Integrand size = 25, antiderivative size = 126 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {(3 A+5 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(A-B) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(3 A+5 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}} \] Output:
1/32*(3*A+5*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^(5/2)/d+1/4*(A-B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)+1/16*( 3*A+5*B)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(3/2)
Time = 0.52 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.63 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {4 (3 A+5 B) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^5\left (\frac {1}{2} (c+d x)\right )+(7 A+B+(3 A+5 B) \cos (c+d x)) \sin (c+d x)}{16 d (a (1+\cos (c+d x)))^{5/2}} \] Input:
Integrate[(A + B*Cos[c + d*x])/(a + a*Cos[c + d*x])^(5/2),x]
Output:
(4*(3*A + 5*B)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^5 + (7*A + B + ( 3*A + 5*B)*Cos[c + d*x])*Sin[c + d*x])/(16*d*(a*(1 + Cos[c + d*x]))^(5/2))
Time = 0.44 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 3229, 3042, 3129, 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.
| \(\displaystyle \int \frac {A+B \cos (c+d x)}{(a \cos (c+d x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3229 |
| \(\displaystyle \frac {(3 A+5 B) \int \frac {1}{(\cos (c+d x) a+a)^{3/2}}dx}{8 a}+\frac {(A-B) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(3 A+5 B) \int \frac {1}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a}+\frac {(A-B) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3129 |
| \(\displaystyle \frac {(3 A+5 B) \left (\frac {\int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx}{4 a}+\frac {\sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\right )}{8 a}+\frac {(A-B) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(3 A+5 B) \left (\frac {\int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a}+\frac {\sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\right )}{8 a}+\frac {(A-B) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {(3 A+5 B) \left (\frac {\sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac {\int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{2 a d}\right )}{8 a}+\frac {(A-B) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {(3 A+5 B) \left (\frac {\text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {\sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\right )}{8 a}+\frac {(A-B) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
Input:
Int[(A + B*Cos[c + d*x])/(a + a*Cos[c + d*x])^(5/2),x]
Output:
((A - B)*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) + ((3*A + 5*B)*(Ar cTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])]/(2*Sqrt[2 ]*a^(3/2)*d) + Sin[c + d*x]/(2*d*(a + a*Cos[c + d*x])^(3/2))))/(8*a)
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[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d*(2*n + 1))), x] + Simp[(n + 1)/(a*(2*n + 1)) Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*((a + b*Sin[e + f* x])^m/(a*f*(2*m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)) I nt[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Leaf count of result is larger than twice the leaf count of optimal. \(291\) vs. \(2(107)=214\).
Time = 2.29 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.32
| method | result | size |
| default | \(\frac {\sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (3 A \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +5 B \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +3 A \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+5 B \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 A \sqrt {a}\, \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-2 B \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\right )}{32 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{\frac {7}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\) | \(292\) |
| parts | \(\frac {A \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (3 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \sqrt {2}\, \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+2 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\right )}{32 a^{\frac {7}{2}} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}+\frac {B \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (5 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+5 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \sqrt {2}\, \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-2 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\right )}{32 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{\frac {7}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\) | \(350\) |
Input:
int((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
1/32/cos(1/2*d*x+1/2*c)^3*(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)^4*a+5*B*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/c os(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^4*a+3*A*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+5*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+2*A*a^(1/2)*2^(1/2)*(a*sin(1 /2*d*x+1/2*c)^2)^(1/2)-2*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)) /a^(7/2)/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
Leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (107) = 214\).
Time = 0.09 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.77 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {2} {\left ({\left (3 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A + 5 \, B\right )} \cos \left (d x + c\right ) + 3 \, A + 5 \, B\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left ({\left (3 \, A + 5 \, B\right )} \cos \left (d x + c\right ) + 7 \, A + B\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:
integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(5/2),x, algorithm="fricas")
Output:
1/64*(sqrt(2)*((3*A + 5*B)*cos(d*x + c)^3 + 3*(3*A + 5*B)*cos(d*x + c)^2 + 3*(3*A + 5*B)*cos(d*x + c) + 3*A + 5*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*((3*A + 5*B)*cos(d*x + c) + 7*A + B)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)
\[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {A + B \cos {\left (c + d x \right )}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))**(5/2),x)
Output:
Integral((A + B*cos(c + d*x))/(a*(cos(c + d*x) + 1))**(5/2), x)
Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")
Output:
Timed out
Exception generated. \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(5/2),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. \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \] Input:
int((A + B*cos(c + d*x))/(a + a*cos(c + d*x))^(5/2),x)
Output:
int((A + B*cos(c + d*x))/(a + a*cos(c + d*x))^(5/2), x)
\[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}}{\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )}{\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1}d x \right ) b \right )}{a^{3}} \] Input:
int((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(5/2),x)
Output:
(sqrt(a)*(int(sqrt(cos(c + d*x) + 1)/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1),x)*a + int((sqrt(cos(c + d*x) + 1)*cos(c + d*x))/(co s(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1),x)*b))/a**3