Integrand size = 34, antiderivative size = 179 \[ \int \frac {(a B+b B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=-\frac {2 b B \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{a \left (a^2-b^2\right ) d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 B \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{a d \sqrt {a+b \cos (c+d x)}}+\frac {2 b^2 B \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}} \] Output:
-2*b*B*(a+b*cos(d*x+c))^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b ))^(1/2))/a/(a^2-b^2)/d/((a+b*cos(d*x+c))/(a+b))^(1/2)+2*B*((a+b*cos(d*x+c ))/(a+b))^(1/2)*EllipticPi(sin(1/2*d*x+1/2*c),2,2^(1/2)*(b/(a+b))^(1/2))/a /d/(a+b*cos(d*x+c))^(1/2)+2*b^2*B*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c) )^(1/2)
Result contains complex when optimal does not.
Time = 6.25 (sec) , antiderivative size = 403, normalized size of antiderivative = 2.25 \[ \int \frac {(a B+b B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\frac {B \left (-\frac {-\frac {4 a b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 \left (2 a^2-3 b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}-\frac {2 i \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {b (1+\cos (c+d x))}{-a+b}} \csc (c+d x) \left (-2 a (a-b) E\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (-2 a \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )+b \operatorname {EllipticPi}\left (\frac {a+b}{a},i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )\right )\right )}{a \sqrt {-\frac {1}{a+b}}}}{(-a+b) (a+b)}+\frac {4 b^2 \sin (c+d x)}{\left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\right )}{2 a d} \] Input:
Integrate[((a*B + b*B*Cos[c + d*x])*Sec[c + d*x])/(a + b*Cos[c + d*x])^(5/ 2),x]
Output:
(B*(-(((-4*a*b*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, ( 2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]] + (2*(2*a^2 - 3*b^2)*Sqrt[(a + b*C os[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]] - ((2*I)*Sqrt[-((b*(-1 + Cos[c + d*x]))/(a + b))]*Sqrt[(b* (1 + Cos[c + d*x]))/(-a + b)]*Csc[c + d*x]*(-2*a*(a - b)*EllipticE[I*ArcSi nh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)] + b*(-2 *a*EllipticF[I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)] + b*EllipticPi[(a + b)/a, I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[ a + b*Cos[c + d*x]]], (a + b)/(a - b)])))/(a*Sqrt[-(a + b)^(-1)]))/((-a + b)*(a + b))) + (4*b^2*Sin[c + d*x])/((a^2 - b^2)*Sqrt[a + b*Cos[c + d*x]]) ))/(2*a*d)
Time = 1.15 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.04, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.441, Rules used = {2011, 3042, 3281, 27, 3042, 3538, 25, 27, 3042, 3134, 3042, 3132, 3286, 3042, 3284}
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.
| \(\displaystyle \int \frac {\sec (c+d x) (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 2011 |
| \(\displaystyle B \int \frac {\sec (c+d x)}{(a+b \cos (c+d x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle B \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3281 |
| \(\displaystyle B \left (\frac {2 \int \frac {\left (a^2-b \cos (c+d x) a-b^2-b^2 \cos ^2(c+d x)\right ) \sec (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle B \left (\frac {\int \frac {\left (a^2-b \cos (c+d x) a-b^2-b^2 \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle B \left (\frac {\int \frac {a^2-b \sin \left (c+d x+\frac {\pi }{2}\right ) a-b^2-b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3538 |
| \(\displaystyle B \left (\frac {-\frac {\int -\frac {b \left (a^2-b^2\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}-b \int \sqrt {a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle B \left (\frac {\frac {\int \frac {b \left (a^2-b^2\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}-b \int \sqrt {a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle B \left (\frac {\left (a^2-b^2\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx-b \int \sqrt {a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle B \left (\frac {\left (a^2-b^2\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-b \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3134 |
| \(\displaystyle B \left (\frac {\left (a^2-b^2\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {b \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle B \left (\frac {\left (a^2-b^2\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {b \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3132 |
| \(\displaystyle B \left (\frac {\left (a^2-b^2\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3286 |
| \(\displaystyle B \left (\frac {\frac {\left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {\sec (c+d x)}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}-\frac {2 b \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle B \left (\frac {\frac {\left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}-\frac {2 b \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle B \left (\frac {2 b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}+\frac {\frac {2 \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}-\frac {2 b \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}\right )\) |
Input:
Int[((a*B + b*B*Cos[c + d*x])*Sec[c + d*x])/(a + b*Cos[c + d*x])^(5/2),x]
Output:
B*(((-2*b*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/ (d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + (2*(a^2 - b^2)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[a + b* Cos[c + d*x]]))/(a*(a^2 - b^2)) + (2*b^2*Sin[c + d*x])/(a*(a^2 - b^2)*d*Sq rt[a + b*Cos[c + d*x]]))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Simp[(b/d)^m Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x ] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x , a + b*x])
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)] Int[Sqrt[a/(a + b) + ( b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 , 0] && !GtQ[a + b, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f* x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2 ))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n + 3)*Si n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[ 2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2* n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0])))
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c , d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt [c + d*Sin[e + f*x]] Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/(c + d))*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !GtQ[c + d, 0]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d) Int[Sqrt[a + b*Sin[e + f*x]], x] , x] - Simp[1/(b*d) Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 ] && NeQ[c^2 - d^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(376\) vs. \(2(179)=358\).
Time = 11.98 (sec) , antiderivative size = 377, normalized size of antiderivative = 2.11
| method | result | size |
| default | \(\frac {2 B \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b^{2}+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-\frac {2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a -b}+\frac {a +b}{a -b}}\, \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2, \sqrt {-\frac {2 b}{a -b}}\right ) a^{2}-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-\frac {2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a -b}+\frac {a +b}{a -b}}\, \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2, \sqrt {-\frac {2 b}{a -b}}\right ) b^{2}+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-\frac {2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a -b}+\frac {a +b}{a -b}}\, b \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a -b^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-\frac {2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a -b}+\frac {a +b}{a -b}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )\right )}{a \left (a -b \right ) \left (a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b}\, d}\) | \(377\) |
| parts | \(\text {Expression too large to display}\) | \(1342\) |
Input:
int((B*a+b*B*cos(d*x+c))*sec(d*x+c)/(a+cos(d*x+c)*b)^(5/2),x,method=_RETUR NVERBOSE)
Output:
2*B*(2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*b^2+(sin(1/2*d*x+1/2*c)^2)^ (1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticPi(cos(1 /2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*a^2-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b /(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticPi(cos(1/2*d*x+1/2* c),2,(-2*b/(a-b))^(1/2))*b^2+(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin( 1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*b*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/( a-b))^(1/2))*a-b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/ 2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2)) )/a/(a-b)/(a+b)/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)^(1/2)/d
Timed out. \[ \int \frac {(a B+b B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((a*B+b*B*cos(d*x+c))*sec(d*x+c)/(a+b*cos(d*x+c))^(5/2),x, algori thm="fricas")
Output:
Timed out
\[ \int \frac {(a B+b B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=B \int \frac {\sec {\left (c + d x \right )}}{a \sqrt {a + b \cos {\left (c + d x \right )}} + b \sqrt {a + b \cos {\left (c + d x \right )}} \cos {\left (c + d x \right )}}\, dx \] Input:
integrate((a*B+b*B*cos(d*x+c))*sec(d*x+c)/(a+b*cos(d*x+c))**(5/2),x)
Output:
B*Integral(sec(c + d*x)/(a*sqrt(a + b*cos(c + d*x)) + b*sqrt(a + b*cos(c + d*x))*cos(c + d*x)), x)
\[ \int \frac {(a B+b B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (B b \cos \left (d x + c\right ) + B a\right )} \sec \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((a*B+b*B*cos(d*x+c))*sec(d*x+c)/(a+b*cos(d*x+c))^(5/2),x, algori thm="maxima")
Output:
integrate((B*b*cos(d*x + c) + B*a)*sec(d*x + c)/(b*cos(d*x + c) + a)^(5/2) , x)
\[ \int \frac {(a B+b B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (B b \cos \left (d x + c\right ) + B a\right )} \sec \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((a*B+b*B*cos(d*x+c))*sec(d*x+c)/(a+b*cos(d*x+c))^(5/2),x, algori thm="giac")
Output:
integrate((B*b*cos(d*x + c) + B*a)*sec(d*x + c)/(b*cos(d*x + c) + a)^(5/2) , x)
Timed out. \[ \int \frac {(a B+b B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int \frac {B\,a+B\,b\,\cos \left (c+d\,x\right )}{\cos \left (c+d\,x\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \] Input:
int((B*a + B*b*cos(c + d*x))/(cos(c + d*x)*(a + b*cos(c + d*x))^(5/2)),x)
Output:
int((B*a + B*b*cos(c + d*x))/(cos(c + d*x)*(a + b*cos(c + d*x))^(5/2)), x)
\[ \int \frac {(a B+b B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )}{\cos \left (d x +c \right )^{2} b^{2}+2 \cos \left (d x +c \right ) a b +a^{2}}d x \right ) b \] Input:
int((a*B+b*B*cos(d*x+c))*sec(d*x+c)/(a+b*cos(d*x+c))^(5/2),x)
Output:
int((sqrt(cos(c + d*x)*b + a)*sec(c + d*x))/(cos(c + d*x)**2*b**2 + 2*cos( c + d*x)*a*b + a**2),x)*b