Optimal. Leaf size=323 \[ \frac{\left (a^2 (-B)+6 a A b-8 b^2 B\right ) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{8 b d}+\frac{\left (6 a^2 A b+a^3 (-B)-24 a b^2 B-16 A b^3\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{8 b^{3/2} d}+\frac{(-b+i a)^{3/2} (A+i B) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{(6 A b-a B) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac{(b+i a)^{3/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d} \]
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Rubi [A] time = 2.55959, antiderivative size = 323, normalized size of antiderivative = 1., number of steps used = 15, 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} \[ \frac{\left (a^2 (-B)+6 a A b-8 b^2 B\right ) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{8 b d}+\frac{\left (6 a^2 A b+a^3 (-B)-24 a b^2 B-16 A b^3\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{8 b^{3/2} d}+\frac{(-b+i a)^{3/2} (A+i B) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{(6 A b-a B) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac{(b+i a)^{3/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d} \]
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 \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx &=\frac{B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}+\frac{\int \frac{(a+b \tan (c+d x))^{3/2} \left (-\frac{a B}{2}-3 b B \tan (c+d x)+\frac{1}{2} (6 A b-a B) \tan ^2(c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx}{3 b}\\ &=\frac{(6 A b-a B) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac{B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}+\frac{\int \frac{\sqrt{a+b \tan (c+d x)} \left (-\frac{3}{4} a (2 A b+a B)-6 b (A b+a B) \tan (c+d x)+\frac{3}{4} \left (6 a A b-a^2 B-8 b^2 B\right ) \tan ^2(c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx}{6 b}\\ &=\frac{\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{8 b d}+\frac{(6 A b-a B) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac{B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}+\frac{\int \frac{-\frac{3}{8} a \left (10 a A b+a^2 B-8 b^2 B\right )-6 b \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)+\frac{3}{8} \left (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 B\right ) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{6 b}\\ &=\frac{\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{8 b d}+\frac{(6 A b-a B) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac{B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{3}{8} a \left (10 a A b+a^2 B-8 b^2 B\right )-6 b \left (2 a A b+a^2 B-b^2 B\right ) x+\frac{3}{8} \left (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 B\right ) x^2}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{6 b d}\\ &=\frac{\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{8 b d}+\frac{(6 A b-a B) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac{B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}+\frac{\operatorname{Subst}\left (\int \left (\frac{3 \left (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 B\right )}{8 \sqrt{x} \sqrt{a+b x}}-\frac{6 \left (b \left (a^2 A-A b^2-2 a b B\right )+b \left (2 a A b+a^2 B-b^2 B\right ) x\right )}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{6 b d}\\ &=\frac{\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{8 b d}+\frac{(6 A b-a B) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac{B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac{\operatorname{Subst}\left (\int \frac{b \left (a^2 A-A b^2-2 a b B\right )+b \left (2 a A b+a^2 B-b^2 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 (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 B\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{16 b d}\\ &=\frac{\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{8 b d}+\frac{(6 A b-a B) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac{B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac{\operatorname{Subst}\left (\int \left (\frac{i b \left (a^2 A-A b^2-2 a b B\right )-b \left (2 a A b+a^2 B-b^2 B\right )}{2 (i-x) \sqrt{x} \sqrt{a+b x}}+\frac{i b \left (a^2 A-A b^2-2 a b B\right )+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 (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 B\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{8 b d}\\ &=\frac{\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{8 b d}+\frac{(6 A b-a B) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac{B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac{\left ((a+i b)^2 (i A-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 ((a-i b)^2 (i A+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 (6 a^2 A b-16 A b^3-a^3 B-24 a 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 )}{8 b d}\\ &=\frac{\left (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 B\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{8 b^{3/2} d}+\frac{\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{8 b d}+\frac{(6 A b-a B) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac{B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac{\left ((a+i b)^2 (i A-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 (i A+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{(i a-b)^{3/2} (A+i B) \tan ^{-1}\left (\frac{\sqrt{i a-b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{\left (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 B\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{8 b^{3/2} d}+\frac{(i a+b)^{3/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{i a+b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{8 b d}+\frac{(6 A b-a B) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac{B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}\\ \end{align*}
Mathematica [A] time = 4.2701, size = 347, normalized size = 1.07 \[ \frac{-3 \left (a^2 B-6 a A b+8 b^2 B\right ) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}-\frac{3 \sqrt{a} \left (-6 a^2 A b+a^3 B+24 a b^2 B+16 A b^3\right ) \sqrt{\frac{b \tan (c+d x)}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{\sqrt{b} \sqrt{a+b \tan (c+d x)}}+2 (6 A b-a B) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}+24 (-1)^{3/4} b (-a-i b)^{3/2} (A+i B) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )+24 \sqrt [4]{-1} b (a-i b)^{3/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )+8 B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{24 b d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.767, size = 2400808, normalized size = 7432.8 \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}} \tan \left (d x + c\right )^{\frac{3}{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(-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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