3.500 \(\int \cos ^4(c+d x) (a+b \sin (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=398 \[ \frac {2 \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)} \left (7 b \left (53 a^2+11 b^2\right ) \sin (c+d x)+a \left (5 a^2+59 b^2\right )\right )}{3003 b d}-\frac {4 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (4 a \left (5 a^4-40 a^2 b^2-93 b^4\right )-3 b \left (5 a^4+430 a^2 b^2+77 b^4\right ) \sin (c+d x)\right )}{15015 b^3 d}+\frac {32 a \left (5 a^6-45 a^4 b^2-53 a^2 b^4+93 b^6\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{15015 b^4 d \sqrt {a+b \sin (c+d x)}}-\frac {8 \left (20 a^6-175 a^4 b^2-1662 a^2 b^4-231 b^6\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{15015 b^4 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {2 b \cos ^5(c+d x) (a+b \sin (c+d x))^{3/2}}{13 d}-\frac {32 a b \cos ^5(c+d x) \sqrt {a+b \sin (c+d x)}}{143 d} \]

[Out]

-2/13*b*cos(d*x+c)^5*(a+b*sin(d*x+c))^(3/2)/d-32/143*a*b*cos(d*x+c)^5*(a+b*sin(d*x+c))^(1/2)/d+2/3003*cos(d*x+
c)^3*(a*(5*a^2+59*b^2)+7*b*(53*a^2+11*b^2)*sin(d*x+c))*(a+b*sin(d*x+c))^(1/2)/b/d-4/15015*cos(d*x+c)*(4*a*(5*a
^4-40*a^2*b^2-93*b^4)-3*b*(5*a^4+430*a^2*b^2+77*b^4)*sin(d*x+c))*(a+b*sin(d*x+c))^(1/2)/b^3/d+8/15015*(20*a^6-
175*a^4*b^2-1662*a^2*b^4-231*b^6)*(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)*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^4/d/((a+b*sin(d*x+c))/(a+b))^(1/2)-32/
15015*a*(5*a^6-45*a^4*b^2-53*a^2*b^4+93*b^6)*(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)*Ell
ipticF(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^4/d/(a+b*sin(d*x+c)
)^(1/2)

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Rubi [A]  time = 0.94, antiderivative size = 398, normalized size of antiderivative = 1.00, number of steps used = 9, 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} \[ \frac {2 \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)} \left (7 b \left (53 a^2+11 b^2\right ) \sin (c+d x)+a \left (5 a^2+59 b^2\right )\right )}{3003 b d}-\frac {4 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (4 a \left (-40 a^2 b^2+5 a^4-93 b^4\right )-3 b \left (430 a^2 b^2+5 a^4+77 b^4\right ) \sin (c+d x)\right )}{15015 b^3 d}+\frac {32 a \left (-45 a^4 b^2-53 a^2 b^4+5 a^6+93 b^6\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{15015 b^4 d \sqrt {a+b \sin (c+d x)}}-\frac {8 \left (-175 a^4 b^2-1662 a^2 b^4+20 a^6-231 b^6\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{15015 b^4 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {2 b \cos ^5(c+d x) (a+b \sin (c+d x))^{3/2}}{13 d}-\frac {32 a b \cos ^5(c+d x) \sqrt {a+b \sin (c+d x)}}{143 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + b*Sin[c + d*x])^(5/2),x]

[Out]

(-32*a*b*Cos[c + d*x]^5*Sqrt[a + b*Sin[c + d*x]])/(143*d) - (2*b*Cos[c + d*x]^5*(a + b*Sin[c + d*x])^(3/2))/(1
3*d) - (8*(20*a^6 - 175*a^4*b^2 - 1662*a^2*b^4 - 231*b^6)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a
+ b*Sin[c + d*x]])/(15015*b^4*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) + (32*a*(5*a^6 - 45*a^4*b^2 - 53*a^2*b^4 +
93*b^6)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(15015*b^4*d*Sqrt[a
+ b*Sin[c + d*x]]) + (2*Cos[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]]*(a*(5*a^2 + 59*b^2) + 7*b*(53*a^2 + 11*b^2)*Si
n[c + d*x]))/(3003*b*d) - (4*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*(4*a*(5*a^4 - 40*a^2*b^2 - 93*b^4) - 3*b*(5
*a^4 + 430*a^2*b^2 + 77*b^4)*Sin[c + d*x]))/(15015*b^3*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 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 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 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]

Rubi steps

\begin {align*} \int \cos ^4(c+d x) (a+b \sin (c+d x))^{5/2} \, dx &=-\frac {2 b \cos ^5(c+d x) (a+b \sin (c+d x))^{3/2}}{13 d}+\frac {2}{13} \int \cos ^4(c+d x) \sqrt {a+b \sin (c+d x)} \left (\frac {13 a^2}{2}+\frac {3 b^2}{2}+8 a b \sin (c+d x)\right ) \, dx\\ &=-\frac {32 a b \cos ^5(c+d x) \sqrt {a+b \sin (c+d x)}}{143 d}-\frac {2 b \cos ^5(c+d x) (a+b \sin (c+d x))^{3/2}}{13 d}+\frac {4}{143} \int \frac {\cos ^4(c+d x) \left (\frac {1}{4} a \left (143 a^2+49 b^2\right )+\frac {3}{4} b \left (53 a^2+11 b^2\right ) \sin (c+d x)\right )}{\sqrt {a+b \sin (c+d x)}} \, dx\\ &=-\frac {32 a b \cos ^5(c+d x) \sqrt {a+b \sin (c+d x)}}{143 d}-\frac {2 b \cos ^5(c+d x) (a+b \sin (c+d x))^{3/2}}{13 d}+\frac {2 \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)} \left (a \left (5 a^2+59 b^2\right )+7 b \left (53 a^2+11 b^2\right ) \sin (c+d x)\right )}{3003 b d}+\frac {16 \int \frac {\cos ^2(c+d x) \left (3 a b^2 \left (47 a^2+17 b^2\right )+\frac {3}{8} b \left (5 a^4+430 a^2 b^2+77 b^4\right ) \sin (c+d x)\right )}{\sqrt {a+b \sin (c+d x)}} \, dx}{3003 b^2}\\ &=-\frac {32 a b \cos ^5(c+d x) \sqrt {a+b \sin (c+d x)}}{143 d}-\frac {2 b \cos ^5(c+d x) (a+b \sin (c+d x))^{3/2}}{13 d}+\frac {2 \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)} \left (a \left (5 a^2+59 b^2\right )+7 b \left (53 a^2+11 b^2\right ) \sin (c+d x)\right )}{3003 b d}-\frac {4 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (4 a \left (5 a^4-40 a^2 b^2-93 b^4\right )-3 b \left (5 a^4+430 a^2 b^2+77 b^4\right ) \sin (c+d x)\right )}{15015 b^3 d}+\frac {64 \int \frac {-\frac {3}{16} a b^2 \left (5 a^4-1450 a^2 b^2-603 b^4\right )-\frac {3}{16} b \left (20 a^6-175 a^4 b^2-1662 a^2 b^4-231 b^6\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx}{45045 b^4}\\ &=-\frac {32 a b \cos ^5(c+d x) \sqrt {a+b \sin (c+d x)}}{143 d}-\frac {2 b \cos ^5(c+d x) (a+b \sin (c+d x))^{3/2}}{13 d}+\frac {2 \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)} \left (a \left (5 a^2+59 b^2\right )+7 b \left (53 a^2+11 b^2\right ) \sin (c+d x)\right )}{3003 b d}-\frac {4 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (4 a \left (5 a^4-40 a^2 b^2-93 b^4\right )-3 b \left (5 a^4+430 a^2 b^2+77 b^4\right ) \sin (c+d x)\right )}{15015 b^3 d}-\frac {\left (4 \left (20 a^6-175 a^4 b^2-1662 a^2 b^4-231 b^6\right )\right ) \int \sqrt {a+b \sin (c+d x)} \, dx}{15015 b^4}+\frac {\left (16 a \left (5 a^6-45 a^4 b^2-53 a^2 b^4+93 b^6\right )\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}} \, dx}{15015 b^4}\\ &=-\frac {32 a b \cos ^5(c+d x) \sqrt {a+b \sin (c+d x)}}{143 d}-\frac {2 b \cos ^5(c+d x) (a+b \sin (c+d x))^{3/2}}{13 d}+\frac {2 \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)} \left (a \left (5 a^2+59 b^2\right )+7 b \left (53 a^2+11 b^2\right ) \sin (c+d x)\right )}{3003 b d}-\frac {4 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (4 a \left (5 a^4-40 a^2 b^2-93 b^4\right )-3 b \left (5 a^4+430 a^2 b^2+77 b^4\right ) \sin (c+d x)\right )}{15015 b^3 d}-\frac {\left (4 \left (20 a^6-175 a^4 b^2-1662 a^2 b^4-231 b^6\right ) \sqrt {a+b \sin (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}} \, dx}{15015 b^4 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\left (16 a \left (5 a^6-45 a^4 b^2-53 a^2 b^4+93 b^6\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}{15015 b^4 \sqrt {a+b \sin (c+d x)}}\\ &=-\frac {32 a b \cos ^5(c+d x) \sqrt {a+b \sin (c+d x)}}{143 d}-\frac {2 b \cos ^5(c+d x) (a+b \sin (c+d x))^{3/2}}{13 d}-\frac {8 \left (20 a^6-175 a^4 b^2-1662 a^2 b^4-231 b^6\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)}}{15015 b^4 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {32 a \left (5 a^6-45 a^4 b^2-53 a^2 b^4+93 b^6\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}}}{15015 b^4 d \sqrt {a+b \sin (c+d x)}}+\frac {2 \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)} \left (a \left (5 a^2+59 b^2\right )+7 b \left (53 a^2+11 b^2\right ) \sin (c+d x)\right )}{3003 b d}-\frac {4 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (4 a \left (5 a^4-40 a^2 b^2-93 b^4\right )-3 b \left (5 a^4+430 a^2 b^2+77 b^4\right ) \sin (c+d x)\right )}{15015 b^3 d}\\ \end {align*}

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Mathematica [A]  time = 1.30, size = 321, normalized size = 0.81 \[ \frac {128 \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \left (b \left (5 a^5 b-1450 a^3 b^3-603 a b^5\right ) F\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )+\left (20 a^6-175 a^4 b^2-1662 a^2 b^4-231 b^6\right ) \left ((a+b) E\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )-a F\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )\right )\right )-b (a+b \sin (c+d x)) \left (-10 a b^2 \left (20 a^2-2599 b^2\right ) \cos (3 (c+d x))+140 b^3 \left (22 b^2-53 a^2\right ) \sin (4 (c+d x))-b \left (480 a^4+56120 a^2 b^2+4697 b^4\right ) \sin (2 (c+d x))+4 a \left (320 a^4-2710 a^2 b^2+6453 b^4\right ) \cos (c+d x)+5670 a b^4 \cos (5 (c+d x))+1155 b^5 \sin (6 (c+d x))\right )}{240240 b^4 d \sqrt {a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + b*Sin[c + d*x])^(5/2),x]

[Out]

(128*(b*(5*a^5*b - 1450*a^3*b^3 - 603*a*b^5)*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)] + (20*a^6 - 175*a
^4*b^2 - 1662*a^2*b^4 - 231*b^6)*((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])*(4*a*(320*a^4 -
2710*a^2*b^2 + 6453*b^4)*Cos[c + d*x] - 10*a*b^2*(20*a^2 - 2599*b^2)*Cos[3*(c + d*x)] + 5670*a*b^4*Cos[5*(c +
d*x)] - b*(480*a^4 + 56120*a^2*b^2 + 4697*b^4)*Sin[2*(c + d*x)] + 140*b^3*(-53*a^2 + 22*b^2)*Sin[4*(c + d*x)]
+ 1155*b^5*Sin[6*(c + d*x)]))/(240240*b^4*d*Sqrt[a + b*Sin[c + d*x]])

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fricas [F]  time = 1.13, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (b^{2} \cos \left (d x + c\right )^{6} - 2 \, a b \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - {\left (a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{4}\right )} \sqrt {b \sin \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral(-(b^2*cos(d*x + c)^6 - 2*a*b*cos(d*x + c)^4*sin(d*x + c) - (a^2 + b^2)*cos(d*x + c)^4)*sqrt(b*sin(d*x
+ c) + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(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)^4, x)

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maple [B]  time = 0.96, size = 1619, normalized size = 4.07 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a+b*sin(d*x+c))^(5/2),x)

[Out]

-2/15015*(40*a^6*b^2-3990*a*b^7*sin(d*x+c)^7+80*((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^7*b-11606*a^2
*b^6*sin(d*x+c)^2+10*a^5*b^3*sin(d*x+c)-4780*a^3*b^5*sin(d*x+c)+2104*a*b^7*sin(d*x+c)-4690*a^2*b^6*sin(d*x+c)^
6-1880*a^3*b^5*sin(d*x+c)^5+11290*a*b^7*sin(d*x+c)^5+5*a^4*b^4*sin(d*x+c)^4+14500*a^2*b^6*sin(d*x+c)^4-10*a^5*
b^3*sin(d*x+c)^3+6660*a^3*b^5*sin(d*x+c)^3-9404*a*b^7*sin(d*x+c)^3-40*a^6*b^2*sin(d*x+c)^2+340*a^4*b^4*sin(d*x
+c)^2-924*((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)*Ellip
ticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^8-1155*b^8*sin(d*x+c)^8+3080*b^8*sin(d*x+c)^6-2233*
b^8*sin(d*x+c)^4+308*b^8*sin(d*x+c)^2-345*a^4*b^4+1796*a^2*b^6+924*((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))*b^8-80*((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^8-60*((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^6*b^2-720*((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)*E
llipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^5*b^3-5100*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(si
n(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^4-848*((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^5+4236*((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^2*b^6+1488*((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^7+780*((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*b^2+5948*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+si
n(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^4*b^4-5724*((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^6)/b^5/cos(d*x+c)/(a+b*sin(d*x+c))^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(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)^4, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\cos \left (c+d\,x\right )}^4\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4*(a + b*sin(c + d*x))^(5/2),x)

[Out]

int(cos(c + d*x)^4*(a + b*sin(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)**4*(a+b*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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