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

Optimal. Leaf size=308 \[ -\frac {2 \left (10 a^2-49 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{315 b d}-\frac {4 a \left (5 a^2-57 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{315 b d}+\frac {4 a \left (5 a^4-62 a^2 b^2+57 b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{315 b^2 d \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (10 a^4-279 a^2 b^2-147 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{315 b^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 \sin (c+d x) (a+b \cos (c+d x))^{7/2}}{9 b d}-\frac {4 a \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{63 b d} \]

[Out]

-2/315*(10*a^2-49*b^2)*(a+b*cos(d*x+c))^(3/2)*sin(d*x+c)/b/d-4/63*a*(a+b*cos(d*x+c))^(5/2)*sin(d*x+c)/b/d+2/9*
(a+b*cos(d*x+c))^(7/2)*sin(d*x+c)/b/d-4/315*a*(5*a^2-57*b^2)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b/d-2/315*(10*a
^4-279*a^2*b^2-147*b^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)*(
b/(a+b))^(1/2))*(a+b*cos(d*x+c))^(1/2)/b^2/d/((a+b*cos(d*x+c))/(a+b))^(1/2)+4/315*a*(5*a^4-62*a^2*b^2+57*b^4)*
(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)/b^2/d/(a+b*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.51, antiderivative size = 308, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2791, 2753, 2752, 2663, 2661, 2655, 2653} \[ -\frac {2 \left (10 a^2-49 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{315 b d}-\frac {4 a \left (5 a^2-57 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{315 b d}+\frac {4 a \left (-62 a^2 b^2+5 a^4+57 b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{315 b^2 d \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (-279 a^2 b^2+10 a^4-147 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{315 b^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 \sin (c+d x) (a+b \cos (c+d x))^{7/2}}{9 b d}-\frac {4 a \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{63 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(10*a^4 - 279*a^2*b^2 - 147*b^4)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(315*b^2*
d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + (4*a*(5*a^4 - 62*a^2*b^2 + 57*b^4)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*
EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(315*b^2*d*Sqrt[a + b*Cos[c + d*x]]) - (4*a*(5*a^2 - 57*b^2)*Sqrt[a + b
*Cos[c + d*x]]*Sin[c + d*x])/(315*b*d) - (2*(10*a^2 - 49*b^2)*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(315*b*
d) - (4*a*(a + b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(63*b*d) + (2*(a + b*Cos[c + d*x])^(7/2)*Sin[c + d*x])/(9*b
*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 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]

Rule 2791

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

Rubi steps

\begin {align*} \int \cos ^2(c+d x) (a+b \cos (c+d x))^{5/2} \, dx &=\frac {2 (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b d}+\frac {2 \int \left (\frac {7 b}{2}-a \cos (c+d x)\right ) (a+b \cos (c+d x))^{5/2} \, dx}{9 b}\\ &=-\frac {4 a (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{63 b d}+\frac {2 (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b d}+\frac {4 \int (a+b \cos (c+d x))^{3/2} \left (\frac {39 a b}{4}-\frac {1}{4} \left (10 a^2-49 b^2\right ) \cos (c+d x)\right ) \, dx}{63 b}\\ &=-\frac {2 \left (10 a^2-49 b^2\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{315 b d}-\frac {4 a (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{63 b d}+\frac {2 (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b d}+\frac {8 \int \sqrt {a+b \cos (c+d x)} \left (\frac {3}{8} b \left (55 a^2+49 b^2\right )-\frac {3}{4} a \left (5 a^2-57 b^2\right ) \cos (c+d x)\right ) \, dx}{315 b}\\ &=-\frac {4 a \left (5 a^2-57 b^2\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{315 b d}-\frac {2 \left (10 a^2-49 b^2\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{315 b d}-\frac {4 a (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{63 b d}+\frac {2 (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b d}+\frac {16 \int \frac {\frac {3}{16} a b \left (155 a^2+261 b^2\right )-\frac {3}{16} \left (10 a^4-279 a^2 b^2-147 b^4\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{945 b}\\ &=-\frac {4 a \left (5 a^2-57 b^2\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{315 b d}-\frac {2 \left (10 a^2-49 b^2\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{315 b d}-\frac {4 a (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{63 b d}+\frac {2 (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b d}-\frac {\left (10 a^4-279 a^2 b^2-147 b^4\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{315 b^2}+\frac {\left (2 a \left (5 a^4-62 a^2 b^2+57 b^4\right )\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{315 b^2}\\ &=-\frac {4 a \left (5 a^2-57 b^2\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{315 b d}-\frac {2 \left (10 a^2-49 b^2\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{315 b d}-\frac {4 a (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{63 b d}+\frac {2 (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b d}-\frac {\left (\left (10 a^4-279 a^2 b^2-147 b^4\right ) \sqrt {a+b \cos (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{315 b^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (2 a \left (5 a^4-62 a^2 b^2+57 b^4\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}{315 b^2 \sqrt {a+b \cos (c+d x)}}\\ &=-\frac {2 \left (10 a^4-279 a^2 b^2-147 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{315 b^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {4 a \left (5 a^4-62 a^2 b^2+57 b^4\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 )}{315 b^2 d \sqrt {a+b \cos (c+d x)}}-\frac {4 a \left (5 a^2-57 b^2\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{315 b d}-\frac {2 \left (10 a^2-49 b^2\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{315 b d}-\frac {4 a (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{63 b d}+\frac {2 (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b d}\\ \end {align*}

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Mathematica [A]  time = 1.38, size = 263, normalized size = 0.85 \[ \frac {16 a \left (5 a^4-62 a^2 b^2+57 b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )+b \sin (c+d x) \left (40 a^4+4 a b \left (160 a^2+619 b^2\right ) \cos (c+d x)+1984 a^2 b^2+8 \left (85 a^2 b^2+42 b^4\right ) \cos (2 (c+d x))+260 a b^3 \cos (3 (c+d x))+35 b^4 \cos (4 (c+d x))+301 b^4\right )-8 \left (10 a^5+10 a^4 b-279 a^3 b^2-279 a^2 b^3-147 a b^4-147 b^5\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{1260 b^2 d \sqrt {a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*(10*a^5 + 10*a^4*b - 279*a^3*b^2 - 279*a^2*b^3 - 147*a*b^4 - 147*b^5)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*E
llipticE[(c + d*x)/2, (2*b)/(a + b)] + 16*a*(5*a^4 - 62*a^2*b^2 + 57*b^4)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*E
llipticF[(c + d*x)/2, (2*b)/(a + b)] + b*(40*a^4 + 1984*a^2*b^2 + 301*b^4 + 4*a*b*(160*a^2 + 619*b^2)*Cos[c +
d*x] + 8*(85*a^2*b^2 + 42*b^4)*Cos[2*(c + d*x)] + 260*a*b^3*Cos[3*(c + d*x)] + 35*b^4*Cos[4*(c + d*x)])*Sin[c
+ d*x])/(1260*b^2*d*Sqrt[a + b*Cos[c + d*x]])

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fricas [F]  time = 1.29, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{2} \cos \left (d x + c\right )^{4} + 2 \, a b \cos \left (d x + c\right )^{3} + a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt {b \cos \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*cos(d*x + c)^4 + 2*a*b*cos(d*x + c)^3 + a^2*cos(d*x + c)^2)*sqrt(b*cos(d*x + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.95, size = 995, normalized size = 3.23 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-1120*b^5*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1
/2*c)^10+(2080*a*b^4+2240*b^5)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-1360*a^2*b^3-3120*a*b^4-2072*b^5)*sin
(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(320*a^3*b^2+1360*a^2*b^3+2408*a*b^4+952*b^5)*sin(1/2*d*x+1/2*c)^4*cos(1/
2*d*x+1/2*c)+(-10*a^4*b-160*a^3*b^2-666*a^2*b^3-684*a*b^4-168*b^5)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+10*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(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))*a^5-124*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*E
llipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3*b^2+114*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(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))*a*b^4-10*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*(-2*b/(a-b)*sin(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
^5+10*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(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^4*b+279*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(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^3*b^2-279*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*s
in(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*b^3+147*(sin(1/2*d
*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(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^4-147*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(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))*b^5)/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*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 \[ \int {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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