Integrand size = 21, antiderivative size = 95 \[ \int \cos ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {1}{8} \left (3 a^2+b^2\right ) x-\frac {a b \cos ^2(c+d x)}{2 d}+\frac {\left (3 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {\cos ^3(c+d x) \sin (c+d x) (a+b \tan (c+d x))^2}{4 d} \] Output:
1/8*(3*a^2+b^2)*x-1/2*a*b*cos(d*x+c)^2/d+1/8*(3*a^2-b^2)*cos(d*x+c)*sin(d* x+c)/d+1/4*cos(d*x+c)^3*sin(d*x+c)*(a+b*tan(d*x+c))^2/d
Leaf count is larger than twice the leaf count of optimal. \(197\) vs. \(2(95)=190\).
Time = 2.14 (sec) , antiderivative size = 197, normalized size of antiderivative = 2.07 \[ \int \cos ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {\left (3 a^2+b^2\right ) \left (2 a b \left (2 a^2+b^2\right )-2 a b \left (a^2+b^2\right ) \cos (2 (c+d x))+\frac {b \left (a^2+b^2\right )^2 \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )}{\sqrt {-b^2}}-\frac {b \left (a^2+b^2\right )^2 \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )}{\sqrt {-b^2}}+\left (a^4-b^4\right ) \sin (2 (c+d x))\right )+4 \left (a^2+b^2\right ) \cos ^4(c+d x) (b+a \tan (c+d x)) (a+b \tan (c+d x))^3}{16 \left (a^2+b^2\right )^2 d} \] Input:
Integrate[Cos[c + d*x]^4*(a + b*Tan[c + d*x])^2,x]
Output:
((3*a^2 + b^2)*(2*a*b*(2*a^2 + b^2) - 2*a*b*(a^2 + b^2)*Cos[2*(c + d*x)] + (b*(a^2 + b^2)^2*Log[Sqrt[-b^2] - b*Tan[c + d*x]])/Sqrt[-b^2] - (b*(a^2 + b^2)^2*Log[Sqrt[-b^2] + b*Tan[c + d*x]])/Sqrt[-b^2] + (a^4 - b^4)*Sin[2*( c + d*x)]) + 4*(a^2 + b^2)*Cos[c + d*x]^4*(b + a*Tan[c + d*x])*(a + b*Tan[ c + d*x])^3)/(16*(a^2 + b^2)^2*d)
Time = 0.30 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.46, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3987, 27, 495, 454, 216}
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.
\(\displaystyle \int \cos ^4(c+d x) (a+b \tan (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \tan (c+d x))^2}{\sec (c+d x)^4}dx\) |
\(\Big \downarrow \) 3987 |
\(\displaystyle \frac {\int \frac {b^6 (a+b \tan (c+d x))^2}{\left (\tan ^2(c+d x) b^2+b^2\right )^3}d(b \tan (c+d x))}{b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b^5 \int \frac {(a+b \tan (c+d x))^2}{\left (\tan ^2(c+d x) b^2+b^2\right )^3}d(b \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 495 |
\(\displaystyle \frac {b^5 \left (\frac {\int \frac {3 a^2+2 b \tan (c+d x) a+b^2}{\left (\tan ^2(c+d x) b^2+b^2\right )^2}d(b \tan (c+d x))}{4 b^2}-\frac {(a+b \tan (c+d x)) \left (b^2-a b \tan (c+d x)\right )}{4 b^2 \left (b^2 \tan ^2(c+d x)+b^2\right )^2}\right )}{d}\) |
\(\Big \downarrow \) 454 |
\(\displaystyle \frac {b^5 \left (\frac {\frac {1}{2} \left (\frac {3 a^2}{b^2}+1\right ) \int \frac {1}{\tan ^2(c+d x) b^2+b^2}d(b \tan (c+d x))-\frac {2 a b^2-b \left (3 a^2+b^2\right ) \tan (c+d x)}{2 b^2 \left (b^2 \tan ^2(c+d x)+b^2\right )}}{4 b^2}-\frac {(a+b \tan (c+d x)) \left (b^2-a b \tan (c+d x)\right )}{4 b^2 \left (b^2 \tan ^2(c+d x)+b^2\right )^2}\right )}{d}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {b^5 \left (\frac {\frac {\left (\frac {3 a^2}{b^2}+1\right ) \arctan (\tan (c+d x))}{2 b}-\frac {2 a b^2-b \left (3 a^2+b^2\right ) \tan (c+d x)}{2 b^2 \left (b^2 \tan ^2(c+d x)+b^2\right )}}{4 b^2}-\frac {(a+b \tan (c+d x)) \left (b^2-a b \tan (c+d x)\right )}{4 b^2 \left (b^2 \tan ^2(c+d x)+b^2\right )^2}\right )}{d}\) |
Input:
Int[Cos[c + d*x]^4*(a + b*Tan[c + d*x])^2,x]
Output:
(b^5*(-1/4*((a + b*Tan[c + d*x])*(b^2 - a*b*Tan[c + d*x]))/(b^2*(b^2 + b^2 *Tan[c + d*x]^2)^2) + (((1 + (3*a^2)/b^2)*ArcTan[Tan[c + d*x]])/(2*b) - (2 *a*b^2 - b*(3*a^2 + b^2)*Tan[c + d*x])/(2*b^2*(b^2 + b^2*Tan[c + d*x]^2))) /(4*b^2)))/d
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)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*d - b*c*x)/(2*a*b*(p + 1)))*(a + b*x^2)^(p + 1), x] + Simp[c*((2*p + 3)/(2*a *(p + 1))) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d}, x] && L tQ[p, -1] && NeQ[p, -3/2]
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*b*(p + 1))), x] - Simp[1/(2*a*b*(p + 1)) Int[(c + d*x)^(n - 2)*(a + b*x^2)^(p + 1)*Simp[a* d^2*(n - 1) - b*c^2*(2*p + 3) - b*c*d*(n + 2*p + 2)*x, x], x], x] /; FreeQ[ {a, b, c, d}, x] && LtQ[p, -1] && GtQ[n, 1] && IntQuadraticQ[a, 0, b, c, d, n, p, x]
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]
Time = 9.25 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.02
method | result | size |
derivativedivides | \(\frac {b^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\frac {a b \cos \left (d x +c \right )^{4}}{2}+a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(97\) |
default | \(\frac {b^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\frac {a b \cos \left (d x +c \right )^{4}}{2}+a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(97\) |
risch | \(\frac {3 a^{2} x}{8}+\frac {b^{2} x}{8}-\frac {a b \cos \left (4 d x +4 c \right )}{16 d}+\frac {\sin \left (4 d x +4 c \right ) a^{2}}{32 d}-\frac {\sin \left (4 d x +4 c \right ) b^{2}}{32 d}-\frac {a b \cos \left (2 d x +2 c \right )}{4 d}+\frac {\sin \left (2 d x +2 c \right ) a^{2}}{4 d}\) | \(97\) |
Input:
int(cos(d*x+c)^4*(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(b^2*(-1/4*sin(d*x+c)*cos(d*x+c)^3+1/8*cos(d*x+c)*sin(d*x+c)+1/8*d*x+1 /8*c)-1/2*a*b*cos(d*x+c)^4+a^2*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+ c)+3/8*d*x+3/8*c))
Time = 0.12 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.79 \[ \int \cos ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=-\frac {4 \, a b \cos \left (d x + c\right )^{4} - {\left (3 \, a^{2} + b^{2}\right )} d x - {\left (2 \, {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, d} \] Input:
integrate(cos(d*x+c)^4*(a+b*tan(d*x+c))^2,x, algorithm="fricas")
Output:
-1/8*(4*a*b*cos(d*x + c)^4 - (3*a^2 + b^2)*d*x - (2*(a^2 - b^2)*cos(d*x + c)^3 + (3*a^2 + b^2)*cos(d*x + c))*sin(d*x + c))/d
\[ \int \cos ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{2} \cos ^{4}{\left (c + d x \right )}\, dx \] Input:
integrate(cos(d*x+c)**4*(a+b*tan(d*x+c))**2,x)
Output:
Integral((a + b*tan(c + d*x))**2*cos(c + d*x)**4, x)
Time = 0.15 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.89 \[ \int \cos ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {{\left (3 \, a^{2} + b^{2}\right )} {\left (d x + c\right )} + \frac {{\left (3 \, a^{2} + b^{2}\right )} \tan \left (d x + c\right )^{3} - 4 \, a b + {\left (5 \, a^{2} - b^{2}\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1}}{8 \, d} \] Input:
integrate(cos(d*x+c)^4*(a+b*tan(d*x+c))^2,x, algorithm="maxima")
Output:
1/8*((3*a^2 + b^2)*(d*x + c) + ((3*a^2 + b^2)*tan(d*x + c)^3 - 4*a*b + (5* a^2 - b^2)*tan(d*x + c))/(tan(d*x + c)^4 + 2*tan(d*x + c)^2 + 1))/d
Time = 0.21 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.94 \[ \int \cos ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {{\left (3 \, a^{2} + b^{2}\right )} {\left (d x + c\right )}}{8 \, d} + \frac {3 \, a^{2} \tan \left (d x + c\right )^{3} + b^{2} \tan \left (d x + c\right )^{3} + 5 \, a^{2} \tan \left (d x + c\right ) - b^{2} \tan \left (d x + c\right ) - 4 \, a b}{8 \, {\left (\tan \left (d x + c\right )^{2} + 1\right )}^{2} d} \] Input:
integrate(cos(d*x+c)^4*(a+b*tan(d*x+c))^2,x, algorithm="giac")
Output:
1/8*(3*a^2 + b^2)*(d*x + c)/d + 1/8*(3*a^2*tan(d*x + c)^3 + b^2*tan(d*x + c)^3 + 5*a^2*tan(d*x + c) - b^2*tan(d*x + c) - 4*a*b)/((tan(d*x + c)^2 + 1 )^2*d)
Time = 0.80 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.87 \[ \int \cos ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=x\,\left (\frac {3\,a^2}{8}+\frac {b^2}{8}\right )+\frac {\left (\frac {3\,a^2}{8}+\frac {b^2}{8}\right )\,{\mathrm {tan}\left (c+d\,x\right )}^3+\left (\frac {5\,a^2}{8}-\frac {b^2}{8}\right )\,\mathrm {tan}\left (c+d\,x\right )-\frac {a\,b}{2}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4+2\,{\mathrm {tan}\left (c+d\,x\right )}^2+1\right )} \] Input:
int(cos(c + d*x)^4*(a + b*tan(c + d*x))^2,x)
Output:
x*((3*a^2)/8 + b^2/8) + (tan(c + d*x)*((5*a^2)/8 - b^2/8) - (a*b)/2 + tan( c + d*x)^3*((3*a^2)/8 + b^2/8))/(d*(2*tan(c + d*x)^2 + tan(c + d*x)^4 + 1) )
Time = 0.15 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.21 \[ \int \cos ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {-2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{2}+2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b^{2}+5 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2}-\cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{2}-4 \sin \left (d x +c \right )^{4} a b +8 \sin \left (d x +c \right )^{2} a b +3 a^{2} d x +b^{2} d x}{8 d} \] Input:
int(cos(d*x+c)^4*(a+b*tan(d*x+c))^2,x)
Output:
( - 2*cos(c + d*x)*sin(c + d*x)**3*a**2 + 2*cos(c + d*x)*sin(c + d*x)**3*b **2 + 5*cos(c + d*x)*sin(c + d*x)*a**2 - cos(c + d*x)*sin(c + d*x)*b**2 - 4*sin(c + d*x)**4*a*b + 8*sin(c + d*x)**2*a*b + 3*a**2*d*x + b**2*d*x)/(8* d)