Optimal. Leaf size=301 \[ \frac {\, _2F_1\left (1,1+m;2+m;\frac {a+b \tan (e+f x)}{a-i b}\right ) (a+b \tan (e+f x))^{1+m}}{2 (i a+b) (c-i d)^2 f (1+m)}-\frac {\, _2F_1\left (1,1+m;2+m;\frac {a+b \tan (e+f x)}{a+i b}\right ) (a+b \tan (e+f x))^{1+m}}{2 (i a-b) (c+i d)^2 f (1+m)}-\frac {d^2 \left (2 a c d-b \left (c^2 (2-m)-d^2 m\right )\right ) \, _2F_1\left (1,1+m;2+m;-\frac {d (a+b \tan (e+f x))}{b c-a d}\right ) (a+b \tan (e+f x))^{1+m}}{(b c-a d)^2 \left (c^2+d^2\right )^2 f (1+m)}+\frac {d^2 (a+b \tan (e+f x))^{1+m}}{(b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))} \]
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Rubi [A]
time = 0.56, antiderivative size = 299, normalized size of antiderivative = 0.99, number of steps
used = 9, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3650, 3734,
3620, 3618, 70, 3715} \begin {gather*} -\frac {d^2 \left (2 a c d-b c^2 (2-m)+b d^2 m\right ) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac {d (a+b \tan (e+f x))}{b c-a d}\right )}{f (m+1) \left (c^2+d^2\right )^2 (b c-a d)^2}+\frac {d^2 (a+b \tan (e+f x))^{m+1}}{f \left (c^2+d^2\right ) (b c-a d) (c+d \tan (e+f x))}+\frac {(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) (c-i d)^2}-\frac {(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) (c+i d)^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 70
Rule 3618
Rule 3620
Rule 3650
Rule 3715
Rule 3734
Rubi steps
\begin {align*} \int \frac {(a+b \tan (e+f x))^m}{(c+d \tan (e+f x))^2} \, dx &=\frac {d^2 (a+b \tan (e+f x))^{1+m}}{(b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))}+\frac {\int \frac {(a+b \tan (e+f x))^m \left (-a c d+b \left (c^2-d^2 m\right )-d (b c-a d) \tan (e+f x)-b d^2 m \tan ^2(e+f x)\right )}{c+d \tan (e+f x)} \, dx}{(b c-a d) \left (c^2+d^2\right )}\\ &=\frac {d^2 (a+b \tan (e+f x))^{1+m}}{(b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))}+\frac {\int (a+b \tan (e+f x))^m \left ((b c-a d) \left (c^2-d^2\right )-2 c d (b c-a d) \tan (e+f x)\right ) \, dx}{(b c-a d) \left (c^2+d^2\right )^2}-\frac {\left (d^2 \left (2 a c d-b c^2 (2-m)+b d^2 m\right )\right ) \int \frac {(a+b \tan (e+f x))^m \left (1+\tan ^2(e+f x)\right )}{c+d \tan (e+f x)} \, dx}{(b c-a d) \left (c^2+d^2\right )^2}\\ &=\frac {d^2 (a+b \tan (e+f x))^{1+m}}{(b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))}+\frac {\int (1+i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx}{2 (c-i d)^2}+\frac {\int (1-i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx}{2 (c+i d)^2}-\frac {\left (d^2 \left (2 a c d-b c^2 (2-m)+b d^2 m\right )\right ) \text {Subst}\left (\int \frac {(a+b x)^m}{c+d x} \, dx,x,\tan (e+f x)\right )}{(b c-a d) \left (c^2+d^2\right )^2 f}\\ &=-\frac {d^2 \left (2 a c d-b c^2 (2-m)+b d^2 m\right ) \, _2F_1\left (1,1+m;2+m;-\frac {d (a+b \tan (e+f x))}{b c-a d}\right ) (a+b \tan (e+f x))^{1+m}}{(b c-a d)^2 \left (c^2+d^2\right )^2 f (1+m)}+\frac {d^2 (a+b \tan (e+f x))^{1+m}}{(b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))}+\frac {i \text {Subst}\left (\int \frac {(a-i b x)^m}{-1+x} \, dx,x,i \tan (e+f x)\right )}{2 (c-i d)^2 f}-\frac {i \text {Subst}\left (\int \frac {(a+i b x)^m}{-1+x} \, dx,x,-i \tan (e+f x)\right )}{2 (c+i d)^2 f}\\ &=\frac {\, _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 (i a+b) (c-i d)^2 f (1+m)}-\frac {\, _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 (i a-b) (c+i d)^2 f (1+m)}-\frac {d^2 \left (2 a c d-b c^2 (2-m)+b d^2 m\right ) \, _2F_1\left (1,1+m;2+m;-\frac {d (a+b \tan (e+f x))}{b c-a d}\right ) (a+b \tan (e+f x))^{1+m}}{(b c-a d)^2 \left (c^2+d^2\right )^2 f (1+m)}+\frac {d^2 (a+b \tan (e+f x))^{1+m}}{(b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))}\\ \end {align*}
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Mathematica [A]
time = 4.33, size = 266, normalized size = 0.88 \begin {gather*} \frac {(a+b \tan (e+f x))^{1+m} \left (-\frac {i \left (\frac {(c+i d)^2 (-b c+a d) \, _2F_1\left (1,1+m;2+m;\frac {a+b \tan (e+f x)}{a-i b}\right )}{a-i b}+\frac {(c-i d)^2 (b c-a d) \, _2F_1\left (1,1+m;2+m;\frac {a+b \tan (e+f x)}{a+i b}\right )}{a+i b}\right )}{\left (c^2+d^2\right ) (1+m)}-\frac {2 d^2 \left (2 a c d+b c^2 (-2+m)+b d^2 m\right ) \, _2F_1\left (1,1+m;2+m;\frac {d (a+b \tan (e+f x))}{-b c+a d}\right )}{(-b c+a d) \left (c^2+d^2\right ) (1+m)}-\frac {2 d^2}{c+d \tan (e+f x)}\right )}{2 (-b c+a d) \left (c^2+d^2\right ) f} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.27, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \tan \left (f x +e \right )\right )^{m}}{\left (c +d \tan \left (f x +e \right )\right )^{2}}\, dx\]
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 \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{m}}{\left (c + d \tan {\left (e + f x \right )}\right )^{2}}\, 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 \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^m}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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