Optimal. Leaf size=383 \[ -\frac {2 d \sqrt {a+b \tan (e+f x)} \left (A \left (a^2 d^2+b^2 \left (c^2+2 d^2\right )\right )+a^2 \left (-B c d+2 c^2 C+C d^2\right )-a b B \left (c^2+d^2\right )+b^2 c (c C-B d)\right )}{f \left (a^2+b^2\right ) \left (c^2+d^2\right ) (b c-a d)^2 \sqrt {c+d \tan (e+f x)}}-\frac {2 \left (A b^2-a (b B-a C)\right )}{f \left (a^2+b^2\right ) (b c-a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {(i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a-i b)^{3/2} (c-i d)^{3/2}}-\frac {(B-i (A-C)) \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a+i b)^{3/2} (c+i d)^{3/2}} \]
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Rubi [A] time = 1.88, antiderivative size = 382, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.102, Rules used = {3649, 3616, 3615, 93, 208} \[ -\frac {2 d \sqrt {a+b \tan (e+f x)} \left (a^2 A d^2+a^2 \left (-B c d+2 c^2 C+C d^2\right )-a b B \left (c^2+d^2\right )+A b^2 \left (c^2+2 d^2\right )+b^2 c (c C-B d)\right )}{f \left (a^2+b^2\right ) \left (c^2+d^2\right ) (b c-a d)^2 \sqrt {c+d \tan (e+f x)}}-\frac {2 \left (A b^2-a (b B-a C)\right )}{f \left (a^2+b^2\right ) (b c-a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {(i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a-i b)^{3/2} (c-i d)^{3/2}}-\frac {(B-i (A-C)) \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a+i b)^{3/2} (c+i d)^{3/2}} \]
Antiderivative was successfully verified.
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Rule 93
Rule 208
Rule 3615
Rule 3616
Rule 3649
Rubi steps
\begin {align*} \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{(a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx &=-\frac {2 \left (A b^2-a (b B-a C)\right )}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {2 \int \frac {\frac {1}{2} \left (2 A b^2 d-a A (b c-a d)-(b B-a C) (b c+a d)\right )+\frac {1}{2} (A b-a B-b C) (b c-a d) \tan (e+f x)+\left (A b^2-a (b B-a C)\right ) d \tan ^2(e+f x)}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx}{\left (a^2+b^2\right ) (b c-a d)}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right )}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {2 d \left (a^2 A d^2+b^2 c (c C-B d)-a b B \left (c^2+d^2\right )+A b^2 \left (c^2+2 d^2\right )+a^2 \left (2 c^2 C-B c d+C d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{\left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}-\frac {4 \int \frac {-\frac {1}{4} (b c-a d)^2 (b B c-b (A-C) d+a (A c-c C+B d))-\frac {1}{4} (b c-a d)^2 (b c C-b B d-A (b c+a d)+a (B c+C d)) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{\left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right )}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right )}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {2 d \left (a^2 A d^2+b^2 c (c C-B d)-a b B \left (c^2+d^2\right )+A b^2 \left (c^2+2 d^2\right )+a^2 \left (2 c^2 C-B c d+C d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{\left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}+\frac {(A-i B-C) \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) (c-i d)}+\frac {(A+i B-C) \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) (c+i d)}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right )}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {2 d \left (a^2 A d^2+b^2 c (c C-B d)-a b B \left (c^2+d^2\right )+A b^2 \left (c^2+2 d^2\right )+a^2 \left (2 c^2 C-B c d+C d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{\left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}+\frac {(A-i B-C) \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) (c-i d) f}+\frac {(A+i B-C) \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) (c+i d) f}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right )}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {2 d \left (a^2 A d^2+b^2 c (c C-B d)-a b B \left (c^2+d^2\right )+A b^2 \left (c^2+2 d^2\right )+a^2 \left (2 c^2 C-B c d+C d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{\left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}+\frac {(A-i B-C) \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) (c-i d) f}+\frac {(A+i B-C) \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) (c+i d) f}\\ &=-\frac {(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 )}{(a-i b)^{3/2} (c-i d)^{3/2} f}-\frac {(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 )}{(a+i b)^{3/2} (c+i d)^{3/2} f}-\frac {2 \left (A b^2-a (b B-a C)\right )}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {2 d \left (a^2 A d^2+b^2 c (c C-B d)-a b B \left (c^2+d^2\right )+A b^2 \left (c^2+2 d^2\right )+a^2 \left (2 c^2 C-B c d+C d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{\left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}\\ \end {align*}
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Mathematica [A] time = 6.72, size = 484, normalized size = 1.26 \[ -\frac {2 \left (A b^2-a (b B-a C)\right )}{f \left (a^2+b^2\right ) (b c-a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {2 \left (-\frac {2 \sqrt {a+b \tan (e+f x)} \left (\frac {1}{2} d^2 \left (-a A (b c-a d)-(b B-a C) (a d+b c)+2 A b^2 d\right )-c \left (\frac {1}{2} d (b c-a d) (-a B+A b-b C)-c d \left (A b^2-a (b B-a C)\right )\right )\right )}{f \left (c^2+d^2\right ) (a d-b c) \sqrt {c+d \tan (e+f x)}}+\frac {(b c-a d)^2 \left (\frac {(a+i b) (c+i d) (i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a+i b} \sqrt {-c+i d}}+\frac {(b+i a) (c-i d) (A+i B-C) \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+i b} \sqrt {c+i d}}\right )}{2 f \left (c^2+d^2\right ) (a d-b c)}\right )}{\left (a^2+b^2\right ) (b c-a d)} \]
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 [F(-1)] time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {A +B \tan \left (f x +e \right )+C \left (\tan ^{2}\left (f x +e \right )\right )}{\left (a +b \tan \left (f x +e \right )\right )^{\frac {3}{2}} \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\, dx \]
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 \[ \int \frac {A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}}{\left (a + b \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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