Integrand size = 43, antiderivative size = 418 \[ \int \frac {(a+b \cos (c+d x))^{5/2} \left (\frac {3 b B}{2 a}+B \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 (a-b) \sqrt {a+b} \left (a^2+3 b^2\right ) B \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a d}-\frac {(a-3 b) \sqrt {a+b} \left (2 a^2-a b+3 b^2\right ) B \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a d}-\frac {b \sqrt {a+b} \left (5 a+\frac {3 b^2}{a}\right ) B \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{d}+\frac {b B (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)} \] Output:
2*(a-b)*(a+b)^(1/2)*(a^2+3*b^2)*B*cot(d*x+c)*EllipticE((a+b*cos(d*x+c))^(1 /2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(-(a+b)/(a-b))^(1/2))*(a*(1-sec(d*x+c))/( a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a/d-(a-3*b)*(a+b)^(1/2)*(2*a^2- a*b+3*b^2)*B*cot(d*x+c)*EllipticF((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d *x+c)^(1/2),(-(a+b)/(a-b))^(1/2))*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec (d*x+c))/(a-b))^(1/2)/a/d-b*(a+b)^(1/2)*(5*a+3*b^2/a)*B*cot(d*x+c)*Ellipti cPi((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(a+b)/b,(-(a+b)/(a -b))^(1/2))*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/ d+b*B*(a+b*cos(d*x+c))^(3/2)*sin(d*x+c)/d/cos(d*x+c)^(3/2)
Result contains complex when optimal does not.
Time = 25.30 (sec) , antiderivative size = 1236, normalized size of antiderivative = 2.96 \[ \int \frac {(a+b \cos (c+d x))^{5/2} \left (\frac {3 b B}{2 a}+B \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:
Integrate[((a + b*Cos[c + d*x])^(5/2)*((3*b*B)/(2*a) + B*Cos[c + d*x]))/Co s[c + d*x]^(5/2),x]
Output:
-1/2*(B*((-4*a*(-5*a^3*b - 3*a*b^3)*Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Co s[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/((a + b)*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]) - 4*a* (2*a^4 + a^2*b^2 - 3*b^4)*((Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sq rt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d* x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[((a + b*Cos[ c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/ 2]^4)/((a + b)*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]) - (Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x )/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x] *EllipticPi[-(a/b), ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/ a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/(b*Sqrt[Cos[c + d*x]]*Sq rt[a + b*Cos[c + d*x]])) + 2*(2*a^3*b + 6*a*b^3)*((I*Cos[(c + d*x)/2]*Sqrt [a + b*Cos[c + d*x]]*EllipticE[I*ArcSinh[Sin[(c + d*x)/2]/Sqrt[Cos[c + d*x ]]], (-2*a)/(-a - b)]*Sec[c + d*x])/(b*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x ]]*Sqrt[((a + b*Cos[c + d*x])*Sec[c + d*x])/(a + b)]) + (2*a*((a*Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d *x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c +...
Time = 1.85 (sec) , antiderivative size = 423, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.256, Rules used = {3042, 3468, 27, 3042, 3470, 3042, 3288, 3477, 3042, 3295, 3473}
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.
| \(\displaystyle \int \frac {(a+b \cos (c+d x))^{5/2} \left (\frac {3 b B}{2 a}+B \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (\frac {3 b B}{2 a}+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3468 |
| \(\displaystyle \frac {2}{3} \int \frac {3 \sqrt {a+b \cos (c+d x)} \left (2 \left (a^2+3 b^2\right ) B+b \left (\frac {3 b^2}{a}+5 a\right ) \cos (c+d x) B\right )}{4 \cos ^{\frac {3}{2}}(c+d x)}dx+\frac {b B \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \int \frac {\sqrt {a+b \cos (c+d x)} \left (2 \left (a^2+3 b^2\right ) B+b \left (\frac {3 b^2}{a}+5 a\right ) \cos (c+d x) B\right )}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {b B \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )} \left (2 \left (a^2+3 b^2\right ) B+b \left (\frac {3 b^2}{a}+5 a\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) B\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {b B \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3470 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {2 a \left (a^2+3 b^2\right ) B+\left (2 b \left (a^2+3 b^2\right ) B+a b \left (\frac {3 b^2}{a}+5 a\right ) B\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx+b^2 B \left (\frac {3 b^2}{a}+5 a\right ) \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}dx\right )+\frac {b B \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {2 a \left (a^2+3 b^2\right ) B+\left (2 b \left (a^2+3 b^2\right ) B+a b \left (\frac {3 b^2}{a}+5 a\right ) B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+b^2 B \left (\frac {3 b^2}{a}+5 a\right ) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {b B \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3288 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {2 a \left (a^2+3 b^2\right ) B+\left (2 b \left (a^2+3 b^2\right ) B+a b \left (\frac {3 b^2}{a}+5 a\right ) B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b B \sqrt {a+b} \left (\frac {3 b^2}{a}+5 a\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}\right )+\frac {b B \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3477 |
| \(\displaystyle \frac {1}{2} \left (2 a B \left (a^2+3 b^2\right ) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-B (a-3 b) \left (2 a^2-a b+3 b^2\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx-\frac {2 b B \sqrt {a+b} \left (\frac {3 b^2}{a}+5 a\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}\right )+\frac {b B \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (-B (a-3 b) \left (2 a^2-a b+3 b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+2 a B \left (a^2+3 b^2\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b B \sqrt {a+b} \left (\frac {3 b^2}{a}+5 a\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}\right )+\frac {b B \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3295 |
| \(\displaystyle \frac {1}{2} \left (2 a B \left (a^2+3 b^2\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 B (a-3 b) \sqrt {a+b} \left (2 a^2-a b+3 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}-\frac {2 b B \sqrt {a+b} \left (\frac {3 b^2}{a}+5 a\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}\right )+\frac {b B \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3473 |
| \(\displaystyle \frac {1}{2} \left (-\frac {2 B (a-3 b) \sqrt {a+b} \left (2 a^2-a b+3 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}+\frac {4 B (a-b) \sqrt {a+b} \left (a^2+3 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a d}-\frac {2 b B \sqrt {a+b} \left (\frac {3 b^2}{a}+5 a\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}\right )+\frac {b B \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{d \cos ^{\frac {3}{2}}(c+d x)}\) |
Input:
Int[((a + b*Cos[c + d*x])^(5/2)*((3*b*B)/(2*a) + B*Cos[c + d*x]))/Cos[c + d*x]^(5/2),x]
Output:
((4*(a - b)*Sqrt[a + b]*(a^2 + 3*b^2)*B*Cot[c + d*x]*EllipticE[ArcSin[Sqrt [a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b)) ]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b) ])/(a*d) - (2*(a - 3*b)*Sqrt[a + b]*(2*a^2 - a*b + 3*b^2)*B*Cot[c + d*x]*E llipticF[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])] , -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Se c[c + d*x]))/(a - b)])/(a*d) - (2*b*Sqrt[a + b]*(5*a + (3*b^2)/a)*B*Cot[c + d*x]*EllipticPi[(a + b)/b, ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]* Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/d)/2 + (b*B*(a + b*Cos[c + d*x] )^(3/2)*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Sqrt[(b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.) *(x_)]], x_Symbol] :> Simp[2*b*(Tan[e + f*x]/(d*f))*Rt[(c + d)/b, 2]*Sqrt[c *((1 + Csc[e + f*x])/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*Ellipti cPi[(c + d)/d, ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] && PosQ[(c + d)/b]
Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f _.)*(x_)]]), x_Symbol] :> Simp[-2*(Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqr t[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]*Elli pticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2] ], -(a + b)/(a - b)], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && PosQ[(a + b)/d]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(-(b*c - a*d))*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c - (A*b + a *B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) - a*(b*c - a*d)*(B*c - A*d)*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2 , 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && LtQ[n, -1]
Int[(((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]])/((b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2), x_Symbol] :> Simp[B*(d /b^2) Int[Sqrt[b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]], x], x] + Int[(A* c + (B*c + A*d)*Sin[e + f*x])/((b*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f* x]]), x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0]
Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)]) ^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*A* (c - d)*(Tan[e + f*x]/(f*b*c^2))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e + f*x] )/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*EllipticE[ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ[A, B] && PosQ[(c + d)/b]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ .)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> S imp[(A - B)/(a - b) Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f* x]]), x], x] - Simp[(A*b - a*B)/(a - b) Int[(1 + Sin[e + f*x])/((a + b*Si n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e , f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && NeQ[A, B]
Leaf count of result is larger than twice the leaf count of optimal. \(1386\) vs. \(2(381)=762\).
Time = 35.36 (sec) , antiderivative size = 1387, normalized size of antiderivative = 3.32
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(1387\) |
| parts | \(\text {Expression too large to display}\) | \(1834\) |
Input:
int((a+cos(d*x+c)*b)^(5/2)*(3/2*b*B/a+B*cos(d*x+c))/cos(d*x+c)^(5/2),x,met hod=_RETURNVERBOSE)
Output:
B/a/d*(((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*(cos(d*x+c)/(cos(d*x+ c)+1))^(1/2)*a^2*b^2*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,(-(a-b)/(a+b))^(1 /2))*(-10*cos(d*x+c)^3-20*cos(d*x+c)^2-10*cos(d*x+c))+((a+cos(d*x+c)*b)/(c os(d*x+c)+1)/(a+b))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*b^4*EllipticPi (cot(d*x+c)-csc(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*(-6*cos(d*x+c)^3-12*cos(d* x+c)^2-6*cos(d*x+c))+((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*(cos(d* x+c)/(cos(d*x+c)+1))^(1/2)*a^4*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+ b))^(1/2))*(2*cos(d*x+c)^3+4*cos(d*x+c)^2+2*cos(d*x+c))+((a+cos(d*x+c)*b)/ (cos(d*x+c)+1)/(a+b))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^3*b*Ellipt icE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(2*cos(d*x+c)^3+4*cos(d*x+ c)^2+2*cos(d*x+c))+((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*(cos(d*x+ c)/(cos(d*x+c)+1))^(1/2)*a^2*b^2*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/( a+b))^(1/2))*(6*cos(d*x+c)^3+12*cos(d*x+c)^2+6*cos(d*x+c))+((a+cos(d*x+c)* b)/(cos(d*x+c)+1)/(a+b))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a*b^3*Ell ipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(6*cos(d*x+c)^3+12*cos( d*x+c)^2+6*cos(d*x+c))+((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*(cos( d*x+c)/(cos(d*x+c)+1))^(1/2)*a^4*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/( a+b))^(1/2))*(-2*cos(d*x+c)^3-4*cos(d*x+c)^2-2*cos(d*x+c))+((a+cos(d*x+c)* b)/(cos(d*x+c)+1)/(a+b))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^3*b*Ell ipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(-7*cos(d*x+c)^3-14*...
\[ \int \frac {(a+b \cos (c+d x))^{5/2} \left (\frac {3 b B}{2 a}+B \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (2 \, B \cos \left (d x + c\right ) + \frac {3 \, B b}{a}\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{2 \, \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^(5/2)*(3/2*b*B/a+B*cos(d*x+c))/cos(d*x+c)^(5/2) ,x, algorithm="fricas")
Output:
integral(1/2*(2*B*a*b^2*cos(d*x + c)^3 + 3*B*a^2*b + (4*B*a^2*b + 3*B*b^3) *cos(d*x + c)^2 + 2*(B*a^3 + 3*B*a*b^2)*cos(d*x + c))*sqrt(b*cos(d*x + c) + a)/(a*cos(d*x + c)^(5/2)), x)
Timed out. \[ \int \frac {(a+b \cos (c+d x))^{5/2} \left (\frac {3 b B}{2 a}+B \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))**(5/2)*(3/2*b*B/a+B*cos(d*x+c))/cos(d*x+c)**(5/ 2),x)
Output:
Timed out
\[ \int \frac {(a+b \cos (c+d x))^{5/2} \left (\frac {3 b B}{2 a}+B \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (2 \, B \cos \left (d x + c\right ) + \frac {3 \, B b}{a}\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{2 \, \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^(5/2)*(3/2*b*B/a+B*cos(d*x+c))/cos(d*x+c)^(5/2) ,x, algorithm="maxima")
Output:
1/2*integrate((2*B*cos(d*x + c) + 3*B*b/a)*(b*cos(d*x + c) + a)^(5/2)/cos( d*x + c)^(5/2), x)
\[ \int \frac {(a+b \cos (c+d x))^{5/2} \left (\frac {3 b B}{2 a}+B \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (2 \, B \cos \left (d x + c\right ) + \frac {3 \, B b}{a}\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{2 \, \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^(5/2)*(3/2*b*B/a+B*cos(d*x+c))/cos(d*x+c)^(5/2) ,x, algorithm="giac")
Output:
integrate(1/2*(2*B*cos(d*x + c) + 3*B*b/a)*(b*cos(d*x + c) + a)^(5/2)/cos( d*x + c)^(5/2), x)
Timed out. \[ \int \frac {(a+b \cos (c+d x))^{5/2} \left (\frac {3 b B}{2 a}+B \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (B\,\cos \left (c+d\,x\right )+\frac {3\,B\,b}{2\,a}\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^{5/2}} \,d x \] Input:
int(((B*cos(c + d*x) + (3*B*b)/(2*a))*(a + b*cos(c + d*x))^(5/2))/cos(c + d*x)^(5/2),x)
Output:
int(((B*cos(c + d*x) + (3*B*b)/(2*a))*(a + b*cos(c + d*x))^(5/2))/cos(c + d*x)^(5/2), x)
\[ \int \frac {(a+b \cos (c+d x))^{5/2} \left (\frac {3 b B}{2 a}+B \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {b \left (4 \left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) a^{2} b +3 \left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) b^{3}+3 \left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right ) a^{2} b +2 \left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) a^{3}+6 \left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) a \,b^{2}+2 \left (\int \sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}d x \right ) a \,b^{2}\right )}{2 a} \] Input:
int((a+b*cos(d*x+c))^(5/2)*(3/2*b*B/a+B*cos(d*x+c))/cos(d*x+c)^(5/2),x)
Output:
(b*(4*int((sqrt(cos(c + d*x)*b + a)*sqrt(cos(c + d*x)))/cos(c + d*x),x)*a* *2*b + 3*int((sqrt(cos(c + d*x)*b + a)*sqrt(cos(c + d*x)))/cos(c + d*x),x) *b**3 + 3*int((sqrt(cos(c + d*x)*b + a)*sqrt(cos(c + d*x)))/cos(c + d*x)** 3,x)*a**2*b + 2*int((sqrt(cos(c + d*x)*b + a)*sqrt(cos(c + d*x)))/cos(c + d*x)**2,x)*a**3 + 6*int((sqrt(cos(c + d*x)*b + a)*sqrt(cos(c + d*x)))/cos( c + d*x)**2,x)*a*b**2 + 2*int(sqrt(cos(c + d*x)*b + a)*sqrt(cos(c + d*x)), x)*a*b**2))/(2*a)