3.6.1 \(\int \cos ^2(c+d x) (a+b \sin (c+d x))^{5/2} \, dx\) [501]

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

[Out]

-2/9*b*cos(d*x+c)^3*(a+b*sin(d*x+c))^(3/2)/d-8/21*a*b*cos(d*x+c)^3*(a+b*sin(d*x+c))^(1/2)/d+2/315*cos(d*x+c)*(
a*(5*a^2+27*b^2)+3*b*(25*a^2+7*b^2)*sin(d*x+c))*(a+b*sin(d*x+c))^(1/2)/b/d-4/315*(5*a^4+102*a^2*b^2+21*b^4)*(s
in(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2)*(b/(a+
b))^(1/2))*(a+b*sin(d*x+c))^(1/2)/b^2/d/((a+b*sin(d*x+c))/(a+b))^(1/2)+4/315*a*(5*a^4+22*a^2*b^2-27*b^4)*(sin(
1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticF(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2)*(b/(a+b))
^(1/2))*((a+b*sin(d*x+c))/(a+b))^(1/2)/b^2/d/(a+b*sin(d*x+c))^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.42, antiderivative size = 299, normalized size of antiderivative = 1.00, 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 = {2771, 2941, 2944, 2831, 2742, 2740, 2734, 2732} \begin {gather*} \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 (5 a^4+22 a^2 b^2-27 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 (5 a^4+102 a^2 b^2+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} \end {gather*}

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 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 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 2771

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 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]

Rule 2941

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] &&
GtQ[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 2944

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]

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 = 1.01, size = 239, normalized size = 0.80 \begin {gather*} \frac {-16 \left (16 b \left (5 a^3 b+3 a b^3\right ) F\left (\frac {1}{4} (-2 c+\pi -2 d x)|\frac {2 b}{a+b}\right )+\left (5 a^4+102 a^2 b^2+21 b^4\right ) \left ((a+b) E\left (\frac {1}{4} (-2 c+\pi -2 d x)|\frac {2 b}{a+b}\right )-a F\left (\frac {1}{4} (-2 c+\pi -2 d x)|\frac {2 b}{a+b}\right )\right )\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}+b (a+b \sin (c+d x)) \left (\left (40 a^3-354 a b^2\right ) \cos (c+d x)+2 b \left (-95 a b \cos (3 (c+d x))+\left (150 a^2+7 b^2-35 b^2 \cos (2 (c+d x))\right ) \sin (2 (c+d x))\right )\right )}{1260 b^2 d \sqrt {a+b \sin (c+d x)}} \end {gather*}

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]])

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1189\) vs. \(2(341)=682\).
time = 2.31, size = 1190, normalized size = 3.98

method	result	size
default	\(\text {Expression too large to display}\)	\(1190\)

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,method=_RETURNVERBOSE)

[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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.14, size = 536, normalized size = 1.79 \begin {gather*} -\frac {2 \, {\left (2 \, \sqrt {2} {\left (5 \, a^{5} - 18 \, a^{3} b^{2} - 51 \, a b^{4}\right )} \sqrt {i \, b} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 i \, a^{3} - 9 i \, 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 i \, a}{3 \, b}\right ) + 2 \, \sqrt {2} {\left (5 \, a^{5} - 18 \, a^{3} b^{2} - 51 \, a b^{4}\right )} \sqrt {-i \, b} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (-8 i \, a^{3} + 9 i \, 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 i \, a}{3 \, b}\right ) + 3 \, \sqrt {2} {\left (5 i \, a^{4} b + 102 i \, a^{2} b^{3} + 21 i \, b^{5}\right )} \sqrt {i \, b} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 i \, a^{3} - 9 i \, 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 i \, a^{3} - 9 i \, 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 i \, a}{3 \, b}\right )\right ) + 3 \, \sqrt {2} {\left (-5 i \, a^{4} b - 102 i \, a^{2} b^{3} - 21 i \, b^{5}\right )} \sqrt {-i \, b} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (-8 i \, a^{3} + 9 i \, 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 i \, a^{3} + 9 i \, 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 i \, a}{3 \, b}\right )\right ) + 3 \, {\left (95 \, a b^{4} \cos \left (d x + c\right )^{3} - {\left (5 \, a^{3} b^{2} + 27 \, a b^{4}\right )} \cos \left (d x + c\right ) + {\left (35 \, b^{5} \cos \left (d x + c\right )^{3} - 3 \, {\left (25 \, a^{2} b^{3} + 7 \, b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )} \sqrt {b \sin \left (d x + c\right ) + a}\right )}}{945 \, b^{3} d} \end {gather*}

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]

-2/945*(2*sqrt(2)*(5*a^5 - 18*a^3*b^2 - 51*a*b^4)*sqrt(I*b)*weierstrassPInverse(-4/3*(4*a^2 - 3*b^2)/b^2, -8/2
7*(8*I*a^3 - 9*I*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) - 2*I*a)/b) + 2*sqrt(2)*(5*a^5 - 18*a^
3*b^2 - 51*a*b^4)*sqrt(-I*b)*weierstrassPInverse(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(-8*I*a^3 + 9*I*a*b^2)/b^3, 1
/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*I*a)/b) + 3*sqrt(2)*(5*I*a^4*b + 102*I*a^2*b^3 + 21*I*b^5)*sqrt(
I*b)*weierstrassZeta(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*I*a^3 - 9*I*a*b^2)/b^3, weierstrassPInverse(-4/3*(4*a^
2 - 3*b^2)/b^2, -8/27*(8*I*a^3 - 9*I*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) - 2*I*a)/b)) + 3*s
qrt(2)*(-5*I*a^4*b - 102*I*a^2*b^3 - 21*I*b^5)*sqrt(-I*b)*weierstrassZeta(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(-8*
I*a^3 + 9*I*a*b^2)/b^3, weierstrassPInverse(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(-8*I*a^3 + 9*I*a*b^2)/b^3, 1/3*(3
*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*I*a)/b)) + 3*(95*a*b^4*cos(d*x + c)^3 - (5*a^3*b^2 + 27*a*b^4)*cos(d*
x + c) + (35*b^5*cos(d*x + c)^3 - 3*(25*a^2*b^3 + 7*b^5)*cos(d*x + c))*sin(d*x + c))*sqrt(b*sin(d*x + c) + a))
/(b^3*d)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sin {\left (c + d x \right )}\right )^{\frac {5}{2}} \cos ^{2}{\left (c + d x \right )}\, dx \end {gather*}

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]

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________