Optimal. Leaf size=533 \[ -\frac{\left (3 a^2 b (A-B)+a^3 (-(A+B))+3 a b^2 (A+B)-b^3 (A-B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}+\frac{\left (3 a^2 b (A-B)+a^3 (-(A+B))+3 a b^2 (A+B)-b^3 (A-B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}+\frac{\left (-26 a^2 A b^3+3 a^4 A b+18 a^3 b^2 B+a^5 B-15 a b^4 B+3 A b^5\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 \sqrt{a} b^{3/2} d \left (a^2+b^2\right )^3}+\frac{a (A b-a B) \sqrt{\tan (c+d x)}}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac{\left (3 a^2 A b+a^3 B+9 a b^2 B-5 A b^3\right ) \sqrt{\tan (c+d x)}}{4 b d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac{\left (3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3}-\frac{\left (3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3} \]
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Rubi [A] time = 1.22818, antiderivative size = 533, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 13, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.394, Rules used = {3605, 3649, 3653, 3534, 1168, 1162, 617, 204, 1165, 628, 3634, 63, 205} \[ -\frac{\left (3 a^2 b (A-B)+a^3 (-(A+B))+3 a b^2 (A+B)-b^3 (A-B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}+\frac{\left (3 a^2 b (A-B)+a^3 (-(A+B))+3 a b^2 (A+B)-b^3 (A-B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )^3}+\frac{\left (-26 a^2 A b^3+3 a^4 A b+18 a^3 b^2 B+a^5 B-15 a b^4 B+3 A b^5\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 \sqrt{a} b^{3/2} d \left (a^2+b^2\right )^3}+\frac{a (A b-a B) \sqrt{\tan (c+d x)}}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac{\left (3 a^2 A b+a^3 B+9 a b^2 B-5 A b^3\right ) \sqrt{\tan (c+d x)}}{4 b d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac{\left (3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3}-\frac{\left (3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^3} \]
Antiderivative was successfully verified.
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Rule 3605
Rule 3649
Rule 3653
Rule 3534
Rule 1168
Rule 1162
Rule 617
Rule 204
Rule 1165
Rule 628
Rule 3634
Rule 63
Rule 205
Rubi steps
\begin{align*} \int \frac{\tan ^{\frac{3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx &=\frac{a (A b-a B) \sqrt{\tan (c+d x)}}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{\int \frac{-\frac{1}{2} a (A b-a B)+2 b (A b-a B) \tan (c+d x)+\frac{1}{2} \left (3 a A b+a^2 B+4 b^2 B\right ) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^2} \, dx}{2 b \left (a^2+b^2\right )}\\ &=\frac{a (A b-a B) \sqrt{\tan (c+d x)}}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{\left (3 a^2 A b-5 A b^3+a^3 B+9 a b^2 B\right ) \sqrt{\tan (c+d x)}}{4 b \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\int \frac{-\frac{1}{4} a \left (5 a^2 A b-3 A b^3-a^3 B+7 a b^2 B\right )+2 a b \left (2 a A b-a^2 B+b^2 B\right ) \tan (c+d x)+\frac{1}{4} a \left (3 a^2 A b-5 A b^3+a^3 B+9 a b^2 B\right ) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{2 a b \left (a^2+b^2\right )^2}\\ &=\frac{a (A b-a B) \sqrt{\tan (c+d x)}}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{\left (3 a^2 A b-5 A b^3+a^3 B+9 a b^2 B\right ) \sqrt{\tan (c+d x)}}{4 b \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\int \frac{-2 a b \left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right )+2 a b \left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right ) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx}{2 a b \left (a^2+b^2\right )^3}+\frac{\left (3 a^4 A b-26 a^2 A b^3+3 A b^5+a^5 B+18 a^3 b^2 B-15 a b^4 B\right ) \int \frac{1+\tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{8 b \left (a^2+b^2\right )^3}\\ &=\frac{a (A b-a B) \sqrt{\tan (c+d x)}}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{\left (3 a^2 A b-5 A b^3+a^3 B+9 a b^2 B\right ) \sqrt{\tan (c+d x)}}{4 b \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\operatorname{Subst}\left (\int \frac{-2 a b \left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right )+2 a b \left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right ) x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{a b \left (a^2+b^2\right )^3 d}+\frac{\left (3 a^4 A b-26 a^2 A b^3+3 A b^5+a^5 B+18 a^3 b^2 B-15 a b^4 B\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} (a+b x)} \, dx,x,\tan (c+d x)\right )}{8 b \left (a^2+b^2\right )^3 d}\\ &=\frac{a (A b-a B) \sqrt{\tan (c+d x)}}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{\left (3 a^2 A b-5 A b^3+a^3 B+9 a b^2 B\right ) \sqrt{\tan (c+d x)}}{4 b \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\left (3 a^4 A b-26 a^2 A b^3+3 A b^5+a^5 B+18 a^3 b^2 B-15 a b^4 B\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{4 b \left (a^2+b^2\right )^3 d}+\frac{\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{\left (a^2+b^2\right )^3 d}-\frac{\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{\left (a^2+b^2\right )^3 d}\\ &=\frac{\left (3 a^4 A b-26 a^2 A b^3+3 A b^5+a^5 B+18 a^3 b^2 B-15 a b^4 B\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 \sqrt{a} b^{3/2} \left (a^2+b^2\right )^3 d}+\frac{a (A b-a B) \sqrt{\tan (c+d x)}}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{\left (3 a^2 A b-5 A b^3+a^3 B+9 a b^2 B\right ) \sqrt{\tan (c+d x)}}{4 b \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right )^3 d}+\frac{\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right )^3 d}+\frac{\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \sqrt{2} \left (a^2+b^2\right )^3 d}+\frac{\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \sqrt{2} \left (a^2+b^2\right )^3 d}\\ &=\frac{\left (3 a^4 A b-26 a^2 A b^3+3 A b^5+a^5 B+18 a^3 b^2 B-15 a b^4 B\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 \sqrt{a} b^{3/2} \left (a^2+b^2\right )^3 d}+\frac{\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right )^3 d}-\frac{\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right )^3 d}+\frac{a (A b-a B) \sqrt{\tan (c+d x)}}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{\left (3 a^2 A b-5 A b^3+a^3 B+9 a b^2 B\right ) \sqrt{\tan (c+d x)}}{4 b \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right )^3 d}-\frac{\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right )^3 d}\\ &=-\frac{\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right )^3 d}+\frac{\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right )^3 d}+\frac{\left (3 a^4 A b-26 a^2 A b^3+3 A b^5+a^5 B+18 a^3 b^2 B-15 a b^4 B\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{4 \sqrt{a} b^{3/2} \left (a^2+b^2\right )^3 d}+\frac{\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right )^3 d}-\frac{\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right )^3 d}+\frac{a (A b-a B) \sqrt{\tan (c+d x)}}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{\left (3 a^2 A b-5 A b^3+a^3 B+9 a b^2 B\right ) \sqrt{\tan (c+d x)}}{4 b \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}\\ \end{align*}
Mathematica [C] time = 5.67479, size = 333, normalized size = 0.62 \[ \frac{\frac{\left (a^2 B+3 a A b+4 b^2 B\right ) \sqrt{\tan (c+d x)}}{a^2+b^2}-\frac{2 (a+b \tan (c+d x)) \left (-\frac{3}{4} a^{5/2} \sqrt{b} \left (a^2+b^2\right ) \left (3 a^2 A b+a^3 B+9 a b^2 B-5 A b^3\right ) \sqrt{\tan (c+d x)}+(a+b \tan (c+d x)) \left (-\frac{3}{4} a^2 \left (-26 a^2 A b^3+3 a^4 A b+18 a^3 b^2 B+a^5 B-15 a b^4 B+3 A b^5\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )-3 \sqrt [4]{-1} a^{5/2} b^{3/2} \left ((a+i b)^3 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )+(a-i b)^3 (A+i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )\right )\right )\right )}{a^{5/2} \sqrt{b} \left (a^2+b^2\right )^3}-4 B \sqrt{\tan (c+d x)}}{6 b d (a+b \tan (c+d x))^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.059, size = 1835, normalized size = 3.4 \begin{align*} \text{result 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(-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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