Optimal. Leaf size=138 \[ \frac{2 (b c-a d)}{f \left (c^2+d^2\right ) \sqrt{c+d \tan (e+f x)}}-\frac{(b+i a) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{3/2}}+\frac{(-b+i a) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{3/2}} \]
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Rubi [A] time = 0.241404, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3529, 3539, 3537, 63, 208} \[ \frac{2 (b c-a d)}{f \left (c^2+d^2\right ) \sqrt{c+d \tan (e+f x)}}-\frac{(b+i a) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{3/2}}+\frac{(-b+i a) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{3/2}} \]
Antiderivative was successfully verified.
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Rule 3529
Rule 3539
Rule 3537
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{a+b \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx &=\frac{2 (b c-a d)}{\left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{\int \frac{a c+b d+(b c-a d) \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{c^2+d^2}\\ &=\frac{2 (b c-a d)}{\left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{(a-i b) \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (c-i d)}+\frac{(a+i b) \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (c+i d)}\\ &=\frac{2 (b c-a d)}{\left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{(i a+b) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 (c-i d) f}-\frac{(i a-b) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 (c+i d) f}\\ &=\frac{2 (b c-a d)}{\left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}-\frac{(a-i b) \operatorname{Subst}\left (\int \frac{1}{-1-\frac{i c}{d}+\frac{i x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{(c-i d) d f}-\frac{(a+i b) \operatorname{Subst}\left (\int \frac{1}{-1+\frac{i c}{d}-\frac{i x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{(c+i d) d f}\\ &=-\frac{(i a+b) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{(c-i d)^{3/2} f}+\frac{(i a-b) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{(c+i d)^{3/2} f}+\frac{2 (b c-a d)}{\left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}\\ \end{align*}
Mathematica [C] time = 0.185347, size = 113, normalized size = 0.82 \[ \frac{i \left (\frac{(a-i b) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{c+d \tan (e+f x)}{c-i d}\right )}{c-i d}-\frac{(a+i b) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{c+d \tan (e+f x)}{c+i d}\right )}{c+i d}\right )}{f \sqrt{c+d \tan (e+f x)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.061, size = 7951, normalized size = 57.6 \begin{align*} \text{output too large to display} \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: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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{a + b \tan{\left (e + f x \right )}}{\left (c + d \tan{\left (e + f x \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{b \tan \left (f x + e\right ) + a}{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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