Optimal. Leaf size=185 \[ -\frac{2 (A b-a B)}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac{2 \left (a^2 (-B)+2 a A b+b^2 B\right )}{d \left (a^2+b^2\right )^2 \sqrt{a+b \tan (c+d x)}}-\frac{(B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d (a-i b)^{5/2}}+\frac{(-B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d (a+i b)^{5/2}} \]
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Rubi [A] time = 0.364245, antiderivative size = 185, normalized size of antiderivative = 1., number of steps used = 9, 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} \[ -\frac{2 (A b-a B)}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac{2 \left (a^2 (-B)+2 a A b+b^2 B\right )}{d \left (a^2+b^2\right )^2 \sqrt{a+b \tan (c+d x)}}-\frac{(B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d (a-i b)^{5/2}}+\frac{(-B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d (a+i b)^{5/2}} \]
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 (c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx &=-\frac{2 (A b-a B)}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{\int \frac{a A+b B-(A b-a B) \tan (c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx}{a^2+b^2}\\ &=-\frac{2 (A b-a B)}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac{2 \left (2 a A b-a^2 B+b^2 B\right )}{\left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}+\frac{\int \frac{a^2 A-A b^2+2 a b B-\left (2 a A b-a^2 B+b^2 B\right ) \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{\left (a^2+b^2\right )^2}\\ &=-\frac{2 (A b-a B)}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac{2 \left (2 a A b-a^2 B+b^2 B\right )}{\left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}+\frac{(A-i B) \int \frac{1+i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{2 (a-i b)^2}+\frac{(A+i B) \int \frac{1-i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{2 (a+i b)^2}\\ &=-\frac{2 (A b-a B)}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac{2 \left (2 a A b-a^2 B+b^2 B\right )}{\left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}-\frac{(i A-B) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a+i b x}} \, dx,x,-i \tan (c+d x)\right )}{2 (a+i b)^2 d}+\frac{(i A+B) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a-i b x}} \, dx,x,i \tan (c+d x)\right )}{2 (a-i b)^2 d}\\ &=-\frac{2 (A b-a B)}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac{2 \left (2 a A b-a^2 B+b^2 B\right )}{\left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}+\frac{(A-i B) \operatorname{Subst}\left (\int \frac{1}{-1-\frac{i a}{b}+\frac{i x^2}{b}} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{b (i a+b)^2 d}+\frac{(A+i B) \operatorname{Subst}\left (\int \frac{1}{-1+\frac{i a}{b}-\frac{i x^2}{b}} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{(i a-b)^2 b d}\\ &=-\frac{(i A+B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{(a-i b)^{5/2} d}+\frac{(i A-B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{(a+i b)^{5/2} d}-\frac{2 (A b-a B)}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac{2 \left (2 a A b-a^2 B+b^2 B\right )}{\left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}\\ \end{align*}
Mathematica [C] time = 0.14876, size = 115, normalized size = 0.62 \[ -\frac{i \left (\frac{(A+i B) \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\frac{a+b \tan (c+d x)}{a+i b}\right )}{a+i b}-\frac{(A-i B) \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\frac{a+b \tan (c+d x)}{a-i b}\right )}{a-i b}\right )}{3 d (a+b \tan (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.107, size = 12836, normalized size = 69.4 \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 (c + d x \right )}}{\left (a + b \tan{\left (c + d 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{B \tan \left (d x + c\right ) + A}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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