Integrand size = 21, antiderivative size = 142 \[ \int \cos ^7(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {8 a \left (2 a^2+b^2\right ) \sin (c+d x)}{35 d}-\frac {3 \cos ^5(c+d x) (b-2 a \tan (c+d x)) (a+b \tan (c+d x))^2}{35 d}+\frac {\cos ^6(c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{7 d}-\frac {2 \cos ^3(c+d x) \left (b \left (6 a^2+b^2\right )-a \left (4 a^2-b^2\right ) \tan (c+d x)\right )}{35 d} \]
8/35*a*(2*a^2+b^2)*sin(d*x+c)/d-3/35*cos(d*x+c)^5*(b-2*a*tan(d*x+c))*(a+b* tan(d*x+c))^2/d+1/7*cos(d*x+c)^6*sin(d*x+c)*(a+b*tan(d*x+c))^3/d-2/35*cos( d*x+c)^3*(b*(6*a^2+b^2)-a*(4*a^2-b^2)*tan(d*x+c))/d
Time = 3.08 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.99 \[ \int \cos ^7(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {-30 a^2 b \cos ^7(c+d x)+b^3 \cos ^5(c+d x) (-9+5 \cos (2 (c+d x)))+4 b^3 \sqrt {\cos ^2(c+d x)} \sec (c+d x)+2 a \sin (c+d x) \left (35 a^2-35 \left (a^2-b^2\right ) \sin ^2(c+d x)+21 \left (a^2-2 b^2\right ) \sin ^4(c+d x)-5 \left (a^2-3 b^2\right ) \sin ^6(c+d x)\right )}{70 d} \]
(-30*a^2*b*Cos[c + d*x]^7 + b^3*Cos[c + d*x]^5*(-9 + 5*Cos[2*(c + d*x)]) + 4*b^3*Sqrt[Cos[c + d*x]^2]*Sec[c + d*x] + 2*a*Sin[c + d*x]*(35*a^2 - 35*( a^2 - b^2)*Sin[c + d*x]^2 + 21*(a^2 - 2*b^2)*Sin[c + d*x]^4 - 5*(a^2 - 3*b ^2)*Sin[c + d*x]^6))/(70*d)
Time = 0.59 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.92, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3991, 3042, 4159, 27, 290, 2009, 4857, 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.
| \(\displaystyle \int \cos ^7(c+d x) (a+b \tan (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^3}{\sec (c+d x)^7}dx\) |
| \(\Big \downarrow \) 3991 |
| \(\displaystyle \int \cos ^7(c+d x) \left (a^3+3 b^2 \tan ^2(c+d x) a\right )dx+\int \cos ^6(c+d x) \sin (c+d x) \left (\tan ^2(c+d x) b^3+3 a^2 b\right )dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a^3+3 b^2 \tan (c+d x)^2 a}{\sec (c+d x)^7}dx+\int \frac {\sin (c+d x) \left (\tan (c+d x)^2 b^3+3 a^2 b\right )}{\sec (c+d x)^6}dx\) |
| \(\Big \downarrow \) 4159 |
| \(\displaystyle \int \frac {\sin (c+d x) \left (\tan (c+d x)^2 b^3+3 a^2 b\right )}{\sec (c+d x)^6}dx+\frac {\int a \left (1-\sin ^2(c+d x)\right )^2 \left (a^2-\left (a^2-3 b^2\right ) \sin ^2(c+d x)\right )d\sin (c+d x)}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {\sin (c+d x) \left (\tan (c+d x)^2 b^3+3 a^2 b\right )}{\sec (c+d x)^6}dx+\frac {a \int \left (1-\sin ^2(c+d x)\right )^2 \left (a^2-\left (a^2-3 b^2\right ) \sin ^2(c+d x)\right )d\sin (c+d x)}{d}\) |
| \(\Big \downarrow \) 290 |
| \(\displaystyle \int \frac {\sin (c+d x) \left (\tan (c+d x)^2 b^3+3 a^2 b\right )}{\sec (c+d x)^6}dx+\frac {a \int \left (-\left (\left (a^2-3 b^2\right ) \sin ^6(c+d x)\right )+3 \left (a^2-2 b^2\right ) \sin ^4(c+d x)-3 \left (a^2-b^2\right ) \sin ^2(c+d x)+a^2\right )d\sin (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \int \frac {\sin (c+d x) \left (\tan (c+d x)^2 b^3+3 a^2 b\right )}{\sec (c+d x)^6}dx+\frac {a \left (-\frac {1}{7} \left (a^2-3 b^2\right ) \sin ^7(c+d x)+\frac {3}{5} \left (a^2-2 b^2\right ) \sin ^5(c+d x)-\left (a^2-b^2\right ) \sin ^3(c+d x)+a^2 \sin (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 4857 |
| \(\displaystyle \frac {a \left (-\frac {1}{7} \left (a^2-3 b^2\right ) \sin ^7(c+d x)+\frac {3}{5} \left (a^2-2 b^2\right ) \sin ^5(c+d x)-\left (a^2-b^2\right ) \sin ^3(c+d x)+a^2 \sin (c+d x)\right )}{d}-\frac {\int \left (3 a^2 b \cos ^6(c+d x)+b^3 \left (1-\cos ^2(c+d x)\right ) \cos ^4(c+d x)\right )d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a \left (-\frac {1}{7} \left (a^2-3 b^2\right ) \sin ^7(c+d x)+\frac {3}{5} \left (a^2-2 b^2\right ) \sin ^5(c+d x)-\left (a^2-b^2\right ) \sin ^3(c+d x)+a^2 \sin (c+d x)\right )}{d}-\frac {\frac {3}{7} a^2 b \cos ^7(c+d x)-\frac {1}{7} b^3 \cos ^7(c+d x)+\frac {1}{5} b^3 \cos ^5(c+d x)}{d}\) |
-(((b^3*Cos[c + d*x]^5)/5 + (3*a^2*b*Cos[c + d*x]^7)/7 - (b^3*Cos[c + d*x] ^7)/7)/d) + (a*(a^2*Sin[c + d*x] - (a^2 - b^2)*Sin[c + d*x]^3 + (3*(a^2 - 2*b^2)*Sin[c + d*x]^5)/5 - ((a^2 - 3*b^2)*Sin[c + d*x]^7)/7))/d
3.6.43.3.1 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[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> I nt[ExpandIntegrand[(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d }, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]
Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Module[{k}, Int[Sec[e + f*x]^m*Sum[Binomial[n, 2*k]*a^(n - 2*k)*b^(2*k)*Tan[e + f*x]^(2*k), {k, 0, n/2}], x] + Int[Sec[e + f*x]^m*Tan [e + f*x]*Sum[Binomial[n, 2*k + 1]*a^(n - 2*k - 1)*b^(2*k + 1)*Tan[e + f*x] ^(2*k), {k, 0, (n - 1)/2}], x]] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 + b^2, 0] && IntegerQ[(m - 1)/2] && IGtQ[n, 0]
Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_ ))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2 *x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f} , x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2] && IntegerQ[p]
Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFacto rs[Cos[c*(a + b*x)], x]}, Simp[-d/(b*c) Subst[Int[SubstFor[1, Cos[c*(a + b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d], x] /; FunctionOfQ[Cos[c*(a + b* x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Sin] || EqQ[F, sin])
Time = 119.54 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.02
| method | result | size |
| derivativedivides | \(\frac {b^{3} \left (-\frac {\left (\cos ^{5}\left (d x +c \right )\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{7}-\frac {2 \left (\cos ^{5}\left (d x +c \right )\right )}{35}\right )+3 a \,b^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )}{7}+\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{35}\right )-\frac {3 a^{2} b \left (\cos ^{7}\left (d x +c \right )\right )}{7}+\frac {a^{3} \left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{7}}{d}\) | \(145\) |
| default | \(\frac {b^{3} \left (-\frac {\left (\cos ^{5}\left (d x +c \right )\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{7}-\frac {2 \left (\cos ^{5}\left (d x +c \right )\right )}{35}\right )+3 a \,b^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )}{7}+\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{35}\right )-\frac {3 a^{2} b \left (\cos ^{7}\left (d x +c \right )\right )}{7}+\frac {a^{3} \left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{7}}{d}\) | \(145\) |
| risch | \(-\frac {15 b \cos \left (d x +c \right ) a^{2}}{64 d}-\frac {3 b^{3} \cos \left (d x +c \right )}{64 d}+\frac {35 a^{3} \sin \left (d x +c \right )}{64 d}+\frac {15 a \sin \left (d x +c \right ) b^{2}}{64 d}-\frac {3 b \cos \left (7 d x +7 c \right ) a^{2}}{448 d}+\frac {b^{3} \cos \left (7 d x +7 c \right )}{448 d}+\frac {a^{3} \sin \left (7 d x +7 c \right )}{448 d}-\frac {3 a \sin \left (7 d x +7 c \right ) b^{2}}{448 d}-\frac {3 b \cos \left (5 d x +5 c \right ) a^{2}}{64 d}+\frac {b^{3} \cos \left (5 d x +5 c \right )}{320 d}+\frac {7 a^{3} \sin \left (5 d x +5 c \right )}{320 d}-\frac {9 a \sin \left (5 d x +5 c \right ) b^{2}}{320 d}-\frac {9 b \cos \left (3 d x +3 c \right ) a^{2}}{64 d}-\frac {b^{3} \cos \left (3 d x +3 c \right )}{64 d}+\frac {7 a^{3} \sin \left (3 d x +3 c \right )}{64 d}-\frac {a \sin \left (3 d x +3 c \right ) b^{2}}{64 d}\) | \(270\) |
1/d*(b^3*(-1/7*cos(d*x+c)^5*sin(d*x+c)^2-2/35*cos(d*x+c)^5)+3*a*b^2*(-1/7* sin(d*x+c)*cos(d*x+c)^6+1/35*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c ))-3/7*a^2*b*cos(d*x+c)^7+1/7*a^3*(16/5+cos(d*x+c)^6+6/5*cos(d*x+c)^4+8/5* cos(d*x+c)^2)*sin(d*x+c))
Time = 0.26 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.87 \[ \int \cos ^7(c+d x) (a+b \tan (c+d x))^3 \, dx=-\frac {7 \, b^{3} \cos \left (d x + c\right )^{5} + 5 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{7} - {\left (5 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{6} + 3 \, {\left (2 \, a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{4} + 16 \, a^{3} + 8 \, a b^{2} + 4 \, {\left (2 \, a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{35 \, d} \]
-1/35*(7*b^3*cos(d*x + c)^5 + 5*(3*a^2*b - b^3)*cos(d*x + c)^7 - (5*(a^3 - 3*a*b^2)*cos(d*x + c)^6 + 3*(2*a^3 + a*b^2)*cos(d*x + c)^4 + 16*a^3 + 8*a *b^2 + 4*(2*a^3 + a*b^2)*cos(d*x + c)^2)*sin(d*x + c))/d
\[ \int \cos ^7(c+d x) (a+b \tan (c+d x))^3 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{3} \cos ^{7}{\left (c + d x \right )}\, dx \]
Time = 0.37 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.89 \[ \int \cos ^7(c+d x) (a+b \tan (c+d x))^3 \, dx=-\frac {15 \, a^{2} b \cos \left (d x + c\right )^{7} + {\left (5 \, \sin \left (d x + c\right )^{7} - 21 \, \sin \left (d x + c\right )^{5} + 35 \, \sin \left (d x + c\right )^{3} - 35 \, \sin \left (d x + c\right )\right )} a^{3} - {\left (15 \, \sin \left (d x + c\right )^{7} - 42 \, \sin \left (d x + c\right )^{5} + 35 \, \sin \left (d x + c\right )^{3}\right )} a b^{2} - {\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} b^{3}}{35 \, d} \]
-1/35*(15*a^2*b*cos(d*x + c)^7 + (5*sin(d*x + c)^7 - 21*sin(d*x + c)^5 + 3 5*sin(d*x + c)^3 - 35*sin(d*x + c))*a^3 - (15*sin(d*x + c)^7 - 42*sin(d*x + c)^5 + 35*sin(d*x + c)^3)*a*b^2 - (5*cos(d*x + c)^7 - 7*cos(d*x + c)^5)* b^3)/d
Leaf count of result is larger than twice the leaf count of optimal. 101962 vs. \(2 (136) = 272\).
Time = 84.49 (sec) , antiderivative size = 101962, normalized size of antiderivative = 718.04 \[ \int \cos ^7(c+d x) (a+b \tan (c+d x))^3 \, dx=\text {Too large to display} \]
1/17920*(945*pi*a^2*b*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^2*t an(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^2 + 2*tan(1/2*c) - 1)*sgn(tan(1/2* d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + tan(1 /2*c)^2 + 2*tan(1/2*d*x) - 1)*tan(1/2*d*x)^14*tan(1/2*c)^14 + 210*pi*b^3*s gn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^2*tan(1/2*c) + tan(1/2*d*x )^2 - tan(1/2*c)^2 + 2*tan(1/2*c) - 1)*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2 *tan(1/2*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + tan(1/2*c)^2 + 2*tan(1/2*d*x ) - 1)*tan(1/2*d*x)^14*tan(1/2*c)^14 + 945*pi*a^2*b*sgn(tan(1/2*d*x)^2*tan (1/2*c)^2 - 2*tan(1/2*d*x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^2 - 2*tan(1/2*c) - 1)*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)*tan(1/2 *c)^2 - tan(1/2*d*x)^2 + tan(1/2*c)^2 - 2*tan(1/2*d*x) - 1)*tan(1/2*d*x)^1 4*tan(1/2*c)^14 + 210*pi*b^3*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d *x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^2 - 2*tan(1/2*c) - 1)*sgn(t an(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + tan(1/2*c)^2 - 2*tan(1/2*d*x) - 1)*tan(1/2*d*x)^14*tan(1/2*c)^14 - 2205* pi*a^2*b*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)*tan(1/2*c)^2 - t an(1/2*d*x)^2 + tan(1/2*c)^2 + 2*tan(1/2*d*x) - 1)*tan(1/2*d*x)^14*tan(1/2 *c)^14 + 1995*pi*b^3*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)*tan( 1/2*c)^2 - tan(1/2*d*x)^2 + tan(1/2*c)^2 + 2*tan(1/2*d*x) - 1)*tan(1/2*d*x )^14*tan(1/2*c)^14 + 2205*pi*a^2*b*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*...
Time = 4.49 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.51 \[ \int \cos ^7(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {16\,a^3\,\sin \left (c+d\,x\right )}{35\,d}-\frac {b^3\,{\cos \left (c+d\,x\right )}^5}{5\,d}+\frac {b^3\,{\cos \left (c+d\,x\right )}^7}{7\,d}-\frac {3\,a^2\,b\,{\cos \left (c+d\,x\right )}^7}{7\,d}+\frac {8\,a^3\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{35\,d}+\frac {6\,a^3\,{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{35\,d}+\frac {a^3\,{\cos \left (c+d\,x\right )}^6\,\sin \left (c+d\,x\right )}{7\,d}+\frac {8\,a\,b^2\,\sin \left (c+d\,x\right )}{35\,d}+\frac {4\,a\,b^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{35\,d}+\frac {3\,a\,b^2\,{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{35\,d}-\frac {3\,a\,b^2\,{\cos \left (c+d\,x\right )}^6\,\sin \left (c+d\,x\right )}{7\,d} \]
(16*a^3*sin(c + d*x))/(35*d) - (b^3*cos(c + d*x)^5)/(5*d) + (b^3*cos(c + d *x)^7)/(7*d) - (3*a^2*b*cos(c + d*x)^7)/(7*d) + (8*a^3*cos(c + d*x)^2*sin( c + d*x))/(35*d) + (6*a^3*cos(c + d*x)^4*sin(c + d*x))/(35*d) + (a^3*cos(c + d*x)^6*sin(c + d*x))/(7*d) + (8*a*b^2*sin(c + d*x))/(35*d) + (4*a*b^2*c os(c + d*x)^2*sin(c + d*x))/(35*d) + (3*a*b^2*cos(c + d*x)^4*sin(c + d*x)) /(35*d) - (3*a*b^2*cos(c + d*x)^6*sin(c + d*x))/(7*d)