Optimal. Leaf size=267 \[ \frac{\left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right ) \tan ^{m+1}(c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+1}{2},\frac{m+3}{2},-\tan ^2(c+d x)\right )}{d (m+1)}+\frac{\left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right ) \tan ^{m+2}(c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+4}{2},-\tan ^2(c+d x)\right )}{d (m+2)}+\frac{b \left (2 a^2 B (m+4)+3 a A b (m+3)-b^2 B (m+3)\right ) \tan ^{m+1}(c+d x)}{d (m+1) (m+3)}+\frac{b^2 (a B (m+5)+A b (m+3)) \tan ^{m+2}(c+d x)}{d (m+2) (m+3)}+\frac{b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3)} \]
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Rubi [A] time = 0.690549, antiderivative size = 267, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {3607, 3637, 3630, 3538, 3476, 364} \[ \frac{\left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right ) \tan ^{m+1}(c+d x) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\tan ^2(c+d x)\right )}{d (m+1)}+\frac{\left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right ) \tan ^{m+2}(c+d x) \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};-\tan ^2(c+d x)\right )}{d (m+2)}+\frac{b \left (2 a^2 B (m+4)+3 a A b (m+3)-b^2 B (m+3)\right ) \tan ^{m+1}(c+d x)}{d (m+1) (m+3)}+\frac{b^2 (a B (m+5)+A b (m+3)) \tan ^{m+2}(c+d x)}{d (m+2) (m+3)}+\frac{b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3)} \]
Antiderivative was successfully verified.
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Rule 3607
Rule 3637
Rule 3630
Rule 3538
Rule 3476
Rule 364
Rubi steps
\begin{align*} \int \tan ^m(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx &=\frac{b B \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^2}{d (3+m)}+\frac{\int \tan ^m(c+d x) (a+b \tan (c+d x)) \left (-a (b B (1+m)-a A (3+m))+\left (2 a A b+a^2 B-b^2 B\right ) (3+m) \tan (c+d x)+b (A b (3+m)+a B (5+m)) \tan ^2(c+d x)\right ) \, dx}{3+m}\\ &=\frac{b^2 (A b (3+m)+a B (5+m)) \tan ^{2+m}(c+d x)}{d (2+m) (3+m)}+\frac{b B \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^2}{d (3+m)}-\frac{\int \tan ^m(c+d x) \left (a^2 (2+m) (b B (1+m)-a A (3+m))-\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) (2+m) (3+m) \tan (c+d x)-b (2+m) \left (3 a A b (3+m)-b^2 B (3+m)+2 a^2 B (4+m)\right ) \tan ^2(c+d x)\right ) \, dx}{6+5 m+m^2}\\ &=\frac{b \left (3 a A b (3+m)-b^2 B (3+m)+2 a^2 B (4+m)\right ) \tan ^{1+m}(c+d x)}{d (1+m) (3+m)}+\frac{b^2 (A b (3+m)+a B (5+m)) \tan ^{2+m}(c+d x)}{d (2+m) (3+m)}+\frac{b B \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^2}{d (3+m)}-\frac{\int \tan ^m(c+d x) \left (-\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) (2+m) (3+m)-\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) (2+m) (3+m) \tan (c+d x)\right ) \, dx}{6+5 m+m^2}\\ &=\frac{b \left (3 a A b (3+m)-b^2 B (3+m)+2 a^2 B (4+m)\right ) \tan ^{1+m}(c+d x)}{d (1+m) (3+m)}+\frac{b^2 (A b (3+m)+a B (5+m)) \tan ^{2+m}(c+d x)}{d (2+m) (3+m)}+\frac{b B \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^2}{d (3+m)}+\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \int \tan ^{1+m}(c+d x) \, dx+\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \int \tan ^m(c+d x) \, dx\\ &=\frac{b \left (3 a A b (3+m)-b^2 B (3+m)+2 a^2 B (4+m)\right ) \tan ^{1+m}(c+d x)}{d (1+m) (3+m)}+\frac{b^2 (A b (3+m)+a B (5+m)) \tan ^{2+m}(c+d x)}{d (2+m) (3+m)}+\frac{b B \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^2}{d (3+m)}+\frac{\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \operatorname{Subst}\left (\int \frac{x^{1+m}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}+\frac{\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \operatorname{Subst}\left (\int \frac{x^m}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{b \left (3 a A b (3+m)-b^2 B (3+m)+2 a^2 B (4+m)\right ) \tan ^{1+m}(c+d x)}{d (1+m) (3+m)}+\frac{\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \, _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{b^2 (A b (3+m)+a B (5+m)) \tan ^{2+m}(c+d x)}{d (2+m) (3+m)}+\frac{\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \, _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)}+\frac{b B \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^2}{d (3+m)}\\ \end{align*}
Mathematica [A] time = 2.52333, size = 232, normalized size = 0.87 \[ \frac{\tan ^{m+1}(c+d x) \left ((m+2) (m+3) \left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right ) \text{Hypergeometric2F1}\left (1,\frac{m+1}{2},\frac{m+3}{2},-\tan ^2(c+d x)\right )+(m+1) (m+3) \left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right ) \tan (c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+4}{2},-\tan ^2(c+d x)\right )+b (m+2) \left (2 a^2 B (m+4)+3 a A b (m+3)-b^2 B (m+3)\right )+b^2 (m+1) \tan (c+d x) (a B (m+5)+A b (m+3))+b B (m+1) (m+2) (a+b \tan (c+d x))^2\right )}{d (m+1) (m+2) (m+3)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.416, 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 ) ^{3} \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 )}^{3} \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^{3} \tan \left (d x + c\right )^{4} + A a^{3} +{\left (3 \, B a b^{2} + A b^{3}\right )} \tan \left (d x + c\right )^{3} + 3 \,{\left (B a^{2} b + A a b^{2}\right )} \tan \left (d x + c\right )^{2} +{\left (B a^{3} + 3 \, A a^{2} 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 )^{3} \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 )}^{3} \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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