Optimal. Leaf size=183 \[ -\frac {a \left (3 b^2-a^2 (1-m)\right ) (d \cos (e+f x))^m \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {3}{2};-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{m/2} \tan (e+f x)}{f (1-m)}+\frac {b (d \cos (e+f x))^m (a+b \tan (e+f x))^2}{f (2-m)}+\frac {b (d \cos (e+f x))^m \left (2 \left (b^2-a^2 (3-m)\right ) (1-m)+a b (4-m) m \tan (e+f x)\right )}{f m \left (2-3 m+m^2\right )} \]
[Out]
________________________________________________________________________________________
Rubi [A]
time = 0.20, antiderivative size = 175, normalized size of antiderivative = 0.96, number of steps
used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3596, 3593,
757, 794, 251} \begin {gather*} \frac {a \left (a^2-\frac {3 b^2}{1-m}\right ) \tan (e+f x) \sec ^2(e+f x)^{m/2} (d \cos (e+f x))^m \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {3}{2};-\tan ^2(e+f x)\right )}{f}+\frac {b (d \cos (e+f x))^m \left (2 (1-m) \left (b^2-a^2 (3-m)\right )+a b (4-m) m \tan (e+f x)\right )}{f m \left (m^2-3 m+2\right )}+\frac {b (a+b \tan (e+f x))^2 (d \cos (e+f x))^m}{f (2-m)} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
Rule 251
Rule 757
Rule 794
Rule 3593
Rule 3596
Rubi steps
\begin {align*} \int (d \cos (e+f x))^m (a+b \tan (e+f x))^3 \, dx &=\left ((d \cos (e+f x))^m (d \sec (e+f x))^m\right ) \int (d \sec (e+f x))^{-m} (a+b \tan (e+f x))^3 \, dx\\ &=\frac {\left ((d \cos (e+f x))^m \sec ^2(e+f x)^{m/2}\right ) \text {Subst}\left (\int (a+x)^3 \left (1+\frac {x^2}{b^2}\right )^{-1-\frac {m}{2}} \, dx,x,b \tan (e+f x)\right )}{b f}\\ &=\frac {b (d \cos (e+f x))^m (a+b \tan (e+f x))^2}{f (2-m)}+\frac {\left (b (d \cos (e+f x))^m \sec ^2(e+f x)^{m/2}\right ) \text {Subst}\left (\int (a+x) \left (-2+\frac {a^2 (2-m)}{b^2}+\frac {a (4-m) x}{b^2}\right ) \left (1+\frac {x^2}{b^2}\right )^{-1-\frac {m}{2}} \, dx,x,b \tan (e+f x)\right )}{f (2-m)}\\ &=\frac {b (d \cos (e+f x))^m (a+b \tan (e+f x))^2}{f (2-m)}+\frac {b (d \cos (e+f x))^m \left (2 \left (b^2-a^2 (3-m)\right ) (1-m)+a b (4-m) m \tan (e+f x)\right )}{f m \left (2-3 m+m^2\right )}-\frac {\left (a \left (3 b^2-a^2 (1-m)\right ) (d \cos (e+f x))^m \sec ^2(e+f x)^{m/2}\right ) \text {Subst}\left (\int \left (1+\frac {x^2}{b^2}\right )^{-1-\frac {m}{2}} \, dx,x,b \tan (e+f x)\right )}{b f (1-m)}\\ &=-\frac {a \left (3 b^2-a^2 (1-m)\right ) (d \cos (e+f x))^m \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {3}{2};-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{m/2} \tan (e+f x)}{f (1-m)}+\frac {b (d \cos (e+f x))^m (a+b \tan (e+f x))^2}{f (2-m)}+\frac {b (d \cos (e+f x))^m \left (2 \left (b^2-a^2 (3-m)\right ) (1-m)+a b (4-m) m \tan (e+f x)\right )}{f m \left (2-3 m+m^2\right )}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A]
time = 3.37, size = 212, normalized size = 1.16 \begin {gather*} \frac {\cos (e+f x) (d \cos (e+f x))^m \left (-\frac {b^3}{-2+m}+\frac {b \left (-3 a^2+b^2\right ) \cos ^2(e+f x)}{m}-\frac {a \left (a^2-3 b^2\right ) \cos ^3(e+f x) \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};\cos ^2(e+f x)\right ) \sin (e+f x)}{(1+m) \sqrt {\sin ^2(e+f x)}}-\frac {3 a b^2 \, _2F_1\left (\frac {1}{2},\frac {1}{2} (-1+m);\frac {1+m}{2};\cos ^2(e+f x)\right ) \sin (2 (e+f x))}{2 (-1+m) \sqrt {\sin ^2(e+f x)}}\right ) (a+b \tan (e+f x))^3}{f (a \cos (e+f x)+b \sin (e+f x))^3} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [F]
time = 0.84, size = 0, normalized size = 0.00 \[\int \left (d \cos \left (f x +e \right )\right )^{m} \left (a +b \tan \left (f x +e \right )\right )^{3}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
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.
[In]
[Out]
________________________________________________________________________________________
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.
[In]
[Out]
________________________________________________________________________________________
Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (d \cos {\left (e + f x \right )}\right )^{m} \left (a + b \tan {\left (e + f x \right )}\right )^{3}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
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.
[In]
[Out]
________________________________________________________________________________________
Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (d\,\cos \left (e+f\,x\right )\right )}^m\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^3 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________