\(\int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3 \, dx\) [160]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [A] (verified)
   Fricas [C] (verification not implemented)
   Sympy [F(-1)]
   Maxima [F]
   Giac [F]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 23, antiderivative size = 147 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3 \, dx=\frac {68 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {44 a^3 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {44 a^3 \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {68 a^3 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {6 a^3 \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {2 a^3 \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{9 d} \]

[Out]

68/15*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+44/21*a^3*(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+68/45*a^3*cos(d*x+c)^(3/
2)*sin(d*x+c)/d+6/7*a^3*cos(d*x+c)^(5/2)*sin(d*x+c)/d+2/9*a^3*cos(d*x+c)^(7/2)*sin(d*x+c)/d+44/21*a^3*sin(d*x+
c)*cos(d*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2836, 2715, 2720, 2719} \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3 \, dx=\frac {44 a^3 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {68 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 a^3 \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}+\frac {6 a^3 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}+\frac {68 a^3 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{45 d}+\frac {44 a^3 \sin (c+d x) \sqrt {\cos (c+d x)}}{21 d} \]

[In]

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

[Out]

(68*a^3*EllipticE[(c + d*x)/2, 2])/(15*d) + (44*a^3*EllipticF[(c + d*x)/2, 2])/(21*d) + (44*a^3*Sqrt[Cos[c + d
*x]]*Sin[c + d*x])/(21*d) + (68*a^3*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(45*d) + (6*a^3*Cos[c + d*x]^(5/2)*Sin[c
+ d*x])/(7*d) + (2*a^3*Cos[c + d*x]^(7/2)*Sin[c + d*x])/(9*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 (a^3 \cos ^{\frac {3}{2}}(c+d x)+3 a^3 \cos ^{\frac {5}{2}}(c+d x)+3 a^3 \cos ^{\frac {7}{2}}(c+d x)+a^3 \cos ^{\frac {9}{2}}(c+d x)\right ) \, dx \\ & = a^3 \int \cos ^{\frac {3}{2}}(c+d x) \, dx+a^3 \int \cos ^{\frac {9}{2}}(c+d x) \, dx+\left (3 a^3\right ) \int \cos ^{\frac {5}{2}}(c+d x) \, dx+\left (3 a^3\right ) \int \cos ^{\frac {7}{2}}(c+d x) \, dx \\ & = \frac {2 a^3 \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {6 a^3 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {6 a^3 \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {2 a^3 \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {1}{3} a^3 \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx+\frac {1}{9} \left (7 a^3\right ) \int \cos ^{\frac {5}{2}}(c+d x) \, dx+\frac {1}{5} \left (9 a^3\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{7} \left (15 a^3\right ) \int \cos ^{\frac {3}{2}}(c+d x) \, dx \\ & = \frac {18 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a^3 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {44 a^3 \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {68 a^3 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {6 a^3 \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {2 a^3 \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {1}{15} \left (7 a^3\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{7} \left (5 a^3\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {68 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {44 a^3 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {44 a^3 \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {68 a^3 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {6 a^3 \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {2 a^3 \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{9 d} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 6.71 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.73 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3 \, dx=\frac {a^3 (1+\cos (c+d x))^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \left (\frac {1428 (3 \cos (c-d x-\arctan (\tan (c)))+\cos (c+d x+\arctan (\tan (c)))) \csc (c) \sec (c)}{\sqrt {\sec ^2(c)}}-2640 \cos (c+d x) \sqrt {\cos ^2(d x-\arctan (\cot (c)))} \sqrt {\csc ^2(c)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec (d x-\arctan (\cot (c))) \sin (c)+\cos (c+d x) (-5712 \cot (c)+2910 \sin (c+d x)+1022 \sin (2 (c+d x))+270 \sin (3 (c+d x))+35 \sin (4 (c+d x)))-2856 \cos (c) \csc (d x+\arctan (\tan (c))) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sqrt {\sec ^2(c)} \sqrt {\sin ^2(d x+\arctan (\tan (c)))}\right )}{10080 d \sqrt {\cos (c+d x)}} \]

[In]

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

[Out]

(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*((1428*(3*Cos[c - d*x - ArcTan[Tan[c]]] + Cos[c + d*x + ArcTan[Ta
n[c]]])*Csc[c]*Sec[c])/Sqrt[Sec[c]^2] - 2640*Cos[c + d*x]*Sqrt[Cos[d*x - ArcTan[Cot[c]]]^2]*Sqrt[Csc[c]^2]*Hyp
ergeometricPFQ[{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]
*(-5712*Cot[c] + 2910*Sin[c + d*x] + 1022*Sin[2*(c + d*x)] + 270*Sin[3*(c + d*x)] + 35*Sin[4*(c + d*x)]) - 285
6*Cos[c]*Csc[d*x + ArcTan[Tan[c]]]*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sqrt[Se
c[c]^2]*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2]))/(10080*d*Sqrt[Cos[c + d*x]])

Maple [A] (verified)

Time = 12.95 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.77

method result size
default \(-\frac {4 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, a^{3} \left (560 \left (\cos ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-600 \left (\cos ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+212 \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+66 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-430 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+165 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-357 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+192 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{315 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) \(260\)
parts \(\text {Expression too large to display}\) \(802\)

[In]

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

[Out]

-4/315*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*(560*cos(1/2*d*x+1/2*c)^11-600*cos(1/2*d*x+
1/2*c)^9+212*cos(1/2*d*x+1/2*c)^7+66*cos(1/2*d*x+1/2*c)^5-430*cos(1/2*d*x+1/2*c)^3+165*(sin(1/2*d*x+1/2*c)^2)^
(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-357*(sin(1/2*d*x+1/2*c)^2)^(1/2)
*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+192*cos(1/2*d*x+1/2*c))/(-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 = 0.10 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.19 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3 \, dx=-\frac {2 \, {\left (165 i \, \sqrt {2} a^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 165 i \, \sqrt {2} a^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 357 i \, \sqrt {2} a^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 357 i \, \sqrt {2} a^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (35 \, a^{3} \cos \left (d x + c\right )^{3} + 135 \, a^{3} \cos \left (d x + c\right )^{2} + 238 \, a^{3} \cos \left (d x + c\right ) + 330 \, a^{3}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{315 \, d} \]

[In]

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

[Out]

-2/315*(165*I*sqrt(2)*a^3*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 165*I*sqrt(2)*a^3*weiers
trassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 357*I*sqrt(2)*a^3*weierstrassZeta(-4, 0, weierstrassPInv
erse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 357*I*sqrt(2)*a^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4,
 0, cos(d*x + c) - I*sin(d*x + c))) - (35*a^3*cos(d*x + c)^3 + 135*a^3*cos(d*x + c)^2 + 238*a^3*cos(d*x + c) +
 330*a^3)*sqrt(cos(d*x + c))*sin(d*x + c))/d

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 15.00 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.40 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3 \, dx=\frac {2\,\left (a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+a^3\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )\right )}{3\,d}-\frac {2\,\left (\frac {33\,a^3\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )}{\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {5\,a^3\,{\cos \left (c+d\,x\right )}^{11/2}\,\sin \left (c+d\,x\right )}{\sqrt {{\sin \left (c+d\,x\right )}^2}}\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{4};\ \frac {15}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{77\,d}-\frac {2\,a^3\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {104\,a^3\,{\cos \left (c+d\,x\right )}^{11/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{4};\ \frac {19}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{385\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

[In]

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

[Out]

(2*(a^3*ellipticF(c/2 + (d*x)/2, 2) + a^3*cos(c + d*x)^(1/2)*sin(c + d*x)))/(3*d) - (2*((33*a^3*cos(c + d*x)^(
7/2)*sin(c + d*x))/(sin(c + d*x)^2)^(1/2) - (5*a^3*cos(c + d*x)^(11/2)*sin(c + d*x))/(sin(c + d*x)^2)^(1/2))*h
ypergeom([1/2, 11/4], 15/4, cos(c + d*x)^2))/(77*d) - (2*a^3*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9
/4], 13/4, cos(c + d*x)^2))/(3*d*(sin(c + d*x)^2)^(1/2)) - (104*a^3*cos(c + d*x)^(11/2)*sin(c + d*x)*hypergeom
([1/2, 11/4], 19/4, cos(c + d*x)^2))/(385*d*(sin(c + d*x)^2)^(1/2))