\(\int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3 \, dx\) [369]

Optimal result

Integrand size = 23, antiderivative size = 91 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {36 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^3 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {2 a^3 \sqrt {\cos (c+d x)} \sin (c+d x)}{d}+\frac {2 a^3 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d} \]

[Out]

36/5*a^3*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+4*a^3*(cos(1/
2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/5*a^3*cos(d*x+c)^(3/2)*sin(
d*x+c)/d+2*a^3*sin(d*x+c)*cos(d*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {4349, 3876, 3854, 3856, 2719, 2720} \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {4 a^3 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {36 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a^3 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 a^3 \sin (c+d x) \sqrt {\cos (c+d x)}}{d} \]

[In]

Int[Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^3,x]

[Out]

(36*a^3*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a^3*EllipticF[(c + d*x)/2, 2])/d + (2*a^3*Sqrt[Cos[c + d*x]]*Sin
[c + d*x])/d + (2*a^3*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d)

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3854

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Csc[c + d*x])^(n + 1)/(b*d*n)), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3876

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[Expand
Trig[(a + b*csc[e + f*x])^m*(d*csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[m, 0] && RationalQ[n]

Rule 4349

Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \sec (c+d x))^3}{\sec ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \left (\frac {a^3}{\sec ^{\frac {5}{2}}(c+d x)}+\frac {3 a^3}{\sec ^{\frac {3}{2}}(c+d x)}+\frac {3 a^3}{\sqrt {\sec (c+d x)}}+a^3 \sqrt {\sec (c+d x)}\right ) \, dx \\ & = \left (a^3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x)} \, dx+\left (a^3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \, dx+\left (3 a^3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)} \, dx+\left (3 a^3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {2 a^3 \sqrt {\cos (c+d x)} \sin (c+d x)}{d}+\frac {2 a^3 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+a^3 \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx+\left (3 a^3\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{5} \left (3 a^3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\left (a^3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \, dx \\ & = \frac {6 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a^3 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {2 a^3 \sqrt {\cos (c+d x)} \sin (c+d x)}{d}+\frac {2 a^3 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {1}{5} \left (3 a^3\right ) \int \sqrt {\cos (c+d x)} \, dx+a^3 \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {36 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^3 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {2 a^3 \sqrt {\cos (c+d x)} \sin (c+d x)}{d}+\frac {2 a^3 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 5.35 (sec) , antiderivative size = 233, normalized size of antiderivative = 2.56 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {a^3 (1+\cos (c+d x))^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \left (\frac {9 (3 \cos (c-d x-\arctan (\tan (c)))+\cos (c+d x+\arctan (\tan (c)))) \csc (c) \sec (c)}{\sqrt {\sec ^2(c)}}-20 \cos (c+d x) \sqrt {\cos ^2(d x-\arctan (\cot (c)))} \sqrt {\csc ^2(c)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec (d x-\arctan (\cot (c))) \sin (c)+\cos (c+d x) (-36 \cot (c)+10 \sin (c+d x)+\sin (2 (c+d x)))-18 \cos (c) \csc (d x+\arctan (\tan (c))) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sqrt {\sec ^2(c)} \sqrt {\sin ^2(d x+\arctan (\tan (c)))}\right )}{40 d \sqrt {\cos (c+d x)}} \]

[In]

Integrate[Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^3,x]

[Out]

(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*((9*(3*Cos[c - d*x - ArcTan[Tan[c]]] + Cos[c + d*x + ArcTan[Tan[c
]]])*Csc[c]*Sec[c])/Sqrt[Sec[c]^2] - 20*Cos[c + d*x]*Sqrt[Cos[d*x - ArcTan[Cot[c]]]^2]*Sqrt[Csc[c]^2]*Hypergeo
metricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[d*x - ArcTan[Cot[c]]]*Sin[c] + Cos[c + d*x]*(-36
*Cot[c] + 10*Sin[c + d*x] + Sin[2*(c + d*x)]) - 18*Cos[c]*Csc[d*x + ArcTan[Tan[c]]]*HypergeometricPFQ[{-1/2, -
1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sqrt[Sec[c]^2]*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2]))/(40*d*Sqrt[Cos[c
+ d*x]])

Maple [A] (verified)

Time = 19.90 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.75

method	result	size
default	\(-\frac {4 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, a^{3} \left (-4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+14 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-6 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-9 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\right )}{5 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(250\)

[In]

int(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

-4/5*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*(-4*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+1
4*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-6*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+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))-9*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Ell
ipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*
c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.11 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.63 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3 \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} a^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} a^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 9 i \, \sqrt {2} a^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 9 i \, \sqrt {2} a^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (a^{3} \cos \left (d x + c\right ) + 5 \, a^{3}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{5 \, d} \]

[In]

integrate(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-2/5*(5*I*sqrt(2)*a^3*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(2)*a^3*weierstrassP
Inverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 9*I*sqrt(2)*a^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4,
0, cos(d*x + c) + I*sin(d*x + c))) + 9*I*sqrt(2)*a^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*
x + c) - I*sin(d*x + c))) - (a^3*cos(d*x + c) + 5*a^3)*sqrt(cos(d*x + c))*sin(d*x + c))/d

Sympy [F(-1)]

Timed out. \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3 \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**(5/2)*(a+a*sec(d*x+c))**3,x)

[Out]

Timed out

Maxima [F]

\[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3 \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac {5}{2}} \,d x } \]

[In]

integrate(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^3*cos(d*x + c)^(5/2), x)

Giac [F]

\[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3 \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac {5}{2}} \,d x } \]

[In]

integrate(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^3*cos(d*x + c)^(5/2), x)

Mupad [B] (verification not implemented)

Time = 14.27 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.14 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {6\,a^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,a^3\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{d}-\frac {2\,a^3\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

[In]

int(cos(c + d*x)^(5/2)*(a + a/cos(c + d*x))^3,x)

[Out]

(6*a^3*ellipticE(c/2 + (d*x)/2, 2))/d + (4*a^3*ellipticF(c/2 + (d*x)/2, 2))/d + (2*a^3*cos(c + d*x)^(1/2)*sin(
c + d*x))/d - (2*a^3*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c
+ d*x)^2)^(1/2))