Integrand size = 34, antiderivative size = 126 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {(5 B+19 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(B-C) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(5 B-13 C) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}} \]
-1/4*(B-C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)+1/16*(5*B-13*C)*sin(d*x+c)/ a/d/(a+a*cos(d*x+c))^(3/2)+1/32*(5*B+19*C)*arctanh(1/2*sin(d*x+c)*a^(1/2)* 2^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)
Time = 0.63 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.69 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {2 (5 B+19 C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right )+(B-9 C+(5 B-13 C) \cos (c+d x)) \tan \left (\frac {1}{2} (c+d x)\right )}{16 a d (a (1+\cos (c+d x)))^{3/2}} \]
(2*(5*B + 19*C)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^3 + (B - 9*C + (5*B - 13*C)*Cos[c + d*x])*Tan[(c + d*x)/2])/(16*a*d*(a*(1 + Cos[c + d*x]) )^(3/2))
Time = 0.54 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {3042, 3498, 27, 3042, 3229, 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.
\(\displaystyle \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{(a \cos (c+d x)+a)^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3498 |
\(\displaystyle -\frac {\int -\frac {5 a (B-C)+8 a C \cos (c+d x)}{2 (\cos (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(B-C) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {5 a (B-C)+8 a C \cos (c+d x)}{(\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(B-C) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {5 a (B-C)+8 a C \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(B-C) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3229 |
\(\displaystyle \frac {\frac {1}{4} (5 B+19 C) \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx+\frac {a (5 B-13 C) \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(B-C) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{4} (5 B+19 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {a (5 B-13 C) \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(B-C) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3128 |
\(\displaystyle \frac {\frac {a (5 B-13 C) \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac {(5 B+19 C) \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{2 d}}{8 a^2}-\frac {(B-C) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {(5 B+19 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} \sqrt {a} d}+\frac {a (5 B-13 C) \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(B-C) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
-1/4*((B - C)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^(5/2)) + (((5*B + 19*C )*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(2*S qrt[2]*Sqrt[a]*d) + (a*(5*B - 13*C)*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x] )^(3/2)))/(8*a^2)
3.3.82.3.1 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)^(-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[(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)]
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. \(291\) vs. \(2(107)=214\).
Time = 7.24 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.32
method | result | size |
default | \(\frac {\sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (5 B \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +19 C \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +5 B \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {2}\, \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-13 C \sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 B \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+2 C \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{32 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{\frac {7}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(292\) |
parts | \(\frac {B \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (5 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+5 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {2}\, \sqrt {a}-2 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{32 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{\frac {7}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}+\frac {C \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (19 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-13 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {2}\, \sqrt {a}+2 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{32 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{\frac {7}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(350\) |
1/32/cos(1/2*d*x+1/2*c)^3*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(5*B*2^(1/2)*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))*cos(1/ 2*d*x+1/2*c)^4*a+19*C*2^(1/2)*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))*cos(1/2*d*x+1/2*c)^4*a+5*B*cos(1/2*d*x+1/2*c)^ 2*2^(1/2)*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-13*C*a^(1/2)*2^(1/2)*(a*s in(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)^2-2*B*2^(1/2)*(a*sin(1/2*d*x +1/2*c)^2)^(1/2)*a^(1/2)+2*C*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 = 0.28 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.77 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {2} {\left ({\left (5 \, B + 19 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, B + 19 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (5 \, B + 19 \, C\right )} \cos \left (d x + c\right ) + 5 \, B + 19 \, C\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left ({\left (5 \, B - 13 \, C\right )} \cos \left (d x + c\right ) + B - 9 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
1/64*(sqrt(2)*((5*B + 19*C)*cos(d*x + c)^3 + 3*(5*B + 19*C)*cos(d*x + c)^2 + 3*(5*B + 19*C)*cos(d*x + c) + 5*B + 19*C)*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*((5*B - 13*C)*cos(d *x + c) + B - 9*C)*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)
Timed out. \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Time = 1.41 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.54 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\frac {\sqrt {2} {\left (5 \, B \sqrt {a} + 19 \, C \sqrt {a}\right )} \log \left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {\sqrt {2} {\left (5 \, B \sqrt {a} + 19 \, C \sqrt {a}\right )} \log \left (-\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, {\left (5 \, \sqrt {2} B \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 13 \, \sqrt {2} C \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, \sqrt {2} B \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 11 \, \sqrt {2} C \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{64 \, d} \]
1/64*(sqrt(2)*(5*B*sqrt(a) + 19*C*sqrt(a))*log(sin(1/2*d*x + 1/2*c) + 1)/( a^3*sgn(cos(1/2*d*x + 1/2*c))) - sqrt(2)*(5*B*sqrt(a) + 19*C*sqrt(a))*log( -sin(1/2*d*x + 1/2*c) + 1)/(a^3*sgn(cos(1/2*d*x + 1/2*c))) - 2*(5*sqrt(2)* B*sin(1/2*d*x + 1/2*c)^3 - 13*sqrt(2)*C*sin(1/2*d*x + 1/2*c)^3 - 3*sqrt(2) *B*sin(1/2*d*x + 1/2*c) + 11*sqrt(2)*C*sin(1/2*d*x + 1/2*c))/((sin(1/2*d*x + 1/2*c)^2 - 1)^2*a^(5/2)*sgn(cos(1/2*d*x + 1/2*c))))/d
Timed out. \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]