Optimal. Leaf size=214 \[ \frac{d^2 (3 b c-a d) (a+b \tan (e+f x))^{m+1}}{b^2 f (m+1)}+\frac{d^3 (a+b \tan (e+f x))^{m+2}}{b^2 f (m+2)}+\frac{(d+i c)^3 (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (a-i b)}-\frac{(-d+i c)^3 (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (a+i b)} \]
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Rubi [A] time = 0.474896, antiderivative size = 234, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3566, 3630, 3539, 3537, 68} \[ -\frac{d^2 (a d-b c (2 m+5)) (a+b \tan (e+f x))^{m+1}}{b^2 f (m+1) (m+2)}+\frac{d^2 (c+d \tan (e+f x)) (a+b \tan (e+f x))^{m+1}}{b f (m+2)}+\frac{(c-i d)^3 (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a)}-\frac{(-d+i c)^3 (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (a+i b)} \]
Antiderivative was successfully verified.
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Rule 3566
Rule 3630
Rule 3539
Rule 3537
Rule 68
Rubi steps
\begin{align*} \int (a+b \tan (e+f x))^m (c+d \tan (e+f x))^3 \, dx &=\frac{d^2 (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))}{b f (2+m)}+\frac{\int (a+b \tan (e+f x))^m \left (b c^3 (2+m)-d^2 (a d+b c (1+m))+b d \left (3 c^2-d^2\right ) (2+m) \tan (e+f x)-d^2 (a d-b c (5+2 m)) \tan ^2(e+f x)\right ) \, dx}{b (2+m)}\\ &=-\frac{d^2 (a d-b c (5+2 m)) (a+b \tan (e+f x))^{1+m}}{b^2 f (1+m) (2+m)}+\frac{d^2 (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))}{b f (2+m)}+\frac{\int (a+b \tan (e+f x))^m \left (b c \left (c^2-3 d^2\right ) (2+m)+b d \left (3 c^2-d^2\right ) (2+m) \tan (e+f x)\right ) \, dx}{b (2+m)}\\ &=-\frac{d^2 (a d-b c (5+2 m)) (a+b \tan (e+f x))^{1+m}}{b^2 f (1+m) (2+m)}+\frac{d^2 (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))}{b f (2+m)}+\frac{1}{2} (c-i d)^3 \int (1+i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx+\frac{1}{2} (c+i d)^3 \int (1-i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx\\ &=-\frac{d^2 (a d-b c (5+2 m)) (a+b \tan (e+f x))^{1+m}}{b^2 f (1+m) (2+m)}+\frac{d^2 (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))}{b f (2+m)}+\frac{(i c-d)^3 \operatorname{Subst}\left (\int \frac{(a+i b x)^m}{-1+x} \, dx,x,-i \tan (e+f x)\right )}{2 f}-\frac{(i c+d)^3 \operatorname{Subst}\left (\int \frac{(a-i b x)^m}{-1+x} \, dx,x,i \tan (e+f x)\right )}{2 f}\\ &=-\frac{d^2 (a d-b c (5+2 m)) (a+b \tan (e+f x))^{1+m}}{b^2 f (1+m) (2+m)}+\frac{(i c+d)^3 \, _2F_1\left (1,1+m;2+m;\frac{a+b \tan (e+f x)}{a-i b}\right ) (a+b \tan (e+f x))^{1+m}}{2 (a-i b) f (1+m)}-\frac{(i c-d)^3 \, _2F_1\left (1,1+m;2+m;\frac{a+b \tan (e+f x)}{a+i b}\right ) (a+b \tan (e+f x))^{1+m}}{2 (a+i b) f (1+m)}+\frac{d^2 (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))}{b f (2+m)}\\ \end{align*}
Mathematica [A] time = 1.90874, size = 189, normalized size = 0.88 \[ \frac{(a+b \tan (e+f x))^{m+1} \left (\frac{2 d^2 (b c (2 m+5)-a d)}{b (m+1)}-\frac{i b (m+2) (c-i d)^3 \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )}{(m+1) (a-i b)}+\frac{i b (m+2) (c+i d)^3 \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )}{(m+1) (a+i b)}+2 d^2 (c+d \tan (e+f x))\right )}{2 b f (m+2)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.327, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\tan \left ( fx+e \right ) \right ) ^{m} \left ( c+d\tan \left ( fx+e \right ) \right ) ^{3}\, 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 (d \tan \left (f x + e\right ) + c\right )}^{3}{\left (b \tan \left (f x + e\right ) + a\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 (d^{3} \tan \left (f x + e\right )^{3} + 3 \, c d^{2} \tan \left (f x + e\right )^{2} + 3 \, c^{2} d \tan \left (f x + e\right ) + c^{3}\right )}{\left (b \tan \left (f x + e\right ) + a\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 (e + f x \right )}\right )^{m} \left (c + d \tan{\left (e + f x \right )}\right )^{3}\, 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 (d \tan \left (f x + e\right ) + c\right )}^{3}{\left (b \tan \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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