Optimal. Leaf size=125 \[ \frac {(a+i a \tan (e+f x))^m \sqrt {\frac {c+d \tan (e+f x)}{c+i d}} F_1\left (m;\frac {3}{2},1;m+1;-\frac {d (i \tan (e+f x)+1)}{i c-d},\frac {1}{2} (i \tan (e+f x)+1)\right )}{2 f m (-d+i c) \sqrt {c+d \tan (e+f x)}} \]
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Rubi [A] time = 0.16, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3564, 137, 136} \[ \frac {(a+i a \tan (e+f x))^m \sqrt {\frac {c+d \tan (e+f x)}{c+i d}} F_1\left (m;\frac {3}{2},1;m+1;-\frac {d (i \tan (e+f x)+1)}{i c-d},\frac {1}{2} (i \tan (e+f x)+1)\right )}{2 f m (-d+i c) \sqrt {c+d \tan (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 136
Rule 137
Rule 3564
Rubi steps
\begin {align*} \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^{3/2}} \, dx &=\frac {\left (i a^2\right ) \operatorname {Subst}\left (\int \frac {(a+x)^{-1+m}}{\left (-a^2+a x\right ) \left (c-\frac {i d x}{a}\right )^{3/2}} \, dx,x,i a \tan (e+f x)\right )}{f}\\ &=\frac {\left (i a^2 \sqrt {\frac {c+d \tan (e+f x)}{c+i d}}\right ) \operatorname {Subst}\left (\int \frac {(a+x)^{-1+m}}{\left (-a^2+a x\right ) \left (\frac {c}{c+i d}-\frac {i d x}{a (c+i d)}\right )^{3/2}} \, dx,x,i a \tan (e+f x)\right )}{(c+i d) f \sqrt {c+d \tan (e+f x)}}\\ &=\frac {F_1\left (m;\frac {3}{2},1;1+m;-\frac {d (1+i \tan (e+f x))}{i c-d},\frac {1}{2} (1+i \tan (e+f x))\right ) (a+i a \tan (e+f x))^m \sqrt {\frac {c+d \tan (e+f x)}{c+i d}}}{2 (i c-d) f m \sqrt {c+d \tan (e+f x)}}\\ \end {align*}
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Mathematica [F] time = 11.39, size = 0, normalized size = 0.00 \[ \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^{3/2}} \, dx \]
Verification is Not applicable to the result.
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fricas [F] time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (\frac {2 \, a e^{\left (2 i \, f x + 2 i \, e\right )}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{m} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}}{c^{2} + 2 i \, c d - d^{2} + {\left (c^{2} - 2 i \, c d - d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (c^{2} + d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}}, x\right ) \]
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] time = 1.35, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +i a \tan \left (f x +e \right )\right )^{m}}{\left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^m}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \]
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 \[ \int \frac {\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{m}}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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