\(\int (a+b \cos (c+d x))^{3/2} \, dx\) [51]

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

Optimal result

Integrand size = 14, antiderivative size = 157 \[ \int (a+b \cos (c+d x))^{3/2} \, dx=\frac {8 a \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{3 d \sqrt {a+b \cos (c+d x)}}+\frac {2 b \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 d} \]

[Out]

2/3*b*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/d+8/3*a*(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)*(b/(a+b))^(1/2))*(a+b*cos(d*x+c))^(1/2)/d/((a+b*cos(d*x+c))/(a+b))^(1/2)-2/3*(a^2-b^2)*
(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)*(b/(a+b))^(1/2))*((a+b*co
s(d*x+c))/(a+b))^(1/2)/d/(a+b*cos(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2735, 2831, 2742, 2740, 2734, 2732} \[ \int (a+b \cos (c+d x))^{3/2} \, dx=-\frac {2 \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{3 d \sqrt {a+b \cos (c+d x)}}+\frac {2 b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}+\frac {8 a \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}} \]

[In]

Int[(a + b*Cos[c + d*x])^(3/2),x]

[Out]

(8*a*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(3*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)])
- (2*(a^2 - b^2)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(3*d*Sqrt[a + b*Cos
[c + d*x]]) + (2*b*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*d)

Rule 2732

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2
+ d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2734

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
 d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
 b^2, 0] &&  !GtQ[a + b, 0]

Rule 2735

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((a + b*Sin[c + d*x])^(n
- 1)/(d*n)), x] + Dist[1/n, Int[(a + b*Sin[c + d*x])^(n - 2)*Simp[a^2*n + b^2*(n - 1) + a*b*(2*n - 1)*Sin[c +
d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]

Rule 2740

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*Sqrt[a + b]))*EllipticF[(1/2)*(c - P
i/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2742

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2831

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
 - a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {2 b \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2}{3} \int \frac {\frac {1}{2} \left (3 a^2+b^2\right )+2 a b \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx \\ & = \frac {2 b \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 d}+\frac {1}{3} (4 a) \int \sqrt {a+b \cos (c+d x)} \, dx+\frac {1}{3} \left (-a^2+b^2\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx \\ & = \frac {2 b \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 d}+\frac {\left (4 a \sqrt {a+b \cos (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{3 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (\left (-a^2+b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}} \, dx}{3 \sqrt {a+b \cos (c+d x)}} \\ & = \frac {8 a \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{3 d \sqrt {a+b \cos (c+d x)}}+\frac {2 b \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.85 \[ \int (a+b \cos (c+d x))^{3/2} \, dx=\frac {8 a (a+b) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-2 \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )+2 b (a+b \cos (c+d x)) \sin (c+d x)}{3 d \sqrt {a+b \cos (c+d x)}} \]

[In]

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

[Out]

(8*a*(a + b)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticE[(c + d*x)/2, (2*b)/(a + b)] - 2*(a^2 - b^2)*Sqrt[(a
+ b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)] + 2*b*(a + b*Cos[c + d*x])*Sin[c + d*x])/(3*d
*Sqrt[a + b*Cos[c + d*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(449\) vs. \(2(199)=398\).

Time = 4.09 (sec) , antiderivative size = 450, normalized size of antiderivative = 2.87

method result size
default \(-\frac {2 \sqrt {\left (2 b \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a -b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (4 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}+2 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a b -6 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}-a^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a -b}{a -b}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )+b^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a -b}{a -b}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )+4 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a -b}{a -b}}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a^{2}-4 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a -b}{a -b}}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a b -2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a b +2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{2}\right )}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +\left (a +b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b}\, d}\) \(450\)

[In]

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

[Out]

-2/3*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*cos(1/2*d*x+1/2*c)^5*b^2+2*cos(1/2*d*x+1/2
*c)^3*a*b-6*cos(1/2*d*x+1/2*c)^3*b^2-a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(
1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c
)^2+a-b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+4*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*co
s(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2-4*(sin(1/2*d*x+1/2*c
)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b-2
*cos(1/2*d*x+1/2*c)*a*b+2*cos(1/2*d*x+1/2*c)*b^2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)
/sin(1/2*d*x+1/2*c)/(-2*b*sin(1/2*d*x+1/2*c)^2+a+b)^(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 = 398, normalized size of antiderivative = 2.54 \[ \int (a+b \cos (c+d x))^{3/2} \, dx=\frac {12 i \, \sqrt {2} a b^{\frac {3}{2}} {\rm weierstrassZeta}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right )\right ) - 12 i \, \sqrt {2} a b^{\frac {3}{2}} {\rm weierstrassZeta}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right )\right ) + 6 \, \sqrt {b \cos \left (d x + c\right ) + a} b^{2} \sin \left (d x + c\right ) + \sqrt {2} {\left (-i \, a^{2} - 3 i \, b^{2}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right ) + \sqrt {2} {\left (i \, a^{2} + 3 i \, b^{2}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right )}{9 \, b d} \]

[In]

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

[Out]

1/9*(12*I*sqrt(2)*a*b^(3/2)*weierstrassZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, weierstrassP
Inverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*a
)/b)) - 12*I*sqrt(2)*a*b^(3/2)*weierstrassZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, weierstra
ssPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) +
2*a)/b)) + 6*sqrt(b*cos(d*x + c) + a)*b^2*sin(d*x + c) + sqrt(2)*(-I*a^2 - 3*I*b^2)*sqrt(b)*weierstrassPInvers
e(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*a)/b) +
 sqrt(2)*(I*a^2 + 3*I*b^2)*sqrt(b)*weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1
/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) + 2*a)/b))/(b*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^{3/2} \, dx=\int {\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2} \,d x \]

[In]

int((a + b*cos(c + d*x))^(3/2),x)

[Out]

int((a + b*cos(c + d*x))^(3/2), x)