Integrand size = 21, antiderivative size = 434 \[ \int \cos ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {b \left (\frac {a \left (5+\frac {3 a^2}{b^2}-2 n\right ) n}{b^2}-\frac {\sqrt {-b^2} \left (3 a^4+a^2 b^2 \left (6-2 n-n^2\right )+b^4 \left (3-4 n+n^2\right )\right )}{b^6}\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{16 \left (1+\frac {a^2}{b^2}\right )^2 \left (a-\sqrt {-b^2}\right ) d (1+n)}+\frac {b \left (\frac {a \left (5+\frac {3 a^2}{b^2}-2 n\right ) n}{b^2}+\frac {\sqrt {-b^2} \left (3 a^4+a^2 b^2 \left (6-2 n-n^2\right )+b^4 \left (3-4 n+n^2\right )\right )}{b^6}\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{16 \left (1+\frac {a^2}{b^2}\right )^2 \left (a+\sqrt {-b^2}\right ) d (1+n)}+\frac {\cos ^4(c+d x) (b+a \tan (c+d x)) (a+b \tan (c+d x))^{1+n}}{4 \left (a^2+b^2\right ) d}+\frac {b \cos ^2(c+d x) (a+b \tan (c+d x))^{1+n} \left (b^2 (3-n)+a^2 (1+n)+a b \left (5+\frac {3 a^2}{b^2}-2 n\right ) \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^2 d} \]
1/16*b*hypergeom([1, 1+n],[2+n],(a+b*tan(d*x+c))/(a-(-b^2)^(1/2)))*(a*(5+3 *a^2/b^2-2*n)*n/b^2-(3*a^4+a^2*b^2*(-n^2-2*n+6)+b^4*(n^2-4*n+3))*(-b^2)^(1 /2)/b^6)*(a+b*tan(d*x+c))^(1+n)/(1+a^2/b^2)^2/d/(1+n)/(a-(-b^2)^(1/2))+1/1 6*b*hypergeom([1, 1+n],[2+n],(a+b*tan(d*x+c))/(a+(-b^2)^(1/2)))*(a*(5+3*a^ 2/b^2-2*n)*n/b^2+(3*a^4+a^2*b^2*(-n^2-2*n+6)+b^4*(n^2-4*n+3))*(-b^2)^(1/2) /b^6)*(a+b*tan(d*x+c))^(1+n)/(1+a^2/b^2)^2/d/(1+n)/(a+(-b^2)^(1/2))+1/4*co s(d*x+c)^4*(b+a*tan(d*x+c))*(a+b*tan(d*x+c))^(1+n)/(a^2+b^2)/d+1/8*b*cos(d *x+c)^2*(a+b*tan(d*x+c))^(1+n)*(b^2*(3-n)+a^2*(1+n)+a*b*(5+3*a^2/b^2-2*n)* tan(d*x+c))/(a^2+b^2)^2/d
Time = 4.71 (sec) , antiderivative size = 360, normalized size of antiderivative = 0.83 \[ \int \cos ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {(a+b \tan (c+d x))^{1+n} \left (\frac {\frac {\left (a b^2 \left (3 a^2+b^2 (5-2 n)\right ) n+\sqrt {-b^2} \left (-3 a^4-b^4 \left (3-4 n+n^2\right )+a^2 b^2 \left (-6+2 n+n^2\right )\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right )}{a-\sqrt {-b^2}}+\frac {\left (a b^2 \left (3 a^2+b^2 (5-2 n)\right ) n+\sqrt {-b^2} \left (3 a^4+b^4 \left (3-4 n+n^2\right )-a^2 b^2 \left (-6+2 n+n^2\right )\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right )}{a+\sqrt {-b^2}}}{\left (a^2+b^2\right ) (1+n)}+4 b \cos ^4(c+d x) (b+a \tan (c+d x))-\frac {2 b \cos ^2(c+d x) \left (b^3 (-3+n)-a^2 b (1+n)-a \left (3 a^2+b^2 (5-2 n)\right ) \tan (c+d x)\right )}{a^2+b^2}\right )}{16 b \left (a^2+b^2\right ) d} \]
((a + b*Tan[c + d*x])^(1 + n)*((((a*b^2*(3*a^2 + b^2*(5 - 2*n))*n + Sqrt[- b^2]*(-3*a^4 - b^4*(3 - 4*n + n^2) + a^2*b^2*(-6 + 2*n + n^2)))*Hypergeome tric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a - Sqrt[-b^2])])/(a - Sqrt [-b^2]) + ((a*b^2*(3*a^2 + b^2*(5 - 2*n))*n + Sqrt[-b^2]*(3*a^4 + b^4*(3 - 4*n + n^2) - a^2*b^2*(-6 + 2*n + n^2)))*Hypergeometric2F1[1, 1 + n, 2 + n , (a + b*Tan[c + d*x])/(a + Sqrt[-b^2])])/(a + Sqrt[-b^2]))/((a^2 + b^2)*( 1 + n)) + 4*b*Cos[c + d*x]^4*(b + a*Tan[c + d*x]) - (2*b*Cos[c + d*x]^2*(b ^3*(-3 + n) - a^2*b*(1 + n) - a*(3*a^2 + b^2*(5 - 2*n))*Tan[c + d*x]))/(a^ 2 + b^2)))/(16*b*(a^2 + b^2)*d)
Time = 0.74 (sec) , antiderivative size = 478, normalized size of antiderivative = 1.10, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3987, 27, 496, 25, 686, 25, 657, 2009}
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.
| \(\displaystyle \int \cos ^4(c+d x) (a+b \tan (c+d x))^n \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^n}{\sec (c+d x)^4}dx\) |
| \(\Big \downarrow \) 3987 |
| \(\displaystyle \frac {\int \frac {b^6 (a+b \tan (c+d x))^n}{\left (\tan ^2(c+d x) b^2+b^2\right )^3}d(b \tan (c+d x))}{b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^5 \int \frac {(a+b \tan (c+d x))^n}{\left (\tan ^2(c+d x) b^2+b^2\right )^3}d(b \tan (c+d x))}{d}\) |
| \(\Big \downarrow \) 496 |
| \(\displaystyle \frac {b^5 \left (\frac {\left (a b \tan (c+d x)+b^2\right ) (a+b \tan (c+d x))^{n+1}}{4 b^2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )^2}-\frac {\int -\frac {(a+b \tan (c+d x))^n \left (3 a^2+b (2-n) \tan (c+d x) a+b^2 (3-n)\right )}{\left (\tan ^2(c+d x) b^2+b^2\right )^2}d(b \tan (c+d x))}{4 b^2 \left (a^2+b^2\right )}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b^5 \left (\frac {\int \frac {(a+b \tan (c+d x))^n \left (3 a^2+b (2-n) \tan (c+d x) a+b^2 (3-n)\right )}{\left (\tan ^2(c+d x) b^2+b^2\right )^2}d(b \tan (c+d x))}{4 b^2 \left (a^2+b^2\right )}+\frac {\left (a b \tan (c+d x)+b^2\right ) (a+b \tan (c+d x))^{n+1}}{4 b^2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 686 |
| \(\displaystyle \frac {b^5 \left (\frac {\frac {(a+b \tan (c+d x))^{n+1} \left (a b \left (3 a^2+b^2 (5-2 n)\right ) \tan (c+d x)+b^2 \left (a^2 (n+1)+b^2 (3-n)\right )\right )}{2 b^2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )}-\frac {\int -\frac {(a+b \tan (c+d x))^n \left (3 a^4+b^2 \left (-n^2-2 n+6\right ) a^2-b \left (3 a^2+b^2 (5-2 n)\right ) n \tan (c+d x) a+b^4 \left (n^2-4 n+3\right )\right )}{\tan ^2(c+d x) b^2+b^2}d(b \tan (c+d x))}{2 b^2 \left (a^2+b^2\right )}}{4 b^2 \left (a^2+b^2\right )}+\frac {\left (a b \tan (c+d x)+b^2\right ) (a+b \tan (c+d x))^{n+1}}{4 b^2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b^5 \left (\frac {\frac {\int \frac {(a+b \tan (c+d x))^n \left (3 a^4+b^2 \left (-n^2-2 n+6\right ) a^2-b \left (3 a^2+b^2 (5-2 n)\right ) n \tan (c+d x) a+b^4 \left (n^2-4 n+3\right )\right )}{\tan ^2(c+d x) b^2+b^2}d(b \tan (c+d x))}{2 b^2 \left (a^2+b^2\right )}+\frac {\left (a b \left (3 a^2+b^2 (5-2 n)\right ) \tan (c+d x)+b^2 \left (a^2 (n+1)+b^2 (3-n)\right )\right ) (a+b \tan (c+d x))^{n+1}}{2 b^2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )}}{4 b^2 \left (a^2+b^2\right )}+\frac {\left (a b \tan (c+d x)+b^2\right ) (a+b \tan (c+d x))^{n+1}}{4 b^2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 657 |
| \(\displaystyle \frac {b^5 \left (\frac {\frac {\int \left (\frac {\left (a \left (3 a^2+b^2 (5-2 n)\right ) n b^2+\sqrt {-b^2} \left (3 a^4+b^2 \left (-n^2-2 n+6\right ) a^2+b^4 \left (n^2-4 n+3\right )\right )\right ) (a+b \tan (c+d x))^n}{2 b^2 \left (\sqrt {-b^2}-b \tan (c+d x)\right )}+\frac {\left (\sqrt {-b^2} \left (3 a^4+b^2 \left (-n^2-2 n+6\right ) a^2+b^4 \left (n^2-4 n+3\right )\right )-a b^2 \left (3 a^2+b^2 (5-2 n)\right ) n\right ) (a+b \tan (c+d x))^n}{2 b^2 \left (b \tan (c+d x)+\sqrt {-b^2}\right )}\right )d(b \tan (c+d x))}{2 b^2 \left (a^2+b^2\right )}+\frac {\left (a b \left (3 a^2+b^2 (5-2 n)\right ) \tan (c+d x)+b^2 \left (a^2 (n+1)+b^2 (3-n)\right )\right ) (a+b \tan (c+d x))^{n+1}}{2 b^2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )}}{4 b^2 \left (a^2+b^2\right )}+\frac {\left (a b \tan (c+d x)+b^2\right ) (a+b \tan (c+d x))^{n+1}}{4 b^2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^5 \left (\frac {\left (a b \tan (c+d x)+b^2\right ) (a+b \tan (c+d x))^{n+1}}{4 b^2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )^2}+\frac {\frac {\left (a b \left (3 a^2+b^2 (5-2 n)\right ) \tan (c+d x)+b^2 \left (a^2 (n+1)+b^2 (3-n)\right )\right ) (a+b \tan (c+d x))^{n+1}}{2 b^2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )}+\frac {\frac {\left (a b^2 n \left (3 a^2+b^2 (5-2 n)\right )-\sqrt {-b^2} \left (3 a^4+a^2 b^2 \left (-n^2-2 n+6\right )+b^4 \left (n^2-4 n+3\right )\right )\right ) (a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right )}{2 b^2 (n+1) \left (a-\sqrt {-b^2}\right )}+\frac {\left (a b^2 n \left (3 a^2+b^2 (5-2 n)\right )+\sqrt {-b^2} \left (3 a^4+a^2 b^2 \left (-n^2-2 n+6\right )+b^4 \left (n^2-4 n+3\right )\right )\right ) (a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right )}{2 b^2 (n+1) \left (a+\sqrt {-b^2}\right )}}{2 b^2 \left (a^2+b^2\right )}}{4 b^2 \left (a^2+b^2\right )}\right )}{d}\) |
(b^5*(((a + b*Tan[c + d*x])^(1 + n)*(b^2 + a*b*Tan[c + d*x]))/(4*b^2*(a^2 + b^2)*(b^2 + b^2*Tan[c + d*x]^2)^2) + (((a + b*Tan[c + d*x])^(1 + n)*(b^2 *(b^2*(3 - n) + a^2*(1 + n)) + a*b*(3*a^2 + b^2*(5 - 2*n))*Tan[c + d*x]))/ (2*b^2*(a^2 + b^2)*(b^2 + b^2*Tan[c + d*x]^2)) + (((a*b^2*(3*a^2 + b^2*(5 - 2*n))*n - Sqrt[-b^2]*(3*a^4 + a^2*b^2*(6 - 2*n - n^2) + b^4*(3 - 4*n + n ^2)))*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a - Sqrt[-b ^2])]*(a + b*Tan[c + d*x])^(1 + n))/(2*b^2*(a - Sqrt[-b^2])*(1 + n)) + ((a *b^2*(3*a^2 + b^2*(5 - 2*n))*n + Sqrt[-b^2]*(3*a^4 + a^2*b^2*(6 - 2*n - n^ 2) + b^4*(3 - 4*n + n^2)))*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a + Sqrt[-b^2])]*(a + b*Tan[c + d*x])^(1 + n))/(2*b^2*(a + Sqrt[ -b^2])*(1 + n)))/(2*b^2*(a^2 + b^2)))/(4*b^2*(a^2 + b^2))))/d
3.7.50.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-(a*d + b*c*x))*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1)*(b*c^2 + a*d^2))), x] + Simp[1/(2*a*(p + 1)*(b*c^2 + a*d^2)) Int[(c + d*x)^n*(a + b*x^2)^(p + 1)*Simp[b*c^2*(2*p + 3) + a*d^2*(n + 2*p + 3) + b*c*d*(n + 2 *p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[p, -1] && IntQuad raticQ[a, 0, b, c, d, n, p, x]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _), x_Symbol] :> Simp[(-(d + e*x)^(m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] + Simp[ 1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)) Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Sim p[f*(c^2*d^2*(2*p + 3) + a*c*e^2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ [p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[1/(b*f) Subst[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && NeQ[a^2 + b^2, 0] && IntegerQ[m/2]
\[\int \left (\cos ^{4}\left (d x +c \right )\right ) \left (a +b \tan \left (d x +c \right )\right )^{n}d x\]
\[ \int \cos ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \cos \left (d x + c\right )^{4} \,d x } \]
\[ \int \cos ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{n} \cos ^{4}{\left (c + d x \right )}\, dx \]
\[ \int \cos ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \cos \left (d x + c\right )^{4} \,d x } \]
\[ \int \cos ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \cos \left (d x + c\right )^{4} \,d x } \]
Timed out. \[ \int \cos ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\int {\cos \left (c+d\,x\right )}^4\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^n \,d x \]