\(\int \cos ^4(c+d x) \sin ^2(c+d x) \sqrt {a+b \sin (c+d x)} \, dx\) [1143]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [C] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 31, antiderivative size = 463 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\frac {16 a \left (160 a^4-279 a^2 b^2+27 b^4\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{45045 b^5 d}-\frac {8 \left (480 a^4-937 a^2 b^2+231 b^4\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{45045 b^5 d}+\frac {8 a \left (40 a^2-81 b^2\right ) \cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{3/2}}{3003 b^4 d}-\frac {10 \left (16 a^2-33 b^2\right ) \cos (c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^{3/2}}{1287 b^3 d}+\frac {20 a \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{143 b^2 d}-\frac {2 \cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^{3/2}}{13 b d}-\frac {8 \left (320 a^6-798 a^4 b^2+435 a^2 b^4-693 b^6\right ) E\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (c+d x)}}{45045 b^6 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {16 a \left (160 a^6-439 a^4 b^2+306 a^2 b^4-27 b^6\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{45045 b^6 d \sqrt {a+b \sin (c+d x)}} \]

[Out]

-8/45045*(480*a^4-937*a^2*b^2+231*b^4)*cos(d*x+c)*(a+b*sin(d*x+c))^(3/2)/b^5/d+8/3003*a*(40*a^2-81*b^2)*cos(d*
x+c)*sin(d*x+c)*(a+b*sin(d*x+c))^(3/2)/b^4/d-10/1287*(16*a^2-33*b^2)*cos(d*x+c)*sin(d*x+c)^2*(a+b*sin(d*x+c))^
(3/2)/b^3/d+20/143*a*cos(d*x+c)*sin(d*x+c)^3*(a+b*sin(d*x+c))^(3/2)/b^2/d-2/13*cos(d*x+c)*sin(d*x+c)^4*(a+b*si
n(d*x+c))^(3/2)/b/d+16/45045*a*(160*a^4-279*a^2*b^2+27*b^4)*cos(d*x+c)*(a+b*sin(d*x+c))^(1/2)/b^5/d+8/45045*(3
20*a^6-798*a^4*b^2+435*a^2*b^4-693*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)*Elliptic
E(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^6/d/((a+b*sin(d*x+c))/(a+b))^(1/
2)-16/45045*a*(160*a^6-439*a^4*b^2+306*a^2*b^4-27*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)*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^6/d/(a+b*
sin(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 463, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {2974, 3128, 3102, 2832, 2831, 2742, 2740, 2734, 2732} \[ \int \cos ^4(c+d x) \sin ^2(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\frac {8 a \left (40 a^2-81 b^2\right ) \sin (c+d x) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{3003 b^4 d}-\frac {10 \left (16 a^2-33 b^2\right ) \sin ^2(c+d x) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{1287 b^3 d}-\frac {8 \left (480 a^4-937 a^2 b^2+231 b^4\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{45045 b^5 d}+\frac {16 a \left (160 a^4-279 a^2 b^2+27 b^4\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{45045 b^5 d}+\frac {16 a \left (160 a^6-439 a^4 b^2+306 a^2 b^4-27 b^6\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),\frac {2 b}{a+b}\right )}{45045 b^6 d \sqrt {a+b \sin (c+d x)}}-\frac {8 \left (320 a^6-798 a^4 b^2+435 a^2 b^4-693 b^6\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{45045 b^6 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {20 a \sin ^3(c+d x) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{143 b^2 d}-\frac {2 \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{13 b d} \]

[In]

Int[Cos[c + d*x]^4*Sin[c + d*x]^2*Sqrt[a + b*Sin[c + d*x]],x]

[Out]

(16*a*(160*a^4 - 279*a^2*b^2 + 27*b^4)*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(45045*b^5*d) - (8*(480*a^4 - 93
7*a^2*b^2 + 231*b^4)*Cos[c + d*x]*(a + b*Sin[c + d*x])^(3/2))/(45045*b^5*d) + (8*a*(40*a^2 - 81*b^2)*Cos[c + d
*x]*Sin[c + d*x]*(a + b*Sin[c + d*x])^(3/2))/(3003*b^4*d) - (10*(16*a^2 - 33*b^2)*Cos[c + d*x]*Sin[c + d*x]^2*
(a + b*Sin[c + d*x])^(3/2))/(1287*b^3*d) + (20*a*Cos[c + d*x]*Sin[c + d*x]^3*(a + b*Sin[c + d*x])^(3/2))/(143*
b^2*d) - (2*Cos[c + d*x]*Sin[c + d*x]^4*(a + b*Sin[c + d*x])^(3/2))/(13*b*d) - (8*(320*a^6 - 798*a^4*b^2 + 435
*a^2*b^4 - 693*b^6)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(45045*b^6*d*Sqrt[(
a + b*Sin[c + d*x])/(a + b)]) + (16*a*(160*a^6 - 439*a^4*b^2 + 306*a^2*b^4 - 27*b^6)*EllipticF[(c - Pi/2 + d*x
)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(45045*b^6*d*Sqrt[a + b*Sin[c + d*x]])

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 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 2832

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)*Sim
p[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 2974

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[a*(n + 3)*Cos[e + f*x]*(d*Sin[e + f*x])^(n + 1)*((a + b*Sin[e + f*x])^(m + 1)/(b^2*d*f*(m
+ n + 3)*(m + n + 4))), x] + (-Dist[1/(b^2*(m + n + 3)*(m + n + 4)), Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x
])^m*Simp[a^2*(n + 1)*(n + 3) - b^2*(m + n + 3)*(m + n + 4) + a*b*m*Sin[e + f*x] - (a^2*(n + 2)*(n + 3) - b^2*
(m + n + 3)*(m + n + 5))*Sin[e + f*x]^2, x], x], x] - Simp[Cos[e + f*x]*(d*Sin[e + f*x])^(n + 2)*((a + b*Sin[e
 + f*x])^(m + 1)/(b*d^2*f*(m + n + 4))), x]) /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[
m, 0] || IntegersQ[2*m, 2*n]) &&  !m < -1 &&  !LtQ[n, -1] && NeQ[m + n + 3, 0] && NeQ[m + n + 4, 0]

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*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[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 3128

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e
+ f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*
x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n +
2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d
^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rubi steps \begin{align*} \text {integral}& = \frac {20 a \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{143 b^2 d}-\frac {2 \cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^{3/2}}{13 b d}-\frac {4 \int \sin ^2(c+d x) \sqrt {a+b \sin (c+d x)} \left (\frac {1}{4} \left (60 a^2-143 b^2\right )+\frac {1}{2} a b \sin (c+d x)-\frac {5}{4} \left (16 a^2-33 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{143 b^2} \\ & = -\frac {10 \left (16 a^2-33 b^2\right ) \cos (c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^{3/2}}{1287 b^3 d}+\frac {20 a \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{143 b^2 d}-\frac {2 \cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^{3/2}}{13 b d}-\frac {8 \int \sin (c+d x) \sqrt {a+b \sin (c+d x)} \left (-\frac {5}{2} a \left (16 a^2-33 b^2\right )-\frac {1}{2} b \left (5 a^2+33 b^2\right ) \sin (c+d x)+\frac {3}{2} a \left (40 a^2-81 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{1287 b^3} \\ & = \frac {8 a \left (40 a^2-81 b^2\right ) \cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{3/2}}{3003 b^4 d}-\frac {10 \left (16 a^2-33 b^2\right ) \cos (c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^{3/2}}{1287 b^3 d}+\frac {20 a \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{143 b^2 d}-\frac {2 \cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^{3/2}}{13 b d}-\frac {16 \int \sqrt {a+b \sin (c+d x)} \left (\frac {3}{2} a^2 \left (40 a^2-81 b^2\right )+5 a b \left (2 a^2-3 b^2\right ) \sin (c+d x)-\frac {1}{4} \left (480 a^4-937 a^2 b^2+231 b^4\right ) \sin ^2(c+d x)\right ) \, dx}{9009 b^4} \\ & = -\frac {8 \left (480 a^4-937 a^2 b^2+231 b^4\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{45045 b^5 d}+\frac {8 a \left (40 a^2-81 b^2\right ) \cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{3/2}}{3003 b^4 d}-\frac {10 \left (16 a^2-33 b^2\right ) \cos (c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^{3/2}}{1287 b^3 d}+\frac {20 a \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{143 b^2 d}-\frac {2 \cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^{3/2}}{13 b d}-\frac {32 \int \sqrt {a+b \sin (c+d x)} \left (-\frac {3}{8} b \left (80 a^4-127 a^2 b^2+231 b^4\right )+\frac {3}{4} a \left (160 a^4-279 a^2 b^2+27 b^4\right ) \sin (c+d x)\right ) \, dx}{45045 b^5} \\ & = \frac {16 a \left (160 a^4-279 a^2 b^2+27 b^4\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{45045 b^5 d}-\frac {8 \left (480 a^4-937 a^2 b^2+231 b^4\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{45045 b^5 d}+\frac {8 a \left (40 a^2-81 b^2\right ) \cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{3/2}}{3003 b^4 d}-\frac {10 \left (16 a^2-33 b^2\right ) \cos (c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^{3/2}}{1287 b^3 d}+\frac {20 a \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{143 b^2 d}-\frac {2 \cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^{3/2}}{13 b d}-\frac {64 \int \frac {\frac {3}{16} a b \left (80 a^4-177 a^2 b^2-639 b^4\right )+\frac {3}{16} \left (320 a^6-798 a^4 b^2+435 a^2 b^4-693 b^6\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx}{135135 b^5} \\ & = \frac {16 a \left (160 a^4-279 a^2 b^2+27 b^4\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{45045 b^5 d}-\frac {8 \left (480 a^4-937 a^2 b^2+231 b^4\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{45045 b^5 d}+\frac {8 a \left (40 a^2-81 b^2\right ) \cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{3/2}}{3003 b^4 d}-\frac {10 \left (16 a^2-33 b^2\right ) \cos (c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^{3/2}}{1287 b^3 d}+\frac {20 a \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{143 b^2 d}-\frac {2 \cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^{3/2}}{13 b d}-\frac {\left (4 \left (320 a^6-798 a^4 b^2+435 a^2 b^4-693 b^6\right )\right ) \int \sqrt {a+b \sin (c+d x)} \, dx}{45045 b^6}+\frac {\left (8 a \left (160 a^6-439 a^4 b^2+306 a^2 b^4-27 b^6\right )\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}} \, dx}{45045 b^6} \\ & = \frac {16 a \left (160 a^4-279 a^2 b^2+27 b^4\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{45045 b^5 d}-\frac {8 \left (480 a^4-937 a^2 b^2+231 b^4\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{45045 b^5 d}+\frac {8 a \left (40 a^2-81 b^2\right ) \cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{3/2}}{3003 b^4 d}-\frac {10 \left (16 a^2-33 b^2\right ) \cos (c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^{3/2}}{1287 b^3 d}+\frac {20 a \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{143 b^2 d}-\frac {2 \cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^{3/2}}{13 b d}-\frac {\left (4 \left (320 a^6-798 a^4 b^2+435 a^2 b^4-693 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}{45045 b^6 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\left (8 a \left (160 a^6-439 a^4 b^2+306 a^2 b^4-27 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}{45045 b^6 \sqrt {a+b \sin (c+d x)}} \\ & = \frac {16 a \left (160 a^4-279 a^2 b^2+27 b^4\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{45045 b^5 d}-\frac {8 \left (480 a^4-937 a^2 b^2+231 b^4\right ) \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{45045 b^5 d}+\frac {8 a \left (40 a^2-81 b^2\right ) \cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{3/2}}{3003 b^4 d}-\frac {10 \left (16 a^2-33 b^2\right ) \cos (c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^{3/2}}{1287 b^3 d}+\frac {20 a \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{143 b^2 d}-\frac {2 \cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^{3/2}}{13 b d}-\frac {8 \left (320 a^6-798 a^4 b^2+435 a^2 b^4-693 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)}}{45045 b^6 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {16 a \left (160 a^6-439 a^4 b^2+306 a^2 b^4-27 b^6\right ) \operatorname {EllipticF}\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}}}{45045 b^6 d \sqrt {a+b \sin (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.96 (sec) , antiderivative size = 327, normalized size of antiderivative = 0.71 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\frac {\sqrt {a+b \sin (c+d x)} \left (128 \left (320 a^6-798 a^4 b^2+435 a^2 b^4-693 b^6\right ) E\left (\frac {1}{4} (-2 c+\pi -2 d x)|\frac {2 b}{a+b}\right )-256 a \left (160 a^5-160 a^4 b-279 a^3 b^2+279 a^2 b^3+27 a b^4-27 b^5\right ) \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),\frac {2 b}{a+b}\right )-2 b \cos (c+d x) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \left (10240 a^5-21056 a^3 b^2+5898 a b^4-1600 \left (2 a^3 b^2-3 a b^4\right ) \cos (2 (c+d x))+630 a b^4 \cos (4 (c+d x))-7680 a^4 b \sin (c+d x)+13592 a^2 b^3 \sin (c+d x)-19866 b^5 \sin (c+d x)+1400 a^2 b^3 \sin (3 (c+d x))+5775 b^5 \sin (3 (c+d x))+3465 b^5 \sin (5 (c+d x))\right )\right )}{720720 b^6 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}} \]

[In]

Integrate[Cos[c + d*x]^4*Sin[c + d*x]^2*Sqrt[a + b*Sin[c + d*x]],x]

[Out]

(Sqrt[a + b*Sin[c + d*x]]*(128*(320*a^6 - 798*a^4*b^2 + 435*a^2*b^4 - 693*b^6)*EllipticE[(-2*c + Pi - 2*d*x)/4
, (2*b)/(a + b)] - 256*a*(160*a^5 - 160*a^4*b - 279*a^3*b^2 + 279*a^2*b^3 + 27*a*b^4 - 27*b^5)*EllipticF[(-2*c
 + Pi - 2*d*x)/4, (2*b)/(a + b)] - 2*b*Cos[c + d*x]*Sqrt[(a + b*Sin[c + d*x])/(a + b)]*(10240*a^5 - 21056*a^3*
b^2 + 5898*a*b^4 - 1600*(2*a^3*b^2 - 3*a*b^4)*Cos[2*(c + d*x)] + 630*a*b^4*Cos[4*(c + d*x)] - 7680*a^4*b*Sin[c
 + d*x] + 13592*a^2*b^3*Sin[c + d*x] - 19866*b^5*Sin[c + d*x] + 1400*a^2*b^3*Sin[3*(c + d*x)] + 5775*b^5*Sin[3
*(c + d*x)] + 3465*b^5*Sin[5*(c + d*x)])))/(720720*b^6*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1618\) vs. \(2(493)=986\).

Time = 2.57 (sec) , antiderivative size = 1619, normalized size of antiderivative = 3.50

method result size
default \(\text {Expression too large to display}\) \(1619\)

[In]

int(cos(d*x+c)^4*sin(d*x+c)^2*(a+b*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/45045*(2772*((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^8-1280*((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^8-2772*((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^8+1280*((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-960*((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^2-3512*((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^3+2484*((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^4+2448*((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^5-4296*((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)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^2*b^6-216*((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+4472*((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^2-4932*((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*si
n(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^4*b^4+4512*((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^
6-3780*a*b^7*sin(d*x+c)^7+35*a^2*b^6*sin(d*x+c)^6-50*a^3*b^5*sin(d*x+c)^5+10470*a*b^7*sin(d*x+c)^5+80*a^4*b^4*
sin(d*x+c)^4-232*a^2*b^6*sin(d*x+c)^4-160*a^5*b^3*sin(d*x+c)^3+454*a^3*b^5*sin(d*x+c)^3-8322*a*b^7*sin(d*x+c)^
3-640*a^6*b^2*sin(d*x+c)^2+1436*a^4*b^4*sin(d*x+c)^2+640*a^6*b^2-1516*a^4*b^4+708*a^2*b^6-511*a^2*b^6*sin(d*x+
c)^2+160*a^5*b^3*sin(d*x+c)-404*a^3*b^5*sin(d*x+c)+1632*a*b^7*sin(d*x+c)-3465*b^8*sin(d*x+c)^8+9240*b^8*sin(d*
x+c)^6-6699*b^8*sin(d*x+c)^4+924*b^8*sin(d*x+c)^2)/b^7/cos(d*x+c)/(a+b*sin(d*x+c))^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.15 (sec) , antiderivative size = 633, normalized size of antiderivative = 1.37 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\frac {2 \, {\left (2 \, \sqrt {2} {\left (640 \, a^{7} - 1836 \, a^{5} b^{2} + 1401 \, a^{3} b^{4} + 531 \, a b^{6}\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 (640 \, a^{7} - 1836 \, a^{5} b^{2} + 1401 \, a^{3} b^{4} + 531 \, a b^{6}\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 ) - 6 \, \sqrt {2} {\left (-320 i \, a^{6} b + 798 i \, a^{4} b^{3} - 435 i \, a^{2} b^{5} + 693 i \, b^{7}\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 ) - 6 \, \sqrt {2} {\left (320 i \, a^{6} b - 798 i \, a^{4} b^{3} + 435 i \, a^{2} b^{5} - 693 i \, b^{7}\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 (315 \, a b^{6} \cos \left (d x + c\right )^{5} - 5 \, {\left (80 \, a^{3} b^{4} - 57 \, a b^{6}\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (160 \, a^{5} b^{2} - 279 \, a^{3} b^{4} + 27 \, a b^{6}\right )} \cos \left (d x + c\right ) + {\left (3465 \, b^{7} \cos \left (d x + c\right )^{5} + 35 \, {\left (10 \, a^{2} b^{5} - 33 \, b^{7}\right )} \cos \left (d x + c\right )^{3} - 6 \, {\left (80 \, a^{4} b^{3} - 127 \, a^{2} b^{5} + 231 \, b^{7}\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 )}}{135135 \, b^{7} d} \]

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2*(a+b*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/135135*(2*sqrt(2)*(640*a^7 - 1836*a^5*b^2 + 1401*a^3*b^4 + 531*a*b^6)*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) + 2*
sqrt(2)*(640*a^7 - 1836*a^5*b^2 + 1401*a^3*b^4 + 531*a*b^6)*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) - 6*sqrt(2)*(-
320*I*a^6*b + 798*I*a^4*b^3 - 435*I*a^2*b^5 + 693*I*b^7)*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)) - 6*sqrt(2)*(320*I*a^6*b - 798*I*a^4*b^3 + 435*I*a^2*
b^5 - 693*I*b^7)*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, weiers
trassPInverse(-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*(315*a*b^6*cos(d*x + c)^5 - 5*(80*a^3*b^4 - 57*a*b^6)*cos(d*x + c)^3 + 4*(160*a^5*b^2
- 279*a^3*b^4 + 27*a*b^6)*cos(d*x + c) + (3465*b^7*cos(d*x + c)^5 + 35*(10*a^2*b^5 - 33*b^7)*cos(d*x + c)^3 -
6*(80*a^4*b^3 - 127*a^2*b^5 + 231*b^7)*cos(d*x + c))*sin(d*x + c))*sqrt(b*sin(d*x + c) + a))/(b^7*d)

Sympy [F]

\[ \int \cos ^4(c+d x) \sin ^2(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\int \sqrt {a + b \sin {\left (c + d x \right )}} \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}\, dx \]

[In]

integrate(cos(d*x+c)**4*sin(d*x+c)**2*(a+b*sin(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(a + b*sin(c + d*x))*sin(c + d*x)**2*cos(c + d*x)**4, x)

Maxima [F]

\[ \int \cos ^4(c+d x) \sin ^2(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\int { \sqrt {b \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )^{2} \,d x } \]

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2*(a+b*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \cos ^4(c+d x) \sin ^2(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\int { \sqrt {b \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )^{2} \,d x } \]

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2*(a+b*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) \sin ^2(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\int {\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^2\,\sqrt {a+b\,\sin \left (c+d\,x\right )} \,d x \]

[In]

int(cos(c + d*x)^4*sin(c + d*x)^2*(a + b*sin(c + d*x))^(1/2),x)

[Out]

int(cos(c + d*x)^4*sin(c + d*x)^2*(a + b*sin(c + d*x))^(1/2), x)