Optimal. Leaf size=195 \[ \frac{4 (b c-a d) (a c+b d)}{f \left (c^2+d^2\right )^2 \sqrt{c+d \tan (e+f x)}}-\frac{2 (b c-a d)^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac{i (a-i b)^2 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{5/2}}+\frac{i (a+i b)^2 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{5/2}} \]
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Rubi [A] time = 0.49837, antiderivative size = 195, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {3542, 3529, 3539, 3537, 63, 208} \[ \frac{4 (b c-a d) (a c+b d)}{f \left (c^2+d^2\right )^2 \sqrt{c+d \tan (e+f x)}}-\frac{2 (b c-a d)^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac{i (a-i b)^2 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{5/2}}+\frac{i (a+i b)^2 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 3542
Rule 3529
Rule 3539
Rule 3537
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{(a+b \tan (e+f x))^2}{(c+d \tan (e+f x))^{5/2}} \, dx &=-\frac{2 (b c-a d)^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{\int \frac{a^2 c-b^2 c+2 a b d+\left (2 a b c-a^2 d+b^2 d\right ) \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx}{c^2+d^2}\\ &=-\frac{2 (b c-a d)^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{4 (b c-a d) (a c+b d)}{\left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{\int \frac{(a c+b c-a d+b d) (a c-b c+a d+b d)+2 (b c-a d) (a c+b d) \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{\left (c^2+d^2\right )^2}\\ &=-\frac{2 (b c-a d)^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{4 (b c-a d) (a c+b d)}{\left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{(a-i b)^2 \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (c-i d)^2}+\frac{(a+i b)^2 \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (c+i d)^2}\\ &=-\frac{2 (b c-a d)^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{4 (b c-a d) (a c+b d)}{\left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{\left (i (a-i b)^2\right ) \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)^2 f}-\frac{\left (i (a+i b)^2\right ) \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)^2 f}\\ &=-\frac{2 (b c-a d)^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{4 (b c-a d) (a c+b d)}{\left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}-\frac{(a-i b)^2 \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)^2 d f}-\frac{(a+i b)^2 \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)^2 d f}\\ &=-\frac{i (a-i b)^2 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{(c-i d)^{5/2} f}+\frac{i (a+i b)^2 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{(c+i d)^{5/2} f}-\frac{2 (b c-a d)^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{4 (b c-a d) (a c+b d)}{\left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}\\ \end{align*}
Mathematica [C] time = 0.190264, size = 127, normalized size = 0.65 \[ \frac{-\frac{(a-i b)^2 \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{c+d \tan (e+f x)}{c-i d}\right )}{d+i c}+\frac{(a+i b)^2 \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{c+d \tan (e+f x)}{c+i d}\right )}{-d+i c}-\frac{2 b^2}{d}}{3 f (c+d \tan (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.095, size = 19430, normalized size = 99.6 \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(-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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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{\left (a + b \tan{\left (e + f x \right )}\right )^{2}}{\left (c + d \tan{\left (e + f x \right )}\right )^{\frac{5}{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 (b \tan \left (f x + e\right ) + a\right )}^{2}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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