Optimal. Leaf size=213 \[ -\frac {i (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a-i b)^{3/2} f}+\frac {i (c+i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{3/2} f}-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{\left (a^2+b^2\right ) f \sqrt {a+b \tan (e+f x)}} \]
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Rubi [A]
time = 0.56, antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3648, 3697,
3696, 95, 214} \begin {gather*} -\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) \sqrt {a+b \tan (e+f x)}}-\frac {i (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 )}{f (a-i b)^{3/2}}+\frac {i (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 )}{f (a+i b)^{3/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 95
Rule 214
Rule 3648
Rule 3696
Rule 3697
Rubi steps
\begin {align*} \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+b \tan (e+f x))^{3/2}} \, dx &=-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{\left (a^2+b^2\right ) f \sqrt {a+b \tan (e+f x)}}-\frac {2 \int \frac {\frac {1}{2} \left (-2 b c d-a \left (c^2-d^2\right )\right )-\frac {1}{2} \left (2 a c d-b \left (c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{a^2+b^2}\\ &=-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{\left (a^2+b^2\right ) f \sqrt {a+b \tan (e+f x)}}+\frac {(c-i d)^2 \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)}+\frac {(c+i d)^2 \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)}\\ &=-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{\left (a^2+b^2\right ) f \sqrt {a+b \tan (e+f x)}}+\frac {(c-i d)^2 \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) f}+\frac {(c+i d)^2 \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) f}\\ &=-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{\left (a^2+b^2\right ) f \sqrt {a+b \tan (e+f x)}}+\frac {(c-i d)^2 \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) f}+\frac {(c+i d)^2 \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) f}\\ &=-\frac {i (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)^{3/2} f}+\frac {i (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)^{3/2} f}-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{\left (a^2+b^2\right ) f \sqrt {a+b \tan (e+f x)}}\\ \end {align*}
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Mathematica [A]
time = 1.93, size = 231, normalized size = 1.08 \begin {gather*} \frac {-\frac {i (-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)^{3/2}}+\frac {\frac {(i a+b) (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)^{3/2}}+\frac {2 (-b c+a d) \sqrt {c+d \tan (e+f x)}}{(a+i b) \sqrt {a+b \tan (e+f x)}}}{a-i b}}{f} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{\left (a +b \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-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 {\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\left (a + b \tan {\left (e + f x \right )}\right )^{\frac {3}{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]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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