\(\int \frac {(a+a \cos (c+d x))^4}{\sqrt {\cos (c+d x)}} \, dx\) [169]

Optimal result

Integrand size = 23, antiderivative size = 121 \[ \int \frac {(a+a \cos (c+d x))^4}{\sqrt {\cos (c+d x)}} \, dx=\frac {64 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {136 a^4 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {94 a^4 \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {8 a^4 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a^4 \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d} \]

[Out]

64/5*a^4*(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+136/21*a^4*(c
os(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+8/5*a^4*cos(d*x+c)^(3/2)
*sin(d*x+c)/d+2/7*a^4*cos(d*x+c)^(5/2)*sin(d*x+c)/d+94/21*a^4*sin(d*x+c)*cos(d*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2836, 2720, 2719, 2715} \[ \int \frac {(a+a \cos (c+d x))^4}{\sqrt {\cos (c+d x)}} \, dx=\frac {136 a^4 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {64 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a^4 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}+\frac {8 a^4 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {94 a^4 \sin (c+d x) \sqrt {\cos (c+d x)}}{21 d} \]

[In]

Int[(a + a*Cos[c + d*x])^4/Sqrt[Cos[c + d*x]],x]

[Out]

(64*a^4*EllipticE[(c + d*x)/2, 2])/(5*d) + (136*a^4*EllipticF[(c + d*x)/2, 2])/(21*d) + (94*a^4*Sqrt[Cos[c + d
*x]]*Sin[c + d*x])/(21*d) + (8*a^4*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d) + (2*a^4*Cos[c + d*x]^(5/2)*Sin[c +
d*x])/(7*d)

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

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 2836

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[Expan
dTrig[(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] &
& IGtQ[m, 0] && RationalQ[n]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {a^4}{\sqrt {\cos (c+d x)}}+4 a^4 \sqrt {\cos (c+d x)}+6 a^4 \cos ^{\frac {3}{2}}(c+d x)+4 a^4 \cos ^{\frac {5}{2}}(c+d x)+a^4 \cos ^{\frac {7}{2}}(c+d x)\right ) \, dx \\ & = a^4 \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx+a^4 \int \cos ^{\frac {7}{2}}(c+d x) \, dx+\left (4 a^4\right ) \int \sqrt {\cos (c+d x)} \, dx+\left (4 a^4\right ) \int \cos ^{\frac {5}{2}}(c+d x) \, dx+\left (6 a^4\right ) \int \cos ^{\frac {3}{2}}(c+d x) \, dx \\ & = \frac {8 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a^4 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {4 a^4 \sqrt {\cos (c+d x)} \sin (c+d x)}{d}+\frac {8 a^4 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a^4 \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {1}{7} \left (5 a^4\right ) \int \cos ^{\frac {3}{2}}(c+d x) \, dx+\left (2 a^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx+\frac {1}{5} \left (12 a^4\right ) \int \sqrt {\cos (c+d x)} \, dx \\ & = \frac {64 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {6 a^4 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {94 a^4 \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {8 a^4 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a^4 \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {1}{21} \left (5 a^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {64 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {136 a^4 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {94 a^4 \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {8 a^4 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a^4 \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 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 = 5.78 (sec) , antiderivative size = 245, normalized size of antiderivative = 2.02 \[ \int \frac {(a+a \cos (c+d x))^4}{\sqrt {\cos (c+d x)}} \, dx=\frac {a^4 (1+\cos (c+d x))^4 \sec ^8\left (\frac {1}{2} (c+d x)\right ) \left (\frac {672 (3 \cos (c-d x-\arctan (\tan (c)))+\cos (c+d x+\arctan (\tan (c)))) \csc (c) \sec (c)}{\sqrt {\sec ^2(c)}}-1360 \cos (c+d x) \sqrt {\cos ^2(d x-\arctan (\cot (c)))} \sqrt {\csc ^2(c)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec (d x-\arctan (\cot (c))) \sin (c)+\cos (c+d x) (-2688 \cot (c)+955 \sin (c+d x)+168 \sin (2 (c+d x))+15 \sin (3 (c+d x)))-1344 \cos (c) \csc (d x+\arctan (\tan (c))) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sqrt {\sec ^2(c)} \sqrt {\sin ^2(d x+\arctan (\tan (c)))}\right )}{3360 d \sqrt {\cos (c+d x)}} \]

[In]

Integrate[(a + a*Cos[c + d*x])^4/Sqrt[Cos[c + d*x]],x]

[Out]

(a^4*(1 + Cos[c + d*x])^4*Sec[(c + d*x)/2]^8*((672*(3*Cos[c - d*x - ArcTan[Tan[c]]] + Cos[c + d*x + ArcTan[Tan
[c]]])*Csc[c]*Sec[c])/Sqrt[Sec[c]^2] - 1360*Cos[c + d*x]*Sqrt[Cos[d*x - ArcTan[Cot[c]]]^2]*Sqrt[Csc[c]^2]*Hype
rgeometricPFQ[{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]*
(-2688*Cot[c] + 955*Sin[c + d*x] + 168*Sin[2*(c + d*x)] + 15*Sin[3*(c + d*x)]) - 1344*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 + Arc
Tan[Tan[c]]]^2]))/(3360*d*Sqrt[Cos[c + d*x]])

Maple [A] (verified)

Time = 9.15 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.25

method	result	size
default	\(-\frac {8 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, a^{4} \left (60 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-258 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+448 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-167 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+85 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-168 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{105 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(272\)
parts	\(\text {Expression too large to display}\)	\(743\)

[In]

int((a+cos(d*x+c)*a)^4/cos(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-8/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^4*(60*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8
-258*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+448*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-167*sin(1/2*d*x+1/2*c
)^2*cos(1/2*d*x+1/2*c)+85*(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))-168*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c)
,2^(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 = 0.09 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.34 \[ \int \frac {(a+a \cos (c+d x))^4}{\sqrt {\cos (c+d x)}} \, dx=-\frac {2 \, {\left (170 i \, \sqrt {2} a^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 170 i \, \sqrt {2} a^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 336 i \, \sqrt {2} a^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 336 i \, \sqrt {2} a^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (15 \, a^{4} \cos \left (d x + c\right )^{2} + 84 \, a^{4} \cos \left (d x + c\right ) + 235 \, a^{4}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{105 \, d} \]

[In]

integrate((a+a*cos(d*x+c))^4/cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

-2/105*(170*I*sqrt(2)*a^4*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 170*I*sqrt(2)*a^4*weiers
trassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 336*I*sqrt(2)*a^4*weierstrassZeta(-4, 0, weierstrassPInv
erse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 336*I*sqrt(2)*a^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4,
0, cos(d*x + c) - I*sin(d*x + c))) - (15*a^4*cos(d*x + c)^2 + 84*a^4*cos(d*x + c) + 235*a^4)*sqrt(cos(d*x + c
))*sin(d*x + c))/d

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \cos (c+d x))^4}{\sqrt {\cos (c+d x)}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*cos(d*x+c))**4/cos(d*x+c)**(1/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+a \cos (c+d x))^4}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (a \cos \left (d x + c\right ) + a\right )}^{4}}{\sqrt {\cos \left (d x + c\right )}} \,d x } \]

[In]

integrate((a+a*cos(d*x+c))^4/cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((a*cos(d*x + c) + a)^4/sqrt(cos(d*x + c)), x)

Giac [F]

\[ \int \frac {(a+a \cos (c+d x))^4}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (a \cos \left (d x + c\right ) + a\right )}^{4}}{\sqrt {\cos \left (d x + c\right )}} \,d x } \]

[In]

integrate((a+a*cos(d*x+c))^4/cos(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((a*cos(d*x + c) + a)^4/sqrt(cos(d*x + c)), x)

Mupad [B] (verification not implemented)

Time = 15.01 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.21 \[ \int \frac {(a+a \cos (c+d x))^4}{\sqrt {\cos (c+d x)}} \, dx=\frac {2\,\left (4\,a^4\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+3\,a^4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+2\,a^4\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )\right )}{d}-\frac {8\,a^4\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,a^4\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

[In]

int((a + a*cos(c + d*x))^4/cos(c + d*x)^(1/2),x)

[Out]

(2*(4*a^4*ellipticE(c/2 + (d*x)/2, 2) + 3*a^4*ellipticF(c/2 + (d*x)/2, 2) + 2*a^4*cos(c + d*x)^(1/2)*sin(c + d
*x)))/d - (8*a^4*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)) - (2*a^4*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(9*d*(sin(c
+ d*x)^2)^(1/2))