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

Optimal. Leaf size=299 \[ \frac{2 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (3 b \left (25 a^2+7 b^2\right ) \sin (c+d x)+a \left (5 a^2+27 b^2\right )\right )}{315 b d}-\frac{4 a \left (22 a^2 b^2+5 a^4-27 b^4\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{315 b^2 d \sqrt{a+b \sin (c+d x)}}+\frac{4 \left (102 a^2 b^2+5 a^4+21 b^4\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{315 b^2 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{2 b \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 d}-\frac{8 a b \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{21 d} \]

[Out]

(-8*a*b*Cos[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]])/(21*d) - (2*b*Cos[c + d*x]^3*(a + b*Sin[c + d*x])^(3/2))/(9*d
) + (4*(5*a^4 + 102*a^2*b^2 + 21*b^4)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(
315*b^2*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) - (4*a*(5*a^4 + 22*a^2*b^2 - 27*b^4)*EllipticF[(c - Pi/2 + d*x)/
2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(315*b^2*d*Sqrt[a + b*Sin[c + d*x]]) + (2*Cos[c + d*x]*S
qrt[a + b*Sin[c + d*x]]*(a*(5*a^2 + 27*b^2) + 3*b*(25*a^2 + 7*b^2)*Sin[c + d*x]))/(315*b*d)

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Rubi [A]  time = 0.669294, antiderivative size = 299, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {2692, 2862, 2865, 2752, 2663, 2661, 2655, 2653} \[ \frac{2 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (3 b \left (25 a^2+7 b^2\right ) \sin (c+d x)+a \left (5 a^2+27 b^2\right )\right )}{315 b d}-\frac{4 a \left (22 a^2 b^2+5 a^4-27 b^4\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{315 b^2 d \sqrt{a+b \sin (c+d x)}}+\frac{4 \left (102 a^2 b^2+5 a^4+21 b^4\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{315 b^2 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{2 b \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 d}-\frac{8 a b \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{21 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*a*b*Cos[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]])/(21*d) - (2*b*Cos[c + d*x]^3*(a + b*Sin[c + d*x])^(3/2))/(9*d
) + (4*(5*a^4 + 102*a^2*b^2 + 21*b^4)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(
315*b^2*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) - (4*a*(5*a^4 + 22*a^2*b^2 - 27*b^4)*EllipticF[(c - Pi/2 + d*x)/
2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(315*b^2*d*Sqrt[a + b*Sin[c + d*x]]) + (2*Cos[c + d*x]*S
qrt[a + b*Sin[c + d*x]]*(a*(5*a^2 + 27*b^2) + 3*b*(25*a^2 + 7*b^2)*Sin[c + d*x]))/(315*b*d)

Rule 2692

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m + p)), x] + Dist[1/(m + p), Int[(g*Cos[e + f*x])^
p*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(m + p) + a*b*(2*m + p - 1)*Sin[e + f*x]), x], x] /; FreeQ[{
a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && NeQ[m + p, 0] && (IntegersQ[2*m, 2*p] || IntegerQ[m
])

Rule 2862

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

Rule 2865

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

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 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 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 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 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]

Rubi steps

\begin{align*} \int \cos ^2(c+d x) (a+b \sin (c+d x))^{5/2} \, dx &=-\frac{2 b \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 d}+\frac{2}{9} \int \cos ^2(c+d x) \sqrt{a+b \sin (c+d x)} \left (\frac{9 a^2}{2}+\frac{3 b^2}{2}+6 a b \sin (c+d x)\right ) \, dx\\ &=-\frac{8 a b \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{21 d}-\frac{2 b \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 d}+\frac{4}{63} \int \frac{\cos ^2(c+d x) \left (\frac{3}{4} a \left (21 a^2+11 b^2\right )+\frac{3}{4} b \left (25 a^2+7 b^2\right ) \sin (c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx\\ &=-\frac{8 a b \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{21 d}-\frac{2 b \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 d}+\frac{2 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (a \left (5 a^2+27 b^2\right )+3 b \left (25 a^2+7 b^2\right ) \sin (c+d x)\right )}{315 b d}+\frac{16 \int \frac{6 a b^2 \left (5 a^2+3 b^2\right )+\frac{3}{8} b \left (5 a^4+102 a^2 b^2+21 b^4\right ) \sin (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx}{945 b^2}\\ &=-\frac{8 a b \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{21 d}-\frac{2 b \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 d}+\frac{2 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (a \left (5 a^2+27 b^2\right )+3 b \left (25 a^2+7 b^2\right ) \sin (c+d x)\right )}{315 b d}-\frac{\left (2 a \left (5 a^4+22 a^2 b^2-27 b^4\right )\right ) \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx}{315 b^2}+\frac{\left (2 \left (5 a^4+102 a^2 b^2+21 b^4\right )\right ) \int \sqrt{a+b \sin (c+d x)} \, dx}{315 b^2}\\ &=-\frac{8 a b \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{21 d}-\frac{2 b \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 d}+\frac{2 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (a \left (5 a^2+27 b^2\right )+3 b \left (25 a^2+7 b^2\right ) \sin (c+d x)\right )}{315 b d}+\frac{\left (2 \left (5 a^4+102 a^2 b^2+21 b^4\right ) \sqrt{a+b \sin (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}} \, dx}{315 b^2 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{\left (2 a \left (5 a^4+22 a^2 b^2-27 b^4\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}\right ) \int \frac{1}{\sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}}} \, dx}{315 b^2 \sqrt{a+b \sin (c+d x)}}\\ &=-\frac{8 a b \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{21 d}-\frac{2 b \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 d}+\frac{4 \left (5 a^4+102 a^2 b^2+21 b^4\right ) E\left (\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )|\frac{2 b}{a+b}\right ) \sqrt{a+b \sin (c+d x)}}{315 b^2 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{4 a \left (5 a^4+22 a^2 b^2-27 b^4\right ) F\left (\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )|\frac{2 b}{a+b}\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}{315 b^2 d \sqrt{a+b \sin (c+d x)}}+\frac{2 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (a \left (5 a^2+27 b^2\right )+3 b \left (25 a^2+7 b^2\right ) \sin (c+d x)\right )}{315 b d}\\ \end{align*}

Mathematica [A]  time = 0.991456, size = 239, normalized size = 0.8 \[ \frac{b (a+b \sin (c+d x)) \left (\left (40 a^3-354 a b^2\right ) \cos (c+d x)+2 b \left (\sin (2 (c+d x)) \left (150 a^2-35 b^2 \cos (2 (c+d x))+7 b^2\right )-95 a b \cos (3 (c+d x))\right )\right )-16 \sqrt{\frac{a+b \sin (c+d x)}{a+b}} \left (16 b \left (5 a^3 b+3 a b^3\right ) F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )+\left (102 a^2 b^2+5 a^4+21 b^4\right ) \left ((a+b) E\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )-a F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )\right )\right )}{1260 b^2 d \sqrt{a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*(16*b*(5*a^3*b + 3*a*b^3)*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)] + (5*a^4 + 102*a^2*b^2 + 21*b^4
)*((a + b)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)] - a*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]
))*Sqrt[(a + b*Sin[c + d*x])/(a + b)] + b*(a + b*Sin[c + d*x])*((40*a^3 - 354*a*b^2)*Cos[c + d*x] + 2*b*(-95*a
*b*Cos[3*(c + d*x)] + (150*a^2 + 7*b^2 - 35*b^2*Cos[2*(c + d*x)])*Sin[2*(c + d*x)])))/(1260*b^2*d*Sqrt[a + b*S
in[c + d*x]])

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Maple [B]  time = 0.488, size = 1190, normalized size = 4. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(-35*b^6*sin(d*x+c)^6-130*a*b^5*sin(d*x+c)^5+10*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))
^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^5*b+150
*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+
b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^4*b^2+44*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(
a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^3*
b^3-108*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*Ellipti
cF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^2*b^4-54*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)
-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2
))*a*b^5-42*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*Ell
ipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^6-10*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)
-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2
))*a^6-194*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*Elli
pticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^4*b^2+162*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*
x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^
(1/2))*a^2*b^4+42*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/
2)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^6-170*a^2*b^4*sin(d*x+c)^4+49*b^6*sin(d*x+c
)^4-80*a^3*b^3*sin(d*x+c)^3+212*a*b^5*sin(d*x+c)^3-5*a^4*b^2*sin(d*x+c)^2+238*a^2*b^4*sin(d*x+c)^2-14*b^6*sin(
d*x+c)^2+80*a^3*b^3*sin(d*x+c)-82*a*b^5*sin(d*x+c)+5*a^4*b^2-68*a^2*b^4)/b^3/cos(d*x+c)/(a+b*sin(d*x+c))^(1/2)
/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cos \left (d x + c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (b^{2} \cos \left (d x + c\right )^{4} - 2 \, a b \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) -{\left (a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt{b \sin \left (d x + c\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cos \left (d x + c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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