Optimal. Leaf size=244 \[ -\frac{2 (A b-a B) \sqrt{\tan (c+d x)}}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac{2 \left (5 a^2 A b-2 a^3 B+4 a b^2 B-A b^3\right ) \sqrt{\tan (c+d x)}}{3 a d \left (a^2+b^2\right )^2 \sqrt{a+b \tan (c+d x)}}-\frac{(-B+i A) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}+\frac{(B+i A) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}} \]
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Rubi [A] time = 1.00876, antiderivative size = 244, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3608, 3649, 3616, 3615, 93, 203, 206} \[ -\frac{2 (A b-a B) \sqrt{\tan (c+d x)}}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac{2 \left (5 a^2 A b-2 a^3 B+4 a b^2 B-A b^3\right ) \sqrt{\tan (c+d x)}}{3 a d \left (a^2+b^2\right )^2 \sqrt{a+b \tan (c+d x)}}-\frac{(-B+i A) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}+\frac{(B+i A) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 3608
Rule 3649
Rule 3616
Rule 3615
Rule 93
Rule 203
Rule 206
Rubi steps
\begin{align*} \int \frac{\sqrt{\tan (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx &=-\frac{2 (A b-a B) \sqrt{\tan (c+d x)}}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac{2 \int \frac{-\frac{1}{2} b (A b-a B)-\frac{3}{2} b (a A+b B) \tan (c+d x)+b (A b-a B) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx}{3 b \left (a^2+b^2\right )}\\ &=-\frac{2 (A b-a B) \sqrt{\tan (c+d x)}}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac{2 \left (5 a^2 A b-A b^3-2 a^3 B+4 a b^2 B\right ) \sqrt{\tan (c+d x)}}{3 a \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}-\frac{4 \int \frac{-\frac{3}{4} a b \left (2 a A b-a^2 B+b^2 B\right )-\frac{3}{4} a b \left (a^2 A-A b^2+2 a b B\right ) \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{3 a b \left (a^2+b^2\right )^2}\\ &=-\frac{2 (A b-a B) \sqrt{\tan (c+d x)}}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac{2 \left (5 a^2 A b-A b^3-2 a^3 B+4 a b^2 B\right ) \sqrt{\tan (c+d x)}}{3 a \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}+\frac{(i A-B) \int \frac{1-i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{2 (a+i b)^2}-\frac{(i A+B) \int \frac{1+i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{2 (a-i b)^2}\\ &=-\frac{2 (A b-a B) \sqrt{\tan (c+d x)}}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac{2 \left (5 a^2 A b-A b^3-2 a^3 B+4 a b^2 B\right ) \sqrt{\tan (c+d x)}}{3 a \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+i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a+i b)^2 d}-\frac{(i A+B) \operatorname{Subst}\left (\int \frac{1}{(1-i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a-i b)^2 d}\\ &=-\frac{2 (A b-a B) \sqrt{\tan (c+d x)}}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac{2 \left (5 a^2 A b-A b^3-2 a^3 B+4 a b^2 B\right ) \sqrt{\tan (c+d x)}}{3 a \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-(-i a+b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{(a+i b)^2 d}-\frac{(i A+B) \operatorname{Subst}\left (\int \frac{1}{1-(i a+b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{(a-i b)^2 d}\\ &=\frac{(i A-B) \tan ^{-1}\left (\frac{\sqrt{i a-b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{i a-b} (a+i b)^2 d}+\frac{(i A+B) \tanh ^{-1}\left (\frac{\sqrt{i a+b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}-\frac{2 (A b-a B) \sqrt{\tan (c+d x)}}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac{2 \left (5 a^2 A b-A b^3-2 a^3 B+4 a b^2 B\right ) \sqrt{\tan (c+d x)}}{3 a \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}\\ \end{align*}
Mathematica [A] time = 3.25869, size = 320, normalized size = 1.31 \[ \frac{\frac{6 b \left (a^2 (-B)+2 a A b+b^2 B\right ) \tan ^{\frac{3}{2}}(c+d x)}{\left (a^2+b^2\right ) \sqrt{a+b \tan (c+d x)}}+\frac{3 \left (2 \left (a^2 B-2 a A b-b^2 B\right ) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}+\frac{\sqrt [4]{-1} a (a-i b)^2 (A+i B) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{-a-i b}}-\frac{\sqrt [4]{-1} a (a+i b)^2 (A-i B) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{a-i b}}\right )}{a^2+b^2}+\frac{2 b (A b-a B) \tan ^{\frac{3}{2}}(c+d x)}{(a+b \tan (c+d x))^{3/2}}}{3 a d \left (a^2+b^2\right )} \]
Antiderivative was successfully verified.
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Maple [B] time = 2.2, size = 2976700, normalized size = 12199.6 \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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \tan \left (d x + c\right ) + A\right )} \sqrt{\tan \left (d x + c\right )}}{{\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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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(-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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Giac [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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