Optimal. Leaf size=455 \[ \frac {\, _2F_1\left (1,1+m;2+m;\frac {a+b \tan (e+f x)}{a-i b}\right ) (a+b \tan (e+f x))^{1+m}}{2 (i a+b) (c-i d)^3 f (1+m)}+\frac {\, _2F_1\left (1,1+m;2+m;\frac {a+b \tan (e+f x)}{a+i b}\right ) (a+b \tan (e+f x))^{1+m}}{2 (a+i b) (i c-d)^3 f (1+m)}+\frac {d^2 \left (2 a^2 d^2 \left (3 c^2-d^2\right )-4 a b c d \left (c^2 (3-m)-d^2 (1+m)\right )-b^2 \left (d^4 (1-m) m+2 c^2 d^2 \left (1+3 m-m^2\right )-c^4 \left (6-5 m+m^2\right )\right )\right ) \, _2F_1\left (1,1+m;2+m;-\frac {d (a+b \tan (e+f x))}{b c-a d}\right ) (a+b \tan (e+f x))^{1+m}}{2 (b c-a d)^3 \left (c^2+d^2\right )^3 f (1+m)}+\frac {d^2 (a+b \tan (e+f x))^{1+m}}{2 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}-\frac {d^2 \left (4 a c d-b \left (d^2 (1-m)+c^2 (5-m)\right )\right ) (a+b \tan (e+f x))^{1+m}}{2 (b c-a d)^2 \left (c^2+d^2\right )^2 f (c+d \tan (e+f x))} \]
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Rubi [A]
time = 1.06, antiderivative size = 455, normalized size of antiderivative = 1.00, number of steps
used = 10, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3650, 3730,
3734, 3620, 3618, 70, 3715} \begin {gather*} \frac {d^2 \left (2 a^2 d^2 \left (3 c^2-d^2\right )-4 a b c d \left (c^2 (3-m)-d^2 (m+1)\right )-\left (b^2 \left (-\left (c^4 \left (m^2-5 m+6\right )\right )+2 c^2 d^2 \left (-m^2+3 m+1\right )+d^4 (1-m) m\right )\right )\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 )}{2 f (m+1) \left (c^2+d^2\right )^3 (b c-a d)^3}-\frac {d^2 \left (4 a c d-b c^2 (5-m)-b d^2 (1-m)\right ) (a+b \tan (e+f x))^{m+1}}{2 f \left (c^2+d^2\right )^2 (b c-a d)^2 (c+d \tan (e+f x))}+\frac {d^2 (a+b \tan (e+f x))^{m+1}}{2 f \left (c^2+d^2\right ) (b c-a d) (c+d \tan (e+f x))^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)^3}+\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) (a+i b) (-d+i c)^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 70
Rule 3618
Rule 3620
Rule 3650
Rule 3715
Rule 3730
Rule 3734
Rubi steps
\begin {align*} \int \frac {(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}}{2 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}+\frac {\int \frac {(a+b \tan (e+f x))^m \left (2 c (b c-a d)+b d^2 (1-m)-2 d (b c-a d) \tan (e+f x)+b d^2 (1-m) \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^2} \, dx}{2 (b c-a d) \left (c^2+d^2\right )}\\ &=\frac {d^2 (a+b \tan (e+f x))^{1+m}}{2 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}-\frac {d^2 \left (4 a c d-b d^2 (1-m)-b c^2 (5-m)\right ) (a+b \tan (e+f x))^{1+m}}{2 (b c-a d)^2 \left (c^2+d^2\right )^2 f (c+d \tan (e+f x))}+\frac {\int \frac {(a+b \tan (e+f x))^m \left (-d^2 (2 a d-b c (3-m)) (a d-b c (1+m))-\left (2 b c^2-2 a c d+b d^2 (1-m)\right ) \left (a c d-b \left (c^2-d^2 m\right )\right )-4 c d (b c-a d)^2 \tan (e+f x)+b d^2 \left (4 a c d-b d^2 (1-m)-b c^2 (5-m)\right ) m \tan ^2(e+f x)\right )}{c+d \tan (e+f x)} \, dx}{2 (b c-a d)^2 \left (c^2+d^2\right )^2}\\ &=\frac {d^2 (a+b \tan (e+f x))^{1+m}}{2 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}-\frac {d^2 \left (4 a c d-b d^2 (1-m)-b c^2 (5-m)\right ) (a+b \tan (e+f x))^{1+m}}{2 (b c-a d)^2 \left (c^2+d^2\right )^2 f (c+d \tan (e+f x))}+\frac {\int (a+b \tan (e+f x))^m \left (2 c (b c-a d)^2 \left (c^2-3 d^2\right )-2 d (b c-a d)^2 \left (3 c^2-d^2\right ) \tan (e+f x)\right ) \, dx}{2 (b c-a d)^2 \left (c^2+d^2\right )^3}+\frac {\left (d^2 \left (2 a^2 d^2 \left (3 c^2-d^2\right )-4 a b c d \left (c^2 (3-m)-d^2 (1+m)\right )-b^2 \left (d^4 (1-m) m+2 c^2 d^2 \left (1+3 m-m^2\right )-c^4 \left (6-5 m+m^2\right )\right )\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}{2 (b c-a d)^2 \left (c^2+d^2\right )^3}\\ &=\frac {d^2 (a+b \tan (e+f x))^{1+m}}{2 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}-\frac {d^2 \left (4 a c d-b d^2 (1-m)-b c^2 (5-m)\right ) (a+b \tan (e+f x))^{1+m}}{2 (b c-a d)^2 \left (c^2+d^2\right )^2 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)^3}+\frac {\int (1-i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx}{2 (c+i d)^3}+\frac {\left (d^2 \left (2 a^2 d^2 \left (3 c^2-d^2\right )-4 a b c d \left (c^2 (3-m)-d^2 (1+m)\right )-b^2 \left (d^4 (1-m) m+2 c^2 d^2 \left (1+3 m-m^2\right )-c^4 \left (6-5 m+m^2\right )\right )\right )\right ) \text {Subst}\left (\int \frac {(a+b x)^m}{c+d x} \, dx,x,\tan (e+f x)\right )}{2 (b c-a d)^2 \left (c^2+d^2\right )^3 f}\\ &=\frac {d^2 \left (2 a^2 d^2 \left (3 c^2-d^2\right )-4 a b c d \left (c^2 (3-m)-d^2 (1+m)\right )-b^2 \left (d^4 (1-m) m+2 c^2 d^2 \left (1+3 m-m^2\right )-c^4 \left (6-5 m+m^2\right )\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}}{2 (b c-a d)^3 \left (c^2+d^2\right )^3 f (1+m)}+\frac {d^2 (a+b \tan (e+f x))^{1+m}}{2 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}-\frac {d^2 \left (4 a c d-b d^2 (1-m)-b c^2 (5-m)\right ) (a+b \tan (e+f x))^{1+m}}{2 (b c-a d)^2 \left (c^2+d^2\right )^2 f (c+d \tan (e+f x))}-\frac {\text {Subst}\left (\int \frac {(a+i b x)^m}{-1+x} \, dx,x,-i \tan (e+f x)\right )}{2 (i c-d)^3 f}+\frac {\text {Subst}\left (\int \frac {(a-i b x)^m}{-1+x} \, dx,x,i \tan (e+f x)\right )}{2 (i c+d)^3 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 (a-i b) (i c+d)^3 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)^3 f (1+m)}+\frac {d^2 \left (2 a^2 d^2 \left (3 c^2-d^2\right )-4 a b c d \left (c^2 (3-m)-d^2 (1+m)\right )-b^2 \left (d^4 (1-m) m+2 c^2 d^2 \left (1+3 m-m^2\right )-c^4 \left (6-5 m+m^2\right )\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}}{2 (b c-a d)^3 \left (c^2+d^2\right )^3 f (1+m)}+\frac {d^2 (a+b \tan (e+f x))^{1+m}}{2 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}-\frac {d^2 \left (4 a c d-b d^2 (1-m)-b c^2 (5-m)\right ) (a+b \tan (e+f x))^{1+m}}{2 (b c-a d)^2 \left (c^2+d^2\right )^2 f (c+d \tan (e+f x))}\\ \end {align*}
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Mathematica [A]
time = 6.29, size = 670, normalized size = 1.47 \begin {gather*} -\frac {d^2 (a+b \tan (e+f x))^{1+m}}{2 (-b c+a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}-\frac {-\frac {\left (d^2 \left (2 c (b c-a d)+b d^2 (1-m)\right )-c \left (-2 d^2 (b c-a d)-b c d^2 (1-m)\right )\right ) (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 {-\frac {\left (4 c^2 d^2 (b c-a d)^2+b c^2 d^2 \left (4 a c d-b d^2 (1-m)-b c^2 (5-m)\right ) m+d^2 \left (-d^2 (2 a d-b c (3-m)) (a d-b c (1+m))-\left (2 b c^2-2 a c d+b d^2 (1-m)\right ) \left (a c d-b \left (c^2-d^2 m\right )\right )\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) \left (c^2+d^2\right ) f (1+m)}+\frac {\frac {i \left (2 c (b c-a d)^2 \left (c^2-3 d^2\right )-2 i d (b c-a d)^2 \left (3 c^2-d^2\right )\right ) \, _2F_1\left (1,1+m;2+m;\frac {-i a-i b \tan (e+f x)}{-i a+b}\right ) (a+b \tan (e+f x))^{1+m}}{2 (a+i b) f (1+m)}-\frac {i \left (2 c (b c-a d)^2 \left (c^2-3 d^2\right )+2 i d (b c-a d)^2 \left (3 c^2-d^2\right )\right ) \, _2F_1\left (1,1+m;2+m;-\frac {i a+i b \tan (e+f x)}{-i a-b}\right ) (a+b \tan (e+f x))^{1+m}}{2 (a-i b) f (1+m)}}{c^2+d^2}}{(-b c+a d) \left (c^2+d^2\right )}}{2 (-b c+a d) \left (c^2+d^2\right )} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.38, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \tan \left (f x +e \right )\right )^{m}}{\left (c +d \tan \left (f x +e \right )\right )^{3}}\, dx\]
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 \frac {\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 \frac {{\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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