Integrand size = 23, antiderivative size = 98 \[ \int \frac {(a+a \cos (c+d x))^4}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {40 a^4 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a^4 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {8 a^4 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 a^4 \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d} \] Output:
40/3*a^4*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d+2/3*a^4*sin(d*x+c)/d/cos (d*x+c)^(3/2)+8*a^4*sin(d*x+c)/d/cos(d*x+c)^(1/2)+2/3*a^4*cos(d*x+c)^(1/2) *sin(d*x+c)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 1.13 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.90 \[ \int \frac {(a+a \cos (c+d x))^4}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 a^4 \csc (c+d x) \left (-\operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {1}{4},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}+\cos (c+d x) \left (-12 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}+\cos (c+d x) \left (-1+\cos ^2(c+d x)+19 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}+4 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )\right )\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)} \] Input:
Integrate[(a + a*Cos[c + d*x])^4/Cos[c + d*x]^(5/2),x]
Output:
(-2*a^4*Csc[c + d*x]*(-(Hypergeometric2F1[-3/4, 1/2, 1/4, Cos[c + d*x]^2]* Sqrt[Sin[c + d*x]^2]) + Cos[c + d*x]*(-12*Hypergeometric2F1[-1/4, 1/2, 3/4 , Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2] + Cos[c + d*x]*(-1 + Cos[c + d*x]^2 + 19*Hypergeometric2F1[1/4, 1/2, 5/4, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2 ] + 4*Cos[c + d*x]*Hypergeometric2F1[1/2, 3/4, 7/4, Cos[c + d*x]^2]*Sqrt[S in[c + d*x]^2]))))/(3*d*Cos[c + d*x]^(3/2))
Time = 0.35 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 3236, 2009}
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.
| \(\displaystyle \int \frac {(a \cos (c+d x)+a)^4}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3236 |
| \(\displaystyle \int \left (a^4 \cos ^{\frac {3}{2}}(c+d x)+\frac {4 a^4}{\cos ^{\frac {3}{2}}(c+d x)}+\frac {a^4}{\cos ^{\frac {5}{2}}(c+d x)}+4 a^4 \sqrt {\cos (c+d x)}+\frac {6 a^4}{\sqrt {\cos (c+d x)}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {40 a^4 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a^4 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a^4 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}+\frac {8 a^4 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\) |
Input:
Int[(a + a*Cos[c + d*x])^4/Cos[c + d*x]^(5/2),x]
Output:
(40*a^4*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a^4*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + (8*a^4*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) + (2*a^4*Sqrt [Cos[c + d*x]]*Sin[c + d*x])/(3*d)
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)])^(m_.), x_Symbol] :> Int[ExpandTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && IGt Q[m, 0] && RationalQ[n]
Leaf count of result is larger than twice the leaf count of optimal. \(291\) vs. \(2(87)=174\).
Time = 15.98 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.98
| method | result | size |
| default | \(\frac {8 \sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, a^{4} \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-14 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+10 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+7 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{3 \left (4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(292\) |
| parts | \(\text {Expression too large to display}\) | \(737\) |
Input:
int((a+a*cos(d*x+c))^4/cos(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
Output:
8/3*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^4/(4*sin(1 /2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1)/sin(1/2*d*x+1/2*c)^3*(2*cos(1/2* d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-14*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c) +10*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Elliptic F(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+7*sin(1/2*d*x+1/2*c)^2* cos(1/2*d*x+1/2*c)-5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2- 1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))*(-2*sin(1/2*d*x+1/2*c)^4+s in(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.28 \[ \int \frac {(a+a \cos (c+d x))^4}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (10 i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 10 i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - {\left (a^{4} \cos \left (d x + c\right )^{2} + 12 \, a^{4} \cos \left (d x + c\right ) + a^{4}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{3 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate((a+a*cos(d*x+c))^4/cos(d*x+c)^(5/2),x, algorithm="fricas")
Output:
-2/3*(10*I*sqrt(2)*a^4*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 10*I*sqrt(2)*a^4*cos(d*x + c)^2*weierstrassPInvers e(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - (a^4*cos(d*x + c)^2 + 12*a^4*cos (d*x + c) + a^4)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2)
Timed out. \[ \int \frac {(a+a \cos (c+d x))^4}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+a*cos(d*x+c))**4/cos(d*x+c)**(5/2),x)
Output:
Timed out
\[ \int \frac {(a+a \cos (c+d x))^4}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (a \cos \left (d x + c\right ) + a\right )}^{4}}{\cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+a*cos(d*x+c))^4/cos(d*x+c)^(5/2),x, algorithm="maxima")
Output:
integrate((a*cos(d*x + c) + a)^4/cos(d*x + c)^(5/2), x)
\[ \int \frac {(a+a \cos (c+d x))^4}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (a \cos \left (d x + c\right ) + a\right )}^{4}}{\cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+a*cos(d*x+c))^4/cos(d*x+c)^(5/2),x, algorithm="giac")
Output:
integrate((a*cos(d*x + c) + a)^4/cos(d*x + c)^(5/2), x)
Time = 41.08 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.48 \[ \int \frac {(a+a \cos (c+d x))^4}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2\,\left (12\,a^4\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+19\,a^4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+a^4\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )\right )}{3\,d}+\frac {8\,a^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,a^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:
int((a + a*cos(c + d*x))^4/cos(c + d*x)^(5/2),x)
Output:
(2*(12*a^4*ellipticE(c/2 + (d*x)/2, 2) + 19*a^4*ellipticF(c/2 + (d*x)/2, 2 ) + a^4*cos(c + d*x)^(1/2)*sin(c + d*x)))/(3*d) + (8*a^4*sin(c + d*x)*hype rgeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d* x)^2)^(1/2)) + (2*a^4*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x )^2))/(3*d*cos(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2))
\[ \int \frac {(a+a \cos (c+d x))^4}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=a^{4} \left (6 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right )+\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x +4 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right )+4 \left (\int \sqrt {\cos \left (d x +c \right )}d x \right )+\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )d x \right ) \] Input:
int((a+a*cos(d*x+c))^4/cos(d*x+c)^(5/2),x)
Output:
a**4*(6*int(sqrt(cos(c + d*x))/cos(c + d*x),x) + int(sqrt(cos(c + d*x))/co s(c + d*x)**3,x) + 4*int(sqrt(cos(c + d*x))/cos(c + d*x)**2,x) + 4*int(sqr t(cos(c + d*x)),x) + int(sqrt(cos(c + d*x))*cos(c + d*x),x))