Integrand size = 19, antiderivative size = 78 \[ \int \cos ^3(a+b x) (d \tan (a+b x))^n \, dx=\frac {\cos ^3(a+b x) \cos ^2(a+b x)^{\frac {1}{2} (-2+n)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-2+n),\frac {1+n}{2},\frac {3+n}{2},\sin ^2(a+b x)\right ) (d \tan (a+b x))^{1+n}}{b d (1+n)} \] Output:
cos(b*x+a)^3*(cos(b*x+a)^2)^(-1+1/2*n)*hypergeom([-1+1/2*n, 1/2+1/2*n],[3/ 2+1/2*n],sin(b*x+a)^2)*(d*tan(b*x+a))^(1+n)/b/d/(1+n)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 4.79 (sec) , antiderivative size = 1313, normalized size of antiderivative = 16.83 \[ \int \cos ^3(a+b x) (d \tan (a+b x))^n \, dx =\text {Too large to display} \] Input:
Integrate[Cos[a + b*x]^3*(d*Tan[a + b*x])^n,x]
Output:
(4*(3 + n)*(AppellF1[(1 + n)/2, n, 1, (3 + n)/2, Tan[(a + b*x)/2]^2, -Tan[ (a + b*x)/2]^2] - 6*AppellF1[(1 + n)/2, n, 2, (3 + n)/2, Tan[(a + b*x)/2]^ 2, -Tan[(a + b*x)/2]^2] + 12*AppellF1[(1 + n)/2, n, 3, (3 + n)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - 8*AppellF1[(1 + n)/2, n, 4, (3 + n)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2])*Cos[(a + b*x)/2]^3*Cos[a + b*x]^ 3*Sin[(a + b*x)/2]*(d*Tan[a + b*x])^n)/(b*(1 + n)*((3 + n)*AppellF1[(1 + n )/2, n, 1, (3 + n)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2]*(1 + Cos[a + b*x]) - 2*(AppellF1[(3 + n)/2, n, 2, (5 + n)/2, Tan[(a + b*x)/2]^2, -Tan [(a + b*x)/2]^2] - 12*AppellF1[(3 + n)/2, n, 3, (5 + n)/2, Tan[(a + b*x)/2 ]^2, -Tan[(a + b*x)/2]^2] + 36*AppellF1[(3 + n)/2, n, 4, (5 + n)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - 32*AppellF1[(3 + n)/2, n, 5, (5 + n)/ 2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - n*AppellF1[(3 + n)/2, 1 + n, 1, (5 + n)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] + 6*n*AppellF1[(3 + n)/2, 1 + n, 2, (5 + n)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - 12 *n*AppellF1[(3 + n)/2, 1 + n, 3, (5 + n)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] + 8*n*AppellF1[(3 + n)/2, 1 + n, 4, (5 + n)/2, Tan[(a + b*x)/2] ^2, -Tan[(a + b*x)/2]^2] + 18*AppellF1[(1 + n)/2, n, 2, (3 + n)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2]*Cos[(a + b*x)/2]^2 + 6*n*AppellF1[(1 + n )/2, n, 2, (3 + n)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2]*Cos[(a + b* x)/2]^2 + 8*(3 + n)*AppellF1[(1 + n)/2, n, 4, (3 + n)/2, Tan[(a + b*x)/...
Time = 0.28 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3042, 3097}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \cos ^3(a+b x) (d \tan (a+b x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(d \tan (a+b x))^n}{\sec (a+b x)^3}dx\) |
\(\Big \downarrow \) 3097 |
\(\displaystyle \frac {\cos ^3(a+b x) \cos ^2(a+b x)^{\frac {n-2}{2}} (d \tan (a+b x))^{n+1} \operatorname {Hypergeometric2F1}\left (\frac {n-2}{2},\frac {n+1}{2},\frac {n+3}{2},\sin ^2(a+b x)\right )}{b d (n+1)}\) |
Input:
Int[Cos[a + b*x]^3*(d*Tan[a + b*x])^n,x]
Output:
(Cos[a + b*x]^3*(Cos[a + b*x]^2)^((-2 + n)/2)*Hypergeometric2F1[(-2 + n)/2 , (1 + n)/2, (3 + n)/2, Sin[a + b*x]^2]*(d*Tan[a + b*x])^(1 + n))/(b*d*(1 + n))
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[(a*Sec[e + f*x])^m*(b*Tan[e + f*x])^(n + 1)*((Cos[e + f*x]^2)^((m + n + 1)/2)/(b*f*(n + 1)))*Hypergeometric2F1[(n + 1)/2, (m + n + 1)/2, (n + 3)/2, Sin[e + f*x]^2], x] /; FreeQ[{a, b, e, f, m, n}, x] && !IntegerQ[(n - 1)/2] && !IntegerQ[m/2]
\[\int \cos \left (b x +a \right )^{3} \left (d \tan \left (b x +a \right )\right )^{n}d x\]
Input:
int(cos(b*x+a)^3*(d*tan(b*x+a))^n,x)
Output:
int(cos(b*x+a)^3*(d*tan(b*x+a))^n,x)
\[ \int \cos ^3(a+b x) (d \tan (a+b x))^n \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{n} \cos \left (b x + a\right )^{3} \,d x } \] Input:
integrate(cos(b*x+a)^3*(d*tan(b*x+a))^n,x, algorithm="fricas")
Output:
integral((d*tan(b*x + a))^n*cos(b*x + a)^3, x)
Timed out. \[ \int \cos ^3(a+b x) (d \tan (a+b x))^n \, dx=\text {Timed out} \] Input:
integrate(cos(b*x+a)**3*(d*tan(b*x+a))**n,x)
Output:
Timed out
\[ \int \cos ^3(a+b x) (d \tan (a+b x))^n \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{n} \cos \left (b x + a\right )^{3} \,d x } \] Input:
integrate(cos(b*x+a)^3*(d*tan(b*x+a))^n,x, algorithm="maxima")
Output:
integrate((d*tan(b*x + a))^n*cos(b*x + a)^3, x)
\[ \int \cos ^3(a+b x) (d \tan (a+b x))^n \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{n} \cos \left (b x + a\right )^{3} \,d x } \] Input:
integrate(cos(b*x+a)^3*(d*tan(b*x+a))^n,x, algorithm="giac")
Output:
integrate((d*tan(b*x + a))^n*cos(b*x + a)^3, x)
Timed out. \[ \int \cos ^3(a+b x) (d \tan (a+b x))^n \, dx=\int {\cos \left (a+b\,x\right )}^3\,{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^n \,d x \] Input:
int(cos(a + b*x)^3*(d*tan(a + b*x))^n,x)
Output:
int(cos(a + b*x)^3*(d*tan(a + b*x))^n, x)
\[ \int \cos ^3(a+b x) (d \tan (a+b x))^n \, dx=d^{n} \left (\int \tan \left (b x +a \right )^{n} \cos \left (b x +a \right )^{3}d x \right ) \] Input:
int(cos(b*x+a)^3*(d*tan(b*x+a))^n,x)
Output:
d**n*int(tan(a + b*x)**n*cos(a + b*x)**3,x)