Integrand size = 31, antiderivative size = 118 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {(3 A-7 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {2 B \sin (c+d x)}{a d \sqrt {a+a \cos (c+d x)}} \] Output:
1/4*(3*A-7*B)*arctanh(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/(a+a*cos(d*x+c))^(1/2 ))*2^(1/2)/a^(3/2)/d-1/2*(A-B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)+2*B*sin (d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)
Time = 0.43 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.88 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {-\left ((3 A-7 B) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^5\left (\frac {1}{2} (c+d x)\right )\right )+\cos ^3\left (\frac {1}{2} (c+d x)\right ) (A-5 B-4 B \cos (c+d x)) \sin \left (\frac {1}{2} (c+d x)\right )}{d (a (1+\cos (c+d x)))^{3/2} \left (-1+\sin ^2\left (\frac {1}{2} (c+d x)\right )\right )} \] Input:
Integrate[(Cos[c + d*x]*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^(3/2),x ]
Output:
(-((3*A - 7*B)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^5) + Cos[(c + d* x)/2]^3*(A - 5*B - 4*B*Cos[c + d*x])*Sin[(c + d*x)/2])/(d*(a*(1 + Cos[c + d*x]))^(3/2)*(-1 + Sin[(c + d*x)/2]^2))
Time = 0.60 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {3042, 3447, 3042, 3498, 27, 3042, 3230, 3042, 3128, 219}
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.
\(\displaystyle \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a \cos (c+d x)+a)^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3447 |
\(\displaystyle \int \frac {A \cos (c+d x)+B \cos ^2(c+d x)}{(a \cos (c+d x)+a)^{3/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A \sin \left (c+d x+\frac {\pi }{2}\right )+B \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3498 |
\(\displaystyle -\frac {\int -\frac {3 a (A-B)+4 a B \cos (c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {(A-B) \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {3 a (A-B)+4 a B \cos (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {3 a (A-B)+4 a B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
\(\Big \downarrow \) 3230 |
\(\displaystyle \frac {a (3 A-7 B) \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx+\frac {8 a B \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a (3 A-7 B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {8 a B \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
\(\Big \downarrow \) 3128 |
\(\displaystyle \frac {\frac {8 a B \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {2 a (3 A-7 B) \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\sqrt {2} \sqrt {a} (3 A-7 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {8 a B \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
Input:
Int[(Cos[c + d*x]*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^(3/2),x]
Output:
-1/2*((A - B)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^(3/2)) + ((Sqrt[2]*Sqr t[a]*(3*A - 7*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/d + (8*a*B*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/(4*a^2)
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]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( f*(m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(b*(m + 1)) Int[(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b - a* B + b*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] + Simp[1 /(a^2*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b *B - a*C) + b*C*(2*m + 1)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(255\) vs. \(2(101)=202\).
Time = 2.40 (sec) , antiderivative size = 256, normalized size of antiderivative = 2.17
method | result | size |
default | \(\frac {\sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (3 A \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -7 B \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +8 B \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-A \sqrt {a}\, \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+B \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\right )}{4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{\frac {5}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\) | \(256\) |
parts | \(\frac {A \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (3 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\right )}{4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{\frac {5}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}+\frac {B \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-7 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \sqrt {2}\, \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\right )}{4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{\frac {5}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\) | \(315\) |
Input:
int(cos(d*x+c)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(3/2),x,method=_RETURNVER BOSE)
Output:
1/4*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(3*A*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2 *c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^2*a-7*B*l n(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^( 1/2)*cos(1/2*d*x+1/2*c)^2*a+8*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^( 1/2)*cos(1/2*d*x+1/2*c)^2-A*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2) +B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/cos(1/2*d*x+1/2*c)/a^(5 /2)/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
Time = 0.09 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.60 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {\sqrt {2} {\left ({\left (3 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, A - 7 \, B\right )} \cos \left (d x + c\right ) + 3 \, A - 7 \, B\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left (4 \, B \cos \left (d x + c\right ) - A + 5 \, B\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:
integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(3/2),x, algorithm= "fricas")
Output:
-1/8*(sqrt(2)*((3*A - 7*B)*cos(d*x + c)^2 + 2*(3*A - 7*B)*cos(d*x + c) + 3 *A - 7*B)*sqrt(a)*log(-(a*cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos (d*x + c) + 1)) - 4*(4*B*cos(d*x + c) - A + 5*B)*sqrt(a*cos(d*x + c) + a)* sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
\[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))**(3/2),x)
Output:
Integral((A + B*cos(c + d*x))*cos(c + d*x)/(a*(cos(c + d*x) + 1))**(3/2), x)
Timed out. \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(3/2),x, algorithm= "maxima")
Output:
Timed out
Exception generated. \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(3/2),x, algorithm= "giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {\cos \left (c+d\,x\right )\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \] Input:
int((cos(c + d*x)*(A + B*cos(c + d*x)))/(a + a*cos(c + d*x))^(3/2),x)
Output:
int((cos(c + d*x)*(A + B*cos(c + d*x)))/(a + a*cos(c + d*x))^(3/2), x)
\[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )}{\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1}d x \right ) b \right )}{a^{2}} \] Input:
int(cos(d*x+c)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(3/2),x)
Output:
(sqrt(a)*(int((sqrt(cos(c + d*x) + 1)*cos(c + d*x))/(cos(c + d*x)**2 + 2*c os(c + d*x) + 1),x)*a + int((sqrt(cos(c + d*x) + 1)*cos(c + d*x)**2)/(cos( c + d*x)**2 + 2*cos(c + d*x) + 1),x)*b))/a**2