Optimal. Leaf size=370 \[ -\frac {(i A+B-i C) \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a-i b)^{5/2} f}-\frac {(B-i (A-C)) \sqrt {c+i d} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{5/2} f}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{3 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{3/2}}-\frac {2 \left (2 a^3 b B d+a^4 C d+b^4 (3 B c+A d)+2 a b^3 (3 A c-3 c C-2 B d)-a^2 b^2 (3 B c+5 A d-7 C d)\right ) \sqrt {c+d \tan (e+f x)}}{3 b \left (a^2+b^2\right )^2 (b c-a d) f \sqrt {a+b \tan (e+f x)}} \]
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Rubi [A]
time = 1.34, antiderivative size = 370, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 6, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {3726, 3730,
3697, 3696, 95, 214} \begin {gather*} -\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{3 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}-\frac {2 \sqrt {c+d \tan (e+f x)} \left (a^4 C d+2 a^3 b B d-a^2 b^2 (5 A d+3 B c-7 C d)+2 a b^3 (3 A c-2 B d-3 c C)+b^4 (A d+3 B c)\right )}{3 b f \left (a^2+b^2\right )^2 (b c-a d) \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)^{5/2}}-\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)^{5/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 95
Rule 214
Rule 3696
Rule 3697
Rule 3726
Rule 3730
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))^{5/2}} \, dx &=-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{3 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{3/2}}+\frac {2 \int \frac {\frac {1}{2} ((b B-a C) (3 b c-a d)+A b (3 a c+b d))-\frac {3}{2} b ((A-C) (b c-a d)-B (a c+b d)) \tan (e+f x)-\frac {1}{2} \left (2 A b^2-2 a b B-a^2 C-3 b^2 C\right ) d \tan ^2(e+f x)}{(a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx}{3 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)}}{3 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{3/2}}-\frac {2 \left (2 a^3 b B d+a^4 C d+b^4 (3 B c+A d)+2 a b^3 (3 A c-3 c C-2 B d)-a^2 b^2 (3 B c+5 A d-7 C d)\right ) \sqrt {c+d \tan (e+f x)}}{3 b \left (a^2+b^2\right )^2 (b c-a d) f \sqrt {a+b \tan (e+f x)}}-\frac {4 \int \frac {-\frac {3}{4} b (b c-a d) \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 {3}{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)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{3 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)}}{3 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{3/2}}-\frac {2 \left (2 a^3 b B d+a^4 C d+b^4 (3 B c+A d)+2 a b^3 (3 A c-3 c C-2 B d)-a^2 b^2 (3 B c+5 A d-7 C d)\right ) \sqrt {c+d \tan (e+f x)}}{3 b \left (a^2+b^2\right )^2 (b c-a d) 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)^2}+\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)^2}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{3 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{3/2}}-\frac {2 \left (2 a^3 b B d+a^4 C d+b^4 (3 B c+A d)+2 a b^3 (3 A c-3 c C-2 B d)-a^2 b^2 (3 B c+5 A d-7 C d)\right ) \sqrt {c+d \tan (e+f x)}}{3 b \left (a^2+b^2\right )^2 (b c-a d) f \sqrt {a+b \tan (e+f x)}}+\frac {((A-i B-C) (c-i d)) \text {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)^2 f}+\frac {((A+i B-C) (c+i d)) \text {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)^2 f}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{3 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{3/2}}-\frac {2 \left (2 a^3 b B d+a^4 C d+b^4 (3 B c+A d)+2 a b^3 (3 A c-3 c C-2 B d)-a^2 b^2 (3 B c+5 A d-7 C d)\right ) \sqrt {c+d \tan (e+f x)}}{3 b \left (a^2+b^2\right )^2 (b c-a d) f \sqrt {a+b \tan (e+f x)}}+\frac {((A-i B-C) (c-i d)) \text {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)^2 f}+\frac {((A+i B-C) (c+i d)) \text {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)^2 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)^{5/2} 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)^{5/2} f}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{3 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{3/2}}-\frac {2 \left (2 a^3 b B d+a^4 C d+b^4 (3 B c+A d)+2 a b^3 (3 A c-3 c C-2 B d)-a^2 b^2 (3 B c+5 A d-7 C d)\right ) \sqrt {c+d \tan (e+f x)}}{3 b \left (a^2+b^2\right )^2 (b c-a d) f \sqrt {a+b \tan (e+f x)}}\\ \end {align*}
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Mathematica [A]
time = 6.15, size = 424, normalized size = 1.15 \begin {gather*} \frac {-\frac {3 C \sqrt {c+d \tan (e+f x)}}{(a+b \tan (e+f x))^{3/2}}+\frac {\left (-2 A b^2+2 a b B+a^2 C+3 b^2 C\right ) \sqrt {c+d \tan (e+f x)}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}+\frac {-3 b \left (\frac {(a+i b)^2 (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 )}{\sqrt {-a+i b}}+\frac {(a-i b)^2 (-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 )}{\sqrt {a+i b}}\right )+\frac {2 \left (2 a^3 b B d+a^4 C d+b^4 (3 B c+A d)+2 a b^3 (3 A c-3 c C-2 B d)+a^2 b^2 (-3 B c-5 A d+7 C d)\right ) \sqrt {c+d \tan (e+f x)}}{(-b c+a d) \sqrt {a+b \tan (e+f x)}}}{\left (a^2+b^2\right )^2}}{3 b f} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\sqrt {c +d \tan \left (f x +e \right )}\, \left (A +B \tan \left (f x +e \right )+C \left (\tan ^{2}\left (f x +e \right )\right )\right )}{\left (a +b \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\, dx\]
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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
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 \begin {gather*} \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 {5}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
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 \begin {gather*} \text {Hanged} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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