Optimal. Leaf size=264 \[ -\frac {i (a-i b)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {c-i d} f}+\frac {i (a+i b)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {c+i d} f}-\frac {b^{3/2} (b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{d^{3/2} f}+\frac {b^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f} \]
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Rubi [A]
time = 1.85, antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps
used = 13, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {3647, 3736,
6857, 65, 223, 212, 95, 214} \begin {gather*} -\frac {b^{3/2} (b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{d^{3/2} f}+\frac {b^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}-\frac {i (a-i b)^{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 \sqrt {c-i d}}+\frac {i (a+i b)^{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 \sqrt {c+i d}} \end {gather*}
Antiderivative was successfully verified.
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Rule 65
Rule 95
Rule 212
Rule 214
Rule 223
Rule 3647
Rule 3736
Rule 6857
Rubi steps
\begin {align*} \int \frac {(a+b \tan (e+f x))^{5/2}}{\sqrt {c+d \tan (e+f x)}} \, dx &=\frac {b^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}+\frac {\int \frac {\frac {1}{2} \left (2 a^3 d-b^2 (b c+a d)\right )+b \left (3 a^2-b^2\right ) d \tan (e+f x)-\frac {1}{2} b^2 (b c-5 a d) \tan ^2(e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{d}\\ &=\frac {b^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}+\frac {\text {Subst}\left (\int \frac {\frac {1}{2} \left (2 a^3 d-b^2 (b c+a d)\right )+b \left (3 a^2-b^2\right ) d x-\frac {1}{2} b^2 (b c-5 a d) x^2}{\sqrt {a+b x} \sqrt {c+d x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{d f}\\ &=\frac {b^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}+\frac {\text {Subst}\left (\int \left (-\frac {b^2 (b c-5 a d)}{2 \sqrt {a+b x} \sqrt {c+d x}}+\frac {a \left (a^2-3 b^2\right ) d+b \left (3 a^2-b^2\right ) d x}{\sqrt {a+b x} \sqrt {c+d x} \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{d f}\\ &=\frac {b^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}+\frac {\text {Subst}\left (\int \frac {a \left (a^2-3 b^2\right ) d+b \left (3 a^2-b^2\right ) d x}{\sqrt {a+b x} \sqrt {c+d x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{d f}-\frac {\left (b^2 (b c-5 a d)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{2 d f}\\ &=\frac {b^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}+\frac {\text {Subst}\left (\int \left (\frac {i a \left (a^2-3 b^2\right ) d-b \left (3 a^2-b^2\right ) d}{2 (i-x) \sqrt {a+b x} \sqrt {c+d x}}+\frac {i a \left (a^2-3 b^2\right ) d+b \left (3 a^2-b^2\right ) d}{2 (i+x) \sqrt {a+b x} \sqrt {c+d x}}\right ) \, dx,x,\tan (e+f x)\right )}{d f}-\frac {(b (b c-5 a d)) \text {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b \tan (e+f x)}\right )}{d f}\\ &=\frac {b^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}-\frac {(i a-b)^3 \text {Subst}\left (\int \frac {1}{(i-x) \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{2 f}-\frac {(i a+b)^3 \text {Subst}\left (\int \frac {1}{(i+x) \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{2 f}-\frac {(b (b c-5 a d)) \text {Subst}\left (\int \frac {1}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}\right )}{d f}\\ &=-\frac {b^{3/2} (b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{d^{3/2} f}+\frac {b^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}-\frac {(i a-b)^3 \text {Subst}\left (\int \frac {1}{a+i b-(c+i d) x^2} \, dx,x,\frac {\sqrt {a+b \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}\right )}{f}-\frac {(i a+b)^3 \text {Subst}\left (\int \frac {1}{-a+i b-(-c+i d) x^2} \, dx,x,\frac {\sqrt {a+b \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}\right )}{f}\\ &=-\frac {i (a-i b)^{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 )}{\sqrt {c-i d} f}+\frac {i (a+i b)^{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 )}{\sqrt {c+i d} f}-\frac {b^{3/2} (b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{d^{3/2} f}+\frac {b^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}\\ \end {align*}
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Mathematica [A]
time = 3.43, size = 438, normalized size = 1.66 \begin {gather*} \frac {\frac {\left (3 a^2 b^2-b^4+a \sqrt {-b^2} \left (a^2-3 b^2\right )\right ) d \tanh ^{-1}\left (\frac {\sqrt {-c+\frac {\sqrt {-b^2} d}{b}} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+\sqrt {-b^2}} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a+\sqrt {-b^2}} \sqrt {-c+\frac {\sqrt {-b^2} d}{b}}}+\frac {\left (-3 a^2 b^2+b^4+a^3 \sqrt {-b^2}+3 a \left (-b^2\right )^{3/2}\right ) d \tanh ^{-1}\left (\frac {\sqrt {c+\frac {\sqrt {-b^2} d}{b}} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+\sqrt {-b^2}} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+\sqrt {-b^2}} \sqrt {c+\frac {\sqrt {-b^2} d}{b}}}+b^3 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}-\frac {b^{5/2} (b c-5 a d) \sqrt {c-\frac {a d}{b}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c-\frac {a d}{b}}}\right ) \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}}}{\sqrt {d} \sqrt {c+d \tan (e+f x)}}}{b d f} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{\sqrt {c +d \tan \left (f x +e \right )}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \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 {\left (a + b \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}{\sqrt {c + d \tan {\left (e + f x \right )}}}\, 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]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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