Optimal. Leaf size=292 \[ -\frac {2 (b c-a d) \left (a^2 d+6 a b c+7 b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{3 b f \left (a^2+b^2\right )^2 \sqrt {a+b \tan (e+f x)}}-\frac {2 (b c-a d)^2 \sqrt {c+d \tan (e+f x)}}{3 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}-\frac {i (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a-i b)^{5/2}}+\frac {i (c+i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a+i b)^{5/2}} \]
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Rubi [A] time = 1.58, antiderivative size = 292, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {3565, 3649, 3616, 3615, 93, 208} \[ -\frac {2 (b c-a d) \left (a^2 d+6 a b c+7 b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{3 b f \left (a^2+b^2\right )^2 \sqrt {a+b \tan (e+f x)}}-\frac {2 (b c-a d)^2 \sqrt {c+d \tan (e+f x)}}{3 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}-\frac {i (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a-i b)^{5/2}}+\frac {i (c+i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a+i b)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 93
Rule 208
Rule 3565
Rule 3615
Rule 3616
Rule 3649
Rubi steps
\begin {align*} \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+b \tan (e+f x))^{5/2}} \, dx &=-\frac {2 (b c-a d)^2 \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} \left (7 b^2 c^2 d+a^2 d^3+a b c \left (3 c^2-5 d^2\right )\right )-\frac {3}{2} b \left (b c^3-3 a c^2 d-3 b c d^2+a d^3\right ) \tan (e+f x)+\frac {1}{2} d \left (\left (a^2+3 b^2\right ) d^2-2 b c (b c-2 a d)\right ) \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 (b c-a d)^2 \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 (b c-a d) \left (6 a b c+a^2 d+7 b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{3 b \left (a^2+b^2\right )^2 f \sqrt {a+b \tan (e+f x)}}-\frac {4 \int \frac {-\frac {3}{4} b (b c-a d) \left (a^2 c^3-b^2 c^3+6 a b c^2 d-3 a^2 c d^2+3 b^2 c d^2-2 a b d^3\right )+\frac {3}{4} b (b c-a d) \left (2 a b c \left (c^2-3 d^2\right )+b^2 d \left (3 c^2-d^2\right )-a^2 \left (3 c^2 d-d^3\right )\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 (b c-a d)^2 \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 (b c-a d) \left (6 a b c+a^2 d+7 b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{3 b \left (a^2+b^2\right )^2 f \sqrt {a+b \tan (e+f x)}}+\frac {(c-i d)^3 \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 {(c+i d)^3 \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 (b c-a d)^2 \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 (b c-a d) \left (6 a b c+a^2 d+7 b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{3 b \left (a^2+b^2\right )^2 f \sqrt {a+b \tan (e+f x)}}+\frac {(c-i d)^3 \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)^2 f}+\frac {(c+i d)^3 \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)^2 f}\\ &=-\frac {2 (b c-a d)^2 \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 (b c-a d) \left (6 a b c+a^2 d+7 b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{3 b \left (a^2+b^2\right )^2 f \sqrt {a+b \tan (e+f x)}}+\frac {(c-i d)^3 \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)^2 f}+\frac {(c+i d)^3 \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)^2 f}\\ &=-\frac {i (c-i d)^{5/2} \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 {i (c+i d)^{5/2} \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 (b c-a d)^2 \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 (b c-a d) \left (6 a b c+a^2 d+7 b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{3 b \left (a^2+b^2\right )^2 f \sqrt {a+b \tan (e+f x)}}\\ \end {align*}
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Mathematica [A] time = 5.83, size = 350, normalized size = 1.20 \[ \frac {\frac {(d+i c) \left (\frac {(c+d \tan (e+f x))^{3/2}}{(a+b \tan (e+f x))^{3/2}}+3 (c-i d) \left (\frac {\sqrt {c+d \tan (e+f x)}}{(a-i b) \sqrt {a+b \tan (e+f x)}}+\frac {\sqrt {-c+i d} \tanh ^{-1}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(-a+i b)^{3/2}}\right )\right )}{a-i b}-(-d+i c) \left (\frac {\sqrt {c+d \tan (e+f x)} ((a d+3 b c+4 i b d) \tan (e+f x)+4 a c+3 i a d+i b c)}{(a+i b)^2 (a+b \tan (e+f x))^{3/2}}-\frac {3 (c+i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{5/2}}\right )}{3 f} \]
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 \[ \int \frac {\left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{\left (a +b \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\, dx \]
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 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}} \,d x \]
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 \[ \int \frac {\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}{\left (a + b \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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