Optimal. Leaf size=127 \[ \frac{(a A-b B) \tan ^{m+1}(c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+1}{2},\frac{m+3}{2},-\tan ^2(c+d x)\right )}{d (m+1)}+\frac{(a B+A b) \tan ^{m+2}(c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+4}{2},-\tan ^2(c+d x)\right )}{d (m+2)}+\frac{b B \tan ^{m+1}(c+d x)}{d (m+1)} \]
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Rubi [A] time = 0.139204, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {3592, 3538, 3476, 364} \[ \frac{(a A-b B) \tan ^{m+1}(c+d x) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\tan ^2(c+d x)\right )}{d (m+1)}+\frac{(a B+A b) \tan ^{m+2}(c+d x) \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};-\tan ^2(c+d x)\right )}{d (m+2)}+\frac{b B \tan ^{m+1}(c+d x)}{d (m+1)} \]
Antiderivative was successfully verified.
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Rule 3592
Rule 3538
Rule 3476
Rule 364
Rubi steps
\begin{align*} \int \tan ^m(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx &=\frac{b B \tan ^{1+m}(c+d x)}{d (1+m)}+\int \tan ^m(c+d x) (a A-b B+(A b+a B) \tan (c+d x)) \, dx\\ &=\frac{b B \tan ^{1+m}(c+d x)}{d (1+m)}+(A b+a B) \int \tan ^{1+m}(c+d x) \, dx+(a A-b B) \int \tan ^m(c+d x) \, dx\\ &=\frac{b B \tan ^{1+m}(c+d x)}{d (1+m)}+\frac{(A b+a B) \operatorname{Subst}\left (\int \frac{x^{1+m}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}+\frac{(a A-b B) \operatorname{Subst}\left (\int \frac{x^m}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{b B \tan ^{1+m}(c+d x)}{d (1+m)}+\frac{(a A-b B) \, _2F_1\left (1,\frac{1+m}{2};\frac{3+m}{2};-\tan ^2(c+d x)\right ) \tan ^{1+m}(c+d x)}{d (1+m)}+\frac{(A b+a B) \, _2F_1\left (1,\frac{2+m}{2};\frac{4+m}{2};-\tan ^2(c+d x)\right ) \tan ^{2+m}(c+d x)}{d (2+m)}\\ \end{align*}
Mathematica [A] time = 0.433015, size = 108, normalized size = 0.85 \[ \frac{\tan ^{m+1}(c+d x) \left (\frac{(a A-b B) \text{Hypergeometric2F1}\left (1,\frac{m+1}{2},\frac{m+3}{2},-\tan ^2(c+d x)\right )}{m+1}+\frac{(a B+A b) \tan (c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+4}{2},-\tan ^2(c+d x)\right )}{m+2}+\frac{b B}{m+1}\right )}{d} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.797, size = 0, normalized size = 0. \begin{align*} \int \left ( \tan \left ( dx+c \right ) \right ) ^{m} \left ( a+b\tan \left ( dx+c \right ) \right ) \left ( A+B\tan \left ( dx+c \right ) \right ) \, dx \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 )} \tan \left (d x + c\right )^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B b \tan \left (d x + c\right )^{2} + A a +{\left (B a + A b\right )} \tan \left (d x + c\right )\right )} \tan \left (d x + c\right )^{m}, x\right ) \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 \left (A + B \tan{\left (c + d x \right )}\right ) \left (a + b \tan{\left (c + d x \right )}\right ) \tan ^{m}{\left (c + d x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [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 )} \tan \left (d x + c\right )^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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