Optimal. Leaf size=268 \[ \frac{\left (3 a^2 B+12 a A b-8 b^2 B\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{4 \sqrt{b} d}+\frac{(a+i b)^2 (-B+i A) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{-b+i a}}+\frac{(5 a B+4 A b) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{(b+i a)^{3/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{b B \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{2 d} \]
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Rubi [A] time = 2.40697, antiderivative size = 268, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 10, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3607, 3647, 3655, 6725, 63, 217, 206, 93, 205, 208} \[ \frac{\left (3 a^2 B+12 a A b-8 b^2 B\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{4 \sqrt{b} d}+\frac{(a+i b)^2 (-B+i A) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{-b+i a}}+\frac{(5 a B+4 A b) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{(b+i a)^{3/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{b B \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{2 d} \]
Antiderivative was successfully verified.
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Rule 3607
Rule 3647
Rule 3655
Rule 6725
Rule 63
Rule 217
Rule 206
Rule 93
Rule 205
Rule 208
Rubi steps
\begin{align*} \int \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx &=\frac{b B \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{2 d}+\frac{1}{2} \int \frac{\sqrt{\tan (c+d x)} \left (\frac{1}{2} a (4 a A-3 b B)+2 \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)+\frac{1}{2} b (4 A b+5 a B) \tan ^2(c+d x)\right )}{\sqrt{a+b \tan (c+d x)}} \, dx\\ &=\frac{(4 A b+5 a B) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{b B \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{2 d}+\frac{\int \frac{-\frac{1}{4} a b (4 A b+5 a B)+2 b \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)+\frac{1}{4} b \left (12 a A b+3 a^2 B-8 b^2 B\right ) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{2 b}\\ &=\frac{(4 A b+5 a B) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{b B \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{2 d}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{1}{4} a b (4 A b+5 a B)+2 b \left (a^2 A-A b^2-2 a b B\right ) x+\frac{1}{4} b \left (12 a A b+3 a^2 B-8 b^2 B\right ) x^2}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{2 b d}\\ &=\frac{(4 A b+5 a B) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{b B \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{2 d}+\frac{\operatorname{Subst}\left (\int \left (\frac{b \left (12 a A b+3 a^2 B-8 b^2 B\right )}{4 \sqrt{x} \sqrt{a+b x}}-\frac{2 \left (b \left (2 a A b+a^2 B-b^2 B\right )-b \left (a^2 A-A b^2-2 a b B\right ) x\right )}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{2 b d}\\ &=\frac{(4 A b+5 a B) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{b B \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{2 d}-\frac{\operatorname{Subst}\left (\int \frac{b \left (2 a A b+a^2 B-b^2 B\right )-b \left (a^2 A-A b^2-2 a b B\right ) x}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{b d}+\frac{\left (12 a A b+3 a^2 B-8 b^2 B\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{8 d}\\ &=\frac{(4 A b+5 a B) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{b B \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{2 d}-\frac{\operatorname{Subst}\left (\int \left (\frac{b \left (a^2 A-A b^2-2 a b B\right )+i b \left (2 a A b+a^2 B-b^2 B\right )}{2 (i-x) \sqrt{x} \sqrt{a+b x}}+\frac{-b \left (a^2 A-A b^2-2 a b B\right )+i b \left (2 a A b+a^2 B-b^2 B\right )}{2 \sqrt{x} (i+x) \sqrt{a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{b d}+\frac{\left (12 a A b+3 a^2 B-8 b^2 B\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{4 d}\\ &=\frac{(4 A b+5 a B) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{b B \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{2 d}+\frac{\left ((a-i b)^2 (A-i B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} (i+x) \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac{\left ((a+i b)^2 (A+i B)\right ) \operatorname{Subst}\left (\int \frac{1}{(i-x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{\left (12 a A b+3 a^2 B-8 b^2 B\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{4 d}\\ &=\frac{\left (12 a A b+3 a^2 B-8 b^2 B\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{4 \sqrt{b} d}+\frac{(4 A b+5 a B) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{b B \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{2 d}+\frac{\left ((a-i b)^2 (A-i B)\right ) \operatorname{Subst}\left (\int \frac{1}{i-(-a+i b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{\left ((a+i b)^2 (A+i B)\right ) \operatorname{Subst}\left (\int \frac{1}{i-(a+i b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}\\ &=\frac{(a+i b)^2 (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} d}+\frac{\left (12 a A b+3 a^2 B-8 b^2 B\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{4 \sqrt{b} d}+\frac{(i a+b)^{3/2} (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{(4 A b+5 a B) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{b B \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{2 d}\\ \end{align*}
Mathematica [A] time = 2.48483, size = 290, normalized size = 1.08 \[ \frac{\frac{\sqrt{a} \left (3 a^2 B+12 a A b-8 b^2 B\right ) \sqrt{\frac{b \tan (c+d x)}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{\sqrt{b} \sqrt{a+b \tan (c+d x)}}+(5 a B+4 A b) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}+4 \sqrt [4]{-1} (-a-i b)^{3/2} (A+i B) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )-4 \sqrt [4]{-1} (a-i b)^{3/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )+2 b B \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{4 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.82, size = 2398581, normalized size = 8949.9 \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{\left (B \tan \left (d x + c\right ) + A\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \sqrt{\tan \left (d x + c\right )}\,{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(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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