Optimal. Leaf size=199 \[ \frac{\sqrt{-b+i a} (-B+i A) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 (3 a B+A b) \sqrt{a+b \tan (c+d x)}}{3 a d \sqrt{\tan (c+d x)}}+\frac{\sqrt{b+i a} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 A \sqrt{a+b \tan (c+d x)}}{3 d \tan ^{\frac{3}{2}}(c+d x)} \]
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Rubi [A] time = 0.757254, antiderivative size = 199, 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} \[ \frac{\sqrt{-b+i a} (-B+i A) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 (3 a B+A b) \sqrt{a+b \tan (c+d x)}}{3 a d \sqrt{\tan (c+d x)}}+\frac{\sqrt{b+i a} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 A \sqrt{a+b \tan (c+d x)}}{3 d \tan ^{\frac{3}{2}}(c+d x)} \]
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{a+b \tan (c+d x)} (A+B \tan (c+d x))}{\tan ^{\frac{5}{2}}(c+d x)} \, dx &=-\frac{2 A \sqrt{a+b \tan (c+d x)}}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2}{3} \int \frac{\frac{1}{2} (-A b-3 a B)+\frac{3}{2} (a A-b B) \tan (c+d x)+A b \tan ^2(c+d x)}{\tan ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{2 A \sqrt{a+b \tan (c+d x)}}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 (A b+3 a B) \sqrt{a+b \tan (c+d x)}}{3 a d \sqrt{\tan (c+d x)}}+\frac{4 \int \frac{-\frac{3}{4} a (a A-b B)-\frac{3}{4} a (A b+a B) \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{3 a}\\ &=-\frac{2 A \sqrt{a+b \tan (c+d x)}}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 (A b+3 a B) \sqrt{a+b \tan (c+d x)}}{3 a d \sqrt{\tan (c+d x)}}-\frac{1}{2} ((a-i b) (A-i B)) \int \frac{1+i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx-\frac{1}{2} ((a+i b) (A+i B)) \int \frac{1-i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{2 A \sqrt{a+b \tan (c+d x)}}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 (A b+3 a B) \sqrt{a+b \tan (c+d x)}}{3 a d \sqrt{\tan (c+d x)}}-\frac{((a-i b) (A-i B)) \operatorname{Subst}\left (\int \frac{1}{(1-i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac{((a+i b) (A+i B)) \operatorname{Subst}\left (\int \frac{1}{(1+i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=-\frac{2 A \sqrt{a+b \tan (c+d x)}}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 (A b+3 a B) \sqrt{a+b \tan (c+d x)}}{3 a d \sqrt{\tan (c+d x)}}-\frac{((a-i b) (A-i 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 )}{d}-\frac{((a+i b) (A+i 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 )}{d}\\ &=\frac{\sqrt{i a-b} (i A-B) \tan ^{-1}\left (\frac{\sqrt{i a-b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{\sqrt{i a+b} (i A+B) \tanh ^{-1}\left (\frac{\sqrt{i a+b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 A \sqrt{a+b \tan (c+d x)}}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 (A b+3 a B) \sqrt{a+b \tan (c+d x)}}{3 a d \sqrt{\tan (c+d x)}}\\ \end{align*}
Mathematica [A] time = 1.39414, size = 194, normalized size = 0.97 \[ \frac{-\frac{2 \sqrt{a+b \tan (c+d x)} ((3 a B+A b) \tan (c+d x)+a A)}{a \tan ^{\frac{3}{2}}(c+d x)}-3 (-1)^{3/4} \sqrt{-a-i b} (A+i B) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )+3 \sqrt [4]{-1} \sqrt{a-i b} (B+i A) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{3 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.704, size = 2178959, normalized size = 10949.5 \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{b \tan \left (d x + c\right ) + a}}{\tan \left (d x + c\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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \tan{\left (c + d x \right )}\right ) \sqrt{a + b \tan{\left (c + d x \right )}}}{\tan ^{\frac{5}{2}}{\left (c + d x \right )}}\, dx \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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