3.503 \(\int (a+b \cos (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=197 \[ -\frac {16 a \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (23 a^2+9 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 b \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}+\frac {16 a b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{15 d} \]

[Out]

2/5*b*(a+b*cos(d*x+c))^(3/2)*sin(d*x+c)/d+16/15*a*b*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/d+2/15*(23*a^2+9*b^2)*(c
os(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)-16/15*a*(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*cos(d*x+c))/(a+b))^(1/2)/d/(a+b*cos(d*x+c))^(1/2
)

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Rubi [A]  time = 0.26, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2656, 2753, 2752, 2663, 2661, 2655, 2653} \[ -\frac {16 a \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (23 a^2+9 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 b \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}+\frac {16 a b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{15 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^(5/2),x]

[Out]

(2*(23*a^2 + 9*b^2)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(15*d*Sqrt[(a + b*Cos[c +
d*x])/(a + b)]) - (16*a*(a^2 - b^2)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/
(15*d*Sqrt[a + b*Cos[c + d*x]]) + (16*a*b*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(15*d) + (2*b*(a + b*Cos[c +
d*x])^(3/2)*Sin[c + d*x])/(5*d)

Rule 2653

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

Rule 2655

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*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2656

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 2661

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

Rule 2663

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*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2752

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]

Rule 2753

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

Rubi steps

\begin {align*} \int (a+b \cos (c+d x))^{5/2} \, dx &=\frac {2 b (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {2}{5} \int \sqrt {a+b \cos (c+d x)} \left (\frac {1}{2} \left (5 a^2+3 b^2\right )+4 a b \cos (c+d x)\right ) \, dx\\ &=\frac {16 a b \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 b (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {4}{15} \int \frac {\frac {1}{4} a \left (15 a^2+17 b^2\right )+\frac {1}{4} b \left (23 a^2+9 b^2\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx\\ &=\frac {16 a b \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 b (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}-\frac {1}{15} \left (8 a \left (a^2-b^2\right )\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx+\frac {1}{15} \left (23 a^2+9 b^2\right ) \int \sqrt {a+b \cos (c+d x)} \, dx\\ &=\frac {16 a b \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 b (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {\left (\left (23 a^2+9 b^2\right ) \sqrt {a+b \cos (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{15 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (8 a \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}{15 \sqrt {a+b \cos (c+d x)}}\\ &=\frac {2 \left (23 a^2+9 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {16 a \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 d \sqrt {a+b \cos (c+d x)}}+\frac {16 a b \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 b (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.77, size = 177, normalized size = 0.90 \[ \frac {b \sin (c+d x) \left (22 a^2+28 a b \cos (c+d x)+3 b^2 \cos (2 (c+d x))+3 b^2\right )-16 a \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )+2 \left (23 a^3+23 a^2 b+9 a b^2+9 b^3\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 d \sqrt {a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(23*a^3 + 23*a^2*b + 9*a*b^2 + 9*b^3)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticE[(c + d*x)/2, (2*b)/(a +
b)] - 16*a*(a^2 - b^2)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)] + b*(22*a^2 +
3*b^2 + 28*a*b*Cos[c + d*x] + 3*b^2*Cos[2*(c + d*x)])*Sin[c + d*x])/(15*d*Sqrt[a + b*Cos[c + d*x]])

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fricas [F]  time = 1.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {b \cos \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)*sqrt(b*cos(d*x + c) + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.86, size = 662, normalized size = 3.36 \[ -\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (24 \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}+56 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \,b^{2}-48 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}+22 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2} b -84 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \,b^{2}+30 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}-8 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b}{a -b}}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a^{3}+8 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b}{a -b}}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a \,b^{2}+23 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b}{a -b}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a^{3}-23 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b}{a -b}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a^{2} b +9 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b}{a -b}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a \,b^{2}-9 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b}{a -b}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) b^{3}-22 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} b +28 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \,b^{2}-6 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{3}\right )}{15 \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 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(24*cos(1/2*d*x+1/2*c)^7*b^3+56*cos(1/2*d*x+
1/2*c)^5*a*b^2-48*cos(1/2*d*x+1/2*c)^5*b^3+22*cos(1/2*d*x+1/2*c)^3*a^2*b-84*cos(1/2*d*x+1/2*c)^3*a*b^2+30*cos(
1/2*d*x+1/2*c)^3*b^3-8*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)*EllipticF(cos
(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3+8*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^
(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^2+23*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1
/2*c)^2*b+a-b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3-23*(sin(1/2*d*x+1/2*c)^2)^(1/
2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b+9*(sin(
1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))
^(1/2))*a*b^2-9*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*
x+1/2*c),(-2*b/(a-b))^(1/2))*b^3-22*cos(1/2*d*x+1/2*c)*a^2*b+28*cos(1/2*d*x+1/2*c)*a*b^2-6*cos(1/2*d*x+1/2*c)*
b^3)/(-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*sin(1/2*d*x+1/2*c)^2*
b+a+b)^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*cos(c + d*x))^(5/2),x)

[Out]

int((a + b*cos(c + d*x))^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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