Integrand size = 25, antiderivative size = 398 \[ \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx=\frac {2 \left (3 a^4-15 a^2 b^2+8 b^4\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^4 (a-b) (a+b)^{3/2} d}-\frac {2 \left (3 a^3+9 a^2 b-6 a b^2-8 b^3\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^3 (a-b) (a+b)^{3/2} d}+\frac {2 b^2 \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}+\frac {8 b^2 \left (2 a^2-b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \] Output:
2/3*(3*a^4-15*a^2*b^2+8*b^4)*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^4/(a-b)/(a+b)^(3/2)/d-2/3*(3*a^3+9 *a^2*b-6*a*b^2-8*b^3)*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^3/(a-b)/(a+b)^(3/2)/d+2/3*b^2*sin(d*x+c)/ a/(a^2-b^2)/d/cos(d*x+c)^(1/2)/(a+b*cos(d*x+c))^(3/2)+8/3*b^2*(2*a^2-b^2)* sin(d*x+c)/a^2/(a^2-b^2)^2/d/cos(d*x+c)^(1/2)/(a+b*cos(d*x+c))^(1/2)
Result contains complex when optimal does not.
Time = 6.64 (sec) , antiderivative size = 1321, normalized size of antiderivative = 3.32 \[ \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:
Integrate[1/(Cos[c + d*x]^(3/2)*(a + b*Cos[c + d*x])^(5/2)),x]
Output:
-1/3*((-4*a*(9*a^4*b - 17*a^2*b^3 + 8*b^5)*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]*EllipticF[ArcSin[Sq rt[((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*(3*a^5 - 15*a^3*b^2 + 8*a*b^4)*((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]*EllipticF[ArcSin[Sqrt[ ((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Si n[(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]]*Sqrt[a + b*Cos[c + d*x]])) + 2*(3*a^4*b - 15*a^2*b^3 + 8*b^5)*(( 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 ...
Time = 1.59 (sec) , antiderivative size = 423, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 3281, 27, 3042, 3534, 27, 3042, 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.
\(\displaystyle \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3281 |
\(\displaystyle \frac {2 \int \frac {3 a^2-3 b \cos (c+d x) a-4 b^2+2 b^2 \cos ^2(c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}dx}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {3 a^2-3 b \cos (c+d x) a-4 b^2+2 b^2 \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}dx}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {3 a^2-3 b \sin \left (c+d x+\frac {\pi }{2}\right ) a-4 b^2+2 b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \frac {\frac {2 \int \frac {3 a^4-15 b^2 a^2-2 b \left (3 a^2-b^2\right ) \cos (c+d x) a+8 b^4}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {8 b^2 \left (2 a^2-b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \frac {3 a^4-15 b^2 a^2-2 b \left (3 a^2-b^2\right ) \cos (c+d x) a+8 b^4}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {8 b^2 \left (2 a^2-b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {3 a^4-15 b^2 a^2-2 b \left (3 a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+8 b^4}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}+\frac {8 b^2 \left (2 a^2-b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3477 |
\(\displaystyle \frac {\frac {\left (3 a^4-15 a^2 b^2+8 b^4\right ) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-(a-b) \left (3 a^3+9 a^2 b-6 a b^2-8 b^3\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {8 b^2 \left (2 a^2-b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\left (3 a^4-15 a^2 b^2+8 b^4\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-(a-b) \left (3 a^3+9 a^2 b-6 a b^2-8 b^3\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}+\frac {8 b^2 \left (2 a^2-b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3295 |
\(\displaystyle \frac {\frac {\left (3 a^4-15 a^2 b^2+8 b^4\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \left (3 a^3+9 a^2 b-6 a b^2-8 b^3\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{a \left (a^2-b^2\right )}+\frac {8 b^2 \left (2 a^2-b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3473 |
\(\displaystyle \frac {2 b^2 \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}+\frac {\frac {8 b^2 \left (2 a^2-b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}+\frac {\frac {2 (a-b) \sqrt {a+b} \left (3 a^4-15 a^2 b^2+8 b^4\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a^2 d}-\frac {2 (a-b) \sqrt {a+b} \left (3 a^3+9 a^2 b-6 a b^2-8 b^3\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{a \left (a^2-b^2\right )}}{3 a \left (a^2-b^2\right )}\) |
Input:
Int[1/(Cos[c + d*x]^(3/2)*(a + b*Cos[c + d*x])^(5/2)),x]
Output:
(2*b^2*Sin[c + d*x])/(3*a*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x])^(3/2)) + (((2*(a - b)*Sqrt[a + b]*(3*a^4 - 15*a^2*b^2 + 8*b^4)*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^2*d) - (2*(a - b)*Sqrt[a + b]*(3*a^3 + 9*a^2*b - 6*a*b^2 - 8*b^3)*Cot[c + d*x]*EllipticF[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))/(a* (a^2 - b^2)) + (8*b^2*(2*a^2 - b^2)*Sin[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[Co s[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]))/(3*a*(a^2 - b^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_.)*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/(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_)])/(((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]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*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[(m + 1)*(b*c - a* d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A *b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ [n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) | | EqQ[a, 0])))
Leaf count of result is larger than twice the leaf count of optimal. \(1696\) vs. \(2(360)=720\).
Time = 24.57 (sec) , antiderivative size = 1697, normalized size of antiderivative = 4.26
Input:
int(1/cos(d*x+c)^(3/2)/(a+cos(d*x+c)*b)^(5/2),x,method=_RETURNVERBOSE)
Output:
2/3/d*((3*cos(d*x+c)^2+6*cos(d*x+c)+3)*((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^6*EllipticE(cot(d*x+c)-csc( d*x+c),(-(a-b)/(a+b))^(1/2))+(3*cos(d*x+c)^3+9*cos(d*x+c)^2+9*cos(d*x+c)+3 )*((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^5*b*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))+(3*cos (d*x+c)^3-9*cos(d*x+c)^2-27*cos(d*x+c)-15)*((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*b^2*EllipticE(cot(d*x +c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))+(-15*cos(d*x+c)^3-45*cos(d*x+c)^2-45* cos(d*x+c)-15)*((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^3*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b) )^(1/2))+(-15*cos(d*x+c)^3-22*cos(d*x+c)^2+cos(d*x+c)+8)*((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^4*Ell ipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))+(8*cos(d*x+c)^3+24*cos( d*x+c)^2+24*cos(d*x+c)+8)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*(c os(d*x+c)/(cos(d*x+c)+1))^(1/2)*a*b^5*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a -b)/(a+b))^(1/2))+((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)*b^6*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b)) ^(1/2))*(8*cos(d*x+c)^3+16*cos(d*x+c)^2+8*cos(d*x+c))+(-3*cos(d*x+c)^2-6*c os(d*x+c)-3)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*(cos(d*x+c)/(co s(d*x+c)+1))^(1/2)*a^6*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(...
\[ \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))^(5/2),x, algorithm="fricas")
Output:
integral(sqrt(b*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(b^3*cos(d*x + c)^5 + 3*a*b^2*cos(d*x + c)^4 + 3*a^2*b*cos(d*x + c)^3 + a^3*cos(d*x + c)^2), x)
Timed out. \[ \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(1/cos(d*x+c)**(3/2)/(a+b*cos(d*x+c))**(5/2),x)
Output:
Timed out
\[ \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))^(5/2),x, algorithm="maxima")
Output:
integrate(1/((b*cos(d*x + c) + a)^(5/2)*cos(d*x + c)^(3/2)), x)
\[ \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))^(5/2),x, algorithm="giac")
Output:
integrate(1/((b*cos(d*x + c) + a)^(5/2)*cos(d*x + c)^(3/2)), x)
Timed out. \[ \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \] Input:
int(1/(cos(c + d*x)^(3/2)*(a + b*cos(c + d*x))^(5/2)),x)
Output:
int(1/(cos(c + d*x)^(3/2)*(a + b*cos(c + d*x))^(5/2)), x)
\[ \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx=\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{5} b^{3}+3 \cos \left (d x +c \right )^{4} a \,b^{2}+3 \cos \left (d x +c \right )^{3} a^{2} b +\cos \left (d x +c \right )^{2} a^{3}}d x \] Input:
int(1/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))^(5/2),x)
Output:
int((sqrt(cos(c + d*x)*b + a)*sqrt(cos(c + d*x)))/(cos(c + d*x)**5*b**3 + 3*cos(c + d*x)**4*a*b**2 + 3*cos(c + d*x)**3*a**2*b + cos(c + d*x)**2*a**3 ),x)