)-2 b^3 (2 c C+B d)-a^2 b (5 c C+3 B d)+a b^2 (B c+4 C d)\right ) \sqrt {c+d \tan (e+f x)}}{b^3 \left (a^2+b^2\right ) f}+\frac {\left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d (c+d \tan (e+f x))^{3/2}}{3 b^2 \left (a^2+b^2\right ) f}-\frac {\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}\\ \end {align*}
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Mathematica [B] time = 6.53, size = 6112, normalized size = 12.92 \[ \text {Result too large to show} \]
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 = 0.87, size = 14119, normalized size = 29.85 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
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(-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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