Optimal. Leaf size=297 \[ -\frac{2 b^{3/2} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{a^{3/2} d \left (a^2+b^2\right )}+\frac{(a (A-B)+b (A+B)) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )}-\frac{(a (A-B)+b (A+B)) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )}+\frac{(b (A-B)-a (A+B)) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )}-\frac{(b (A-B)-a (A+B)) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )}-\frac{2 A}{a d \sqrt{\tan (c+d x)}} \]
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Rubi [A] time = 0.636983, antiderivative size = 297, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 12, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {3609, 3653, 3534, 1168, 1162, 617, 204, 1165, 628, 3634, 63, 205} \[ -\frac{2 b^{3/2} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{a^{3/2} d \left (a^2+b^2\right )}+\frac{(a (A-B)+b (A+B)) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )}-\frac{(a (A-B)+b (A+B)) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )}+\frac{(b (A-B)-a (A+B)) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )}-\frac{(b (A-B)-a (A+B)) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )}-\frac{2 A}{a d \sqrt{\tan (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 3609
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{A+B \tan (c+d x)}{\tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))} \, dx &=-\frac{2 A}{a d \sqrt{\tan (c+d x)}}-\frac{2 \int \frac{\frac{1}{2} (A b-a B)+\frac{1}{2} a A \tan (c+d x)+\frac{1}{2} A b \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{a}\\ &=-\frac{2 A}{a d \sqrt{\tan (c+d x)}}-\frac{2 \int \frac{\frac{1}{2} a (A b-a B)+\frac{1}{2} a (a A+b B) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx}{a \left (a^2+b^2\right )}-\frac{\left (b^2 (A b-a B)\right ) \int \frac{1+\tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{a \left (a^2+b^2\right )}\\ &=-\frac{2 A}{a d \sqrt{\tan (c+d x)}}-\frac{4 \operatorname{Subst}\left (\int \frac{\frac{1}{2} a (A b-a B)+\frac{1}{2} a (a A+b B) x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{a \left (a^2+b^2\right ) d}-\frac{\left (b^2 (A b-a B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} (a+b x)} \, dx,x,\tan (c+d x)\right )}{a \left (a^2+b^2\right ) d}\\ &=-\frac{2 A}{a d \sqrt{\tan (c+d x)}}-\frac{\left (2 b^2 (A b-a B)\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{a \left (a^2+b^2\right ) d}-\frac{(b (A-B)-a (A+B)) \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 ) d}-\frac{(a (A-B)+b (A+B)) \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 ) d}\\ &=-\frac{2 b^{3/2} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{a^{3/2} \left (a^2+b^2\right ) d}-\frac{2 A}{a d \sqrt{\tan (c+d x)}}+\frac{(b (A-B)-a (A+B)) \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 ) d}+\frac{(b (A-B)-a (A+B)) \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 ) d}-\frac{(a (A-B)+b (A+B)) \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 ) d}-\frac{(a (A-B)+b (A+B)) \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 ) d}\\ &=-\frac{2 b^{3/2} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{a^{3/2} \left (a^2+b^2\right ) d}+\frac{(b (A-B)-a (A+B)) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right ) d}-\frac{(b (A-B)-a (A+B)) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right ) d}-\frac{2 A}{a d \sqrt{\tan (c+d x)}}-\frac{(a (A-B)+b (A+B)) \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 ) d}+\frac{(a (A-B)+b (A+B)) \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 ) d}\\ &=\frac{(a (A-B)+b (A+B)) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right ) d}-\frac{(a (A-B)+b (A+B)) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right ) d}-\frac{2 b^{3/2} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{a^{3/2} \left (a^2+b^2\right ) d}+\frac{(b (A-B)-a (A+B)) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right ) d}-\frac{(b (A-B)-a (A+B)) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right ) d}-\frac{2 A}{a d \sqrt{\tan (c+d x)}}\\ \end{align*}
Mathematica [C] time = 0.566075, size = 153, normalized size = 0.52 \[ \frac{\frac{2 b^{3/2} (a B-A b) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{\sqrt{a} \left (a^2+b^2\right )}+\frac{\sqrt [4]{-1} a \left ((b-i a) (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )+(b+i a) (A+i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )\right )}{a^2+b^2}-\frac{2 A}{\sqrt{\tan (c+d x)}}}{a d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.05, size = 628, normalized size = 2.1 \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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B \tan{\left (c + d x \right )}}{\left (a + b \tan{\left (c + d x \right )}\right ) \tan ^{\frac{3}{2}}{\left (c + d x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 2.3635, size = 425, normalized size = 1.43 \begin{align*} -\frac{{\left (\sqrt{2} A a - \sqrt{2} B a + \sqrt{2} A b + \sqrt{2} B b\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right )}{2 \,{\left (a^{2} d + b^{2} d\right )}} - \frac{{\left (\sqrt{2} A a - \sqrt{2} B a + \sqrt{2} A b + \sqrt{2} B b\right )} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right )}{2 \,{\left (a^{2} d + b^{2} d\right )}} + \frac{{\left (\sqrt{2} A a + \sqrt{2} B a - \sqrt{2} A b + \sqrt{2} B b\right )} \log \left (\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{4 \,{\left (a^{2} d + b^{2} d\right )}} - \frac{{\left (\sqrt{2} A a + \sqrt{2} B a - \sqrt{2} A b + \sqrt{2} B b\right )} \log \left (-\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{4 \,{\left (a^{2} d + b^{2} d\right )}} + \frac{2 \,{\left (B a b^{2} - A b^{3}\right )} \arctan \left (\frac{b \sqrt{\tan \left (d x + c\right )}}{\sqrt{a b}}\right )}{{\left (a^{3} d + a b^{2} d\right )} \sqrt{a b}} - \frac{2 \, A}{a d \sqrt{\tan \left (d x + c\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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