Optimal. Leaf size=214 \[ \frac {d^2 (3 b c-a d) (a+b \tan (e+f x))^{1+m}}{b^2 f (1+m)}+\frac {(i c+d)^3 \, _2F_1\left (1,1+m;2+m;\frac {a+b \tan (e+f x)}{a-i b}\right ) (a+b \tan (e+f x))^{1+m}}{2 (a-i b) f (1+m)}-\frac {(i c-d)^3 \, _2F_1\left (1,1+m;2+m;\frac {a+b \tan (e+f x)}{a+i b}\right ) (a+b \tan (e+f x))^{1+m}}{2 (a+i b) f (1+m)}+\frac {d^3 (a+b \tan (e+f x))^{2+m}}{b^2 f (2+m)} \]
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Rubi [A]
time = 0.33, antiderivative size = 234, normalized size of antiderivative = 1.09, number of steps
used = 7, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3647, 3711,
3620, 3618, 70} \begin {gather*} -\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)} \end {gather*}
Antiderivative was successfully verified.
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Rule 70
Rule 3618
Rule 3620
Rule 3647
Rule 3711
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 \text {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 \text {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*}
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Mathematica [A]
time = 2.07, size = 189, normalized size = 0.88 \begin {gather*} \frac {(a+b \tan (e+f x))^{1+m} \left (\frac {2 d^2 (-a d+b c (5+2 m))}{b (1+m)}-\frac {i b (c-i d)^3 (2+m) \, _2F_1\left (1,1+m;2+m;\frac {a+b \tan (e+f x)}{a-i b}\right )}{(a-i b) (1+m)}+\frac {i b (c+i d)^3 (2+m) \, _2F_1\left (1,1+m;2+m;\frac {a+b \tan (e+f x)}{a+i b}\right )}{(a+i b) (1+m)}+2 d^2 (c+d \tan (e+f x))\right )}{2 b f (2+m)} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.22, size = 0, normalized size = 0.00 \[\int \left (a +b \tan \left (f x +e \right )\right )^{m} \left (c +d \tan \left (f x +e \right )\right )^{3}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^m\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^3 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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