Optimal. Leaf size=91 \[ \frac{\sqrt{1+i \tan (e+f x)} F_1\left (n+1;\frac{5}{2},1;n+2;-i \tan (e+f x),i \tan (e+f x)\right ) (d \tan (e+f x))^{n+1}}{a d f (n+1) \sqrt{a+i a \tan (e+f x)}} \]
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Rubi [A] time = 0.129167, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {3564, 135, 133} \[ \frac{\sqrt{1+i \tan (e+f x)} F_1\left (n+1;\frac{5}{2},1;n+2;-i \tan (e+f x),i \tan (e+f x)\right ) (d \tan (e+f x))^{n+1}}{a d f (n+1) \sqrt{a+i a \tan (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 3564
Rule 135
Rule 133
Rubi steps
\begin{align*} \int \frac{(d \tan (e+f x))^n}{(a+i a \tan (e+f x))^{3/2}} \, dx &=\frac{\left (i a^2\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{i d x}{a}\right )^n}{(a+x)^{5/2} \left (-a^2+a x\right )} \, dx,x,i a \tan (e+f x)\right )}{f}\\ &=\frac{\left (i \sqrt{1+i \tan (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{i d x}{a}\right )^n}{\left (1+\frac{x}{a}\right )^{5/2} \left (-a^2+a x\right )} \, dx,x,i a \tan (e+f x)\right )}{f \sqrt{a+i a \tan (e+f x)}}\\ &=\frac{F_1\left (1+n;\frac{5}{2},1;2+n;-i \tan (e+f x),i \tan (e+f x)\right ) \sqrt{1+i \tan (e+f x)} (d \tan (e+f x))^{1+n}}{a d f (1+n) \sqrt{a+i a \tan (e+f x)}}\\ \end{align*}
Mathematica [F] time = 33.5813, size = 0, normalized size = 0. \[ \int \frac{(d \tan (e+f x))^n}{(a+i a \tan (e+f x))^{3/2}} \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.347, size = 0, normalized size = 0. \begin{align*} \int{ \left ( d\tan \left ( fx+e \right ) \right ) ^{n} \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{2} \left (\frac{-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{n} \sqrt{\frac{a}{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 )} e^{\left (-3 i \, f x - 3 i \, e\right )}}{4 \, a^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \tan{\left (e + f x \right )}\right )^{n}}{\left (a \left (i \tan{\left (e + f x \right )} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \tan \left (f x + e\right )\right )^{n}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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