Integrand size = 25, antiderivative size = 445 \[ \int \frac {(a+b \cos (c+d x))^{5/2}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(a-b) \sqrt {a+b} \left (2 a^2-b^2\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a d}-\frac {\sqrt {a+b} \left (2 a^2-6 a b-b^2\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{d}-\frac {5 a b \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{d}+\frac {2 a^2 \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {\left (2 a^2-b^2\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \] Output:
(a-b)*(a+b)^(1/2)*(2*a^2-b^2)*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+b)^(1/2)*(2*a^2-6*a*b-b^2)*c ot(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)/d-5*a*b*(a+b)^(1/2)*cot(d*x+c)*EllipticPi((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+2*a^2*(a+b*cos(d*x+c))^(1/ 2)*sin(d*x+c)/d/cos(d*x+c)^(1/2)-(2*a^2-b^2)*(a+b*cos(d*x+c))^(1/2)*sin(d* x+c)/d/cos(d*x+c)^(1/2)
Result contains complex when optimal does not.
Time = 18.87 (sec) , antiderivative size = 1185, normalized size of antiderivative = 2.66 \[ \int \frac {(a+b \cos (c+d x))^{5/2}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:
Integrate[(a + b*Cos[c + d*x])^(5/2)/Cos[c + d*x]^(3/2),x]
Output:
(2*a^2*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) + ((4 *a*(-4*a^2*b - 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*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]]) + 4*a*(2*a^3 - 6*a*b^2 )*((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)]*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), ArcS in[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*(2*a^2*b - b^3)*((I*Cos[(c + d*x)/2]*Sqrt[a + b*Cos[c + d*x]]*Ellipti cE[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 + ...
Time = 2.04 (sec) , antiderivative size = 458, normalized size of antiderivative = 1.03, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 3271, 27, 3042, 3540, 3042, 3532, 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.
\(\displaystyle \int \frac {(a+b \cos (c+d x))^{5/2}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3271 |
\(\displaystyle 2 \int \frac {3 b a^2-\left (a^2-3 b^2\right ) \cos (c+d x) a-b \left (2 a^2-b^2\right ) \cos ^2(c+d x)}{2 \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx+\frac {2 a^2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {3 b a^2-\left (a^2-3 b^2\right ) \cos (c+d x) a-b \left (2 a^2-b^2\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx+\frac {2 a^2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {3 b a^2-\left (a^2-3 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a-b \left (2 a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3540 |
\(\displaystyle \frac {\int \frac {5 a \cos ^2(c+d x) b^3+6 a^2 \cos (c+d x) b^2+a \left (2 a^2-b^2\right ) b}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{2 b}-\frac {\left (2 a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {2 a^2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {5 a \sin \left (c+d x+\frac {\pi }{2}\right )^2 b^3+6 a^2 \sin \left (c+d x+\frac {\pi }{2}\right ) b^2+a \left (2 a^2-b^2\right ) b}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}-\frac {\left (2 a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {2 a^2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3532 |
\(\displaystyle \frac {\int \frac {6 a^2 \cos (c+d x) b^2+a \left (2 a^2-b^2\right ) b}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx+5 a b^3 \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}dx}{2 b}-\frac {\left (2 a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {2 a^2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {6 a^2 \sin \left (c+d x+\frac {\pi }{2}\right ) b^2+a \left (2 a^2-b^2\right ) b}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+5 a b^3 \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}-\frac {\left (2 a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {2 a^2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3288 |
\(\displaystyle \frac {\int \frac {6 a^2 \sin \left (c+d x+\frac {\pi }{2}\right ) b^2+a \left (2 a^2-b^2\right ) b}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {10 a b^2 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}-\frac {\left (2 a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {2 a^2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3477 |
\(\displaystyle \frac {a b \left (2 a^2-b^2\right ) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-a b \left (2 a^2-6 a b-b^2\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx-\frac {10 a b^2 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}-\frac {\left (2 a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {2 a^2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {-a b \left (2 a^2-6 a b-b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+a b \left (2 a^2-b^2\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {10 a b^2 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}-\frac {\left (2 a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {2 a^2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3295 |
\(\displaystyle \frac {a b \left (2 a^2-b^2\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b \sqrt {a+b} \left (2 a^2-6 a b-b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}-\frac {10 a b^2 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}-\frac {\left (2 a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {2 a^2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3473 |
\(\displaystyle \frac {-\frac {2 b \sqrt {a+b} \left (2 a^2-6 a b-b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}+\frac {2 b (a-b) \sqrt {a+b} \left (2 a^2-b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a d}-\frac {10 a b^2 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}-\frac {\left (2 a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {2 a^2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\) |
Input:
Int[(a + b*Cos[c + d*x])^(5/2)/Cos[c + d*x]^(3/2),x]
Output:
((2*(a - b)*b*Sqrt[a + b]*(2*a^2 - b^2)*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*b*Sqrt[a + b]*(2*a^2 - 6*a*b - b^2)*Cot[c + d*x]*EllipticF[A rcSin[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 - (10*a*b^2*Sqrt[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) + (2*a^2*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(d* Sqrt[Cos[c + d*x]]) - ((2*a^2 - b^2)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x] )/(d*Sqrt[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[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Co s[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((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 - 3)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^ 2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 + d*n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, 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[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
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_)])/(((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_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_.) + (d_.)*sin[(e _.) + (f_.)*(x_)]]), x_Symbol] :> Simp[C/b^2 Int[Sqrt[a + b*Sin[e + f*x]] /Sqrt[c + d*Sin[e + f*x]], x], x] + Simp[1/b^2 Int[(A*b^2 - a^2*C + b*(b* B - 2*a*C)*Sin[e + f*x])/((a + b*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x ]]), x], x] /; FreeQ[{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]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(Sqrt[c + d*Sin[e + f *x]]/(d*f*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[1/(2*d) Int[(1/((a + b*Si n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]))*Simp[2*a*A*d - C*(b*c - a*d) - 2*(a*c*C - d*(A*b + a*B))*Sin[e + f*x] + (2*b*B*d - C*(b*c + a*d))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{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. \(933\) vs. \(2(404)=808\).
Time = 21.16 (sec) , antiderivative size = 934, normalized size of antiderivative = 2.10
Input:
int((a+cos(d*x+c)*b)^(5/2)/cos(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/d*((-10*cos(d*x+c)^2-20*cos(d*x+c)-10)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2) *((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*a*b^2*EllipticPi(cot(d*x+c) -csc(d*x+c),-1,(-(a-b)/(a+b))^(1/2))+(2+2*cos(d*x+c)^2+4*cos(d*x+c))*(cos( d*x+c)/(cos(d*x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2) *a^3*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))+(2+2*cos(d*x+c) ^2+4*cos(d*x+c))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/(cos( d*x+c)+1)/(a+b))^(1/2)*a^2*b*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b) )^(1/2))+(-cos(d*x+c)^2-2*cos(d*x+c)-1)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)* ((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*a*b^2*EllipticE(cot(d*x+c)-c sc(d*x+c),(-(a-b)/(a+b))^(1/2))+(-cos(d*x+c)^2-2*cos(d*x+c)-1)*(cos(d*x+c) /(cos(d*x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*b^3*E llipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))+(-2*cos(d*x+c)^2-4*co s(d*x+c)-2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/(cos(d*x+c )+1)/(a+b))^(1/2)*a^3*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2) )+(-6*cos(d*x+c)^2-12*cos(d*x+c)-6)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*((a+ cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*a^2*b*EllipticF(cot(d*x+c)-csc(d *x+c),(-(a-b)/(a+b))^(1/2))+(6*cos(d*x+c)^2+12*cos(d*x+c)+6)*(cos(d*x+c)/( cos(d*x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*a*b^2*E llipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))+sin(d*x+c)*cos(d*x+c) ^2*b^3+2*a^2*b*cos(d*x+c)*sin(d*x+c)+a*b^2*cos(d*x+c)*sin(d*x+c)+2*sin(...
\[ \int \frac {(a+b \cos (c+d x))^{5/2}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^(5/2)/cos(d*x+c)^(3/2),x, algorithm="fricas")
Output:
integral((b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)*sqrt(b*cos(d*x + c) + a)/cos(d*x + c)^(3/2), x)
Timed out. \[ \int \frac {(a+b \cos (c+d x))^{5/2}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))**(5/2)/cos(d*x+c)**(3/2),x)
Output:
Timed out
\[ \int \frac {(a+b \cos (c+d x))^{5/2}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^(5/2)/cos(d*x+c)^(3/2),x, algorithm="maxima")
Output:
integrate((b*cos(d*x + c) + a)^(5/2)/cos(d*x + c)^(3/2), x)
\[ \int \frac {(a+b \cos (c+d x))^{5/2}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^(5/2)/cos(d*x+c)^(3/2),x, algorithm="giac")
Output:
integrate((b*cos(d*x + c) + a)^(5/2)/cos(d*x + c)^(3/2), x)
Timed out. \[ \int \frac {(a+b \cos (c+d x))^{5/2}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^{3/2}} \,d x \] Input:
int((a + b*cos(c + d*x))^(5/2)/cos(c + d*x)^(3/2),x)
Output:
int((a + b*cos(c + d*x))^(5/2)/cos(c + d*x)^(3/2), x)
\[ \int \frac {(a+b \cos (c+d x))^{5/2}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=2 \left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) a b +\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) a^{2}+\left (\int \sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}d x \right ) b^{2} \] Input:
int((a+b*cos(d*x+c))^(5/2)/cos(d*x+c)^(3/2),x)
Output:
2*int((sqrt(cos(c + d*x)*b + a)*sqrt(cos(c + d*x)))/cos(c + d*x),x)*a*b + int((sqrt(cos(c + d*x)*b + a)*sqrt(cos(c + d*x)))/cos(c + d*x)**2,x)*a**2 + int(sqrt(cos(c + d*x)*b + a)*sqrt(cos(c + d*x)),x)*b**2