3.6.12 \(\int \frac {\cos ^4(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx\) [512]

Optimal. Leaf size=247 \[ \frac {2 \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 b d}-\frac {32 a \left (a^2-2 b^2\right ) E\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (c+d x)}}{35 b^4 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {8 \left (4 a^4-9 a^2 b^2+5 b^4\right ) F\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{35 b^4 d \sqrt {a+b \sin (c+d x)}}-\frac {4 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (4 a^2-5 b^2-3 a b \sin (c+d x)\right )}{35 b^3 d} \]

[Out]

2/7*cos(d*x+c)^3*(a+b*sin(d*x+c))^(1/2)/b/d-4/35*cos(d*x+c)*(4*a^2-5*b^2-3*a*b*sin(d*x+c))*(a+b*sin(d*x+c))^(1
/2)/b^3/d+32/35*a*(a^2-2*b^2)*(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)-8/35*(4
*a^4-9*a^2*b^2+5*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^4/d/(a+b*sin(d*x+c))^(1/2)

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Rubi [A]
time = 0.24, antiderivative size = 247, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2774, 2944, 2831, 2742, 2740, 2734, 2732} \begin {gather*} -\frac {32 a \left (a^2-2 b^2\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 )}{35 b^4 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {4 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (4 a^2-3 a b \sin (c+d x)-5 b^2\right )}{35 b^3 d}+\frac {8 \left (4 a^4-9 a^2 b^2+5 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 )}{35 b^4 d \sqrt {a+b \sin (c+d x)}}+\frac {2 \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]])/(7*b*d) - (32*a*(a^2 - 2*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/
(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(35*b^4*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) + (8*(4*a^4 - 9*a^2*b^2 + 5*b
^4)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(35*b^4*d*Sqrt[a + b*Sin[
c + d*x]]) - (4*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*(4*a^2 - 5*b^2 - 3*a*b*Sin[c + d*x]))/(35*b^3*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 2774

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

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 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 \frac {\cos ^4(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx &=\frac {2 \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 b d}+\frac {6 \int \frac {\cos ^2(c+d x) (b+a \sin (c+d x))}{\sqrt {a+b \sin (c+d x)}} \, dx}{7 b}\\ &=\frac {2 \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 b d}-\frac {4 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (4 a^2-5 b^2-3 a b \sin (c+d x)\right )}{35 b^3 d}+\frac {8 \int \frac {-\frac {1}{2} b \left (a^2-5 b^2\right )-2 a \left (a^2-2 b^2\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx}{35 b^3}\\ &=\frac {2 \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 b d}-\frac {4 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (4 a^2-5 b^2-3 a b \sin (c+d x)\right )}{35 b^3 d}-\frac {\left (16 a \left (a^2-2 b^2\right )\right ) \int \sqrt {a+b \sin (c+d x)} \, dx}{35 b^4}+\frac {\left (4 \left (4 a^4-9 a^2 b^2+5 b^4\right )\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}} \, dx}{35 b^4}\\ &=\frac {2 \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 b d}-\frac {4 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (4 a^2-5 b^2-3 a b \sin (c+d x)\right )}{35 b^3 d}-\frac {\left (16 a \left (a^2-2 b^2\right ) \sqrt {a+b \sin (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}} \, dx}{35 b^4 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\left (4 \left (4 a^4-9 a^2 b^2+5 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}{35 b^4 \sqrt {a+b \sin (c+d x)}}\\ &=\frac {2 \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 b d}-\frac {32 a \left (a^2-2 b^2\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)}}{35 b^4 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {8 \left (4 a^4-9 a^2 b^2+5 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}}}{35 b^4 d \sqrt {a+b \sin (c+d x)}}-\frac {4 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (4 a^2-5 b^2-3 a b \sin (c+d x)\right )}{35 b^3 d}\\ \end {align*}

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Mathematica [A]
time = 1.07, size = 219, normalized size = 0.89 \begin {gather*} \frac {64 a \left (a^3+a^2 b-2 a b^2-2 b^3\right ) E\left (\frac {1}{4} (-2 c+\pi -2 d x)|\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}-16 \left (4 a^4-9 a^2 b^2+5 b^4\right ) F\left (\frac {1}{4} (-2 c+\pi -2 d x)|\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}+b \cos (c+d x) \left (-32 a^3+62 a b^2-2 a b^2 \cos (2 (c+d x))+\left (-8 a^2 b+45 b^3\right ) \sin (c+d x)+5 b^3 \sin (3 (c+d x))\right )}{70 b^4 d \sqrt {a+b \sin (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(64*a*(a^3 + a^2*b - 2*a*b^2 - 2*b^3)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x]
)/(a + b)] - 16*(4*a^4 - 9*a^2*b^2 + 5*b^4)*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c
+ d*x])/(a + b)] + b*Cos[c + d*x]*(-32*a^3 + 62*a*b^2 - 2*a*b^2*Cos[2*(c + d*x)] + (-8*a^2*b + 45*b^3)*Sin[c +
d*x] + 5*b^3*Sin[3*(c + d*x)]))/(70*b^4*d*Sqrt[a + b*Sin[c + d*x]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(941\) vs. \(2(293)=586\).
time = 2.11, size = 942, normalized size = 3.81

method	result	size
default	\(-\frac {2 \left (-5 b^{5} \left (\sin ^{5}\left (d x +c \right )\right )+16 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticF \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{4} b -12 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticF \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{3} b^{2}-36 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticF \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{2} b^{3}+12 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticF \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a \,b^{4}+20 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticF \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b^{5}-16 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticE \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{5}+48 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticE \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{3} b^{2}-32 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticE \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a \,b^{4}+a \,b^{4} \left (\sin ^{4}\left (d x +c \right )\right )-2 a^{2} b^{3} \left (\sin ^{3}\left (d x +c \right )\right )+20 b^{5} \left (\sin ^{3}\left (d x +c \right )\right )-8 a^{3} b^{2} \left (\sin ^{2}\left (d x +c \right )\right )+14 a \,b^{4} \left (\sin ^{2}\left (d x +c \right )\right )+2 a^{2} b^{3} \sin \left (d x +c \right )-15 b^{5} \sin \left (d x +c \right )+8 a^{3} b^{2}-15 a \,b^{4}\right )}{35 b^{5} \cos \left (d x +c \right ) \sqrt {a +b \sin \left (d x +c \right )}\, d}\)	\(942\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/35*(-5*b^5*sin(d*x+c)^5+16*((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-12*((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^2-36*((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^3+12*((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^4+20*((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^5-16*((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^5+48*((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^3*b^2-32*((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*b^4+a*b^4*sin(d*x+c)^4-2*a^2*b^3*sin(d*x+c)^3+20*b^5*sin(d*x+c)^3-8*a^3*b^2*s
in(d*x+c)^2+14*a*b^4*sin(d*x+c)^2+2*a^2*b^3*sin(d*x+c)-15*b^5*sin(d*x+c)+8*a^3*b^2-15*a*b^4)/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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.14, size = 493, normalized size = 2.00 \begin {gather*} \frac {2 \, {\left (2 \, \sqrt {2} {\left (8 \, a^{4} - 19 \, a^{2} b^{2} + 15 \, 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 (8 \, a^{4} - 19 \, a^{2} b^{2} + 15 \, 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 ) - 24 \, \sqrt {2} {\left (-i \, a^{3} b + 2 i \, a b^{3}\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 ) - 24 \, \sqrt {2} {\left (i \, a^{3} b - 2 i \, a b^{3}\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 (5 \, b^{4} \cos \left (d x + c\right )^{3} + 6 \, a b^{3} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, {\left (4 \, a^{2} b^{2} - 5 \, b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt {b \sin \left (d x + c\right ) + a}\right )}}{105 \, b^{5} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(2*sqrt(2)*(8*a^4 - 19*a^2*b^2 + 15*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) + 2*sqrt(2)*(8*a^4 - 19*a^2*b
^2 + 15*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) - 24*sqrt(2)*(-I*a^3*b + 2*I*a*b^3)*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, -8/2
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)) - 24*sqrt(2)*(I*a^3*b - 2
*I*a*b^3)*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, weierstrassPI
nverse(-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*(5*b^4*cos(d*x + c)^3 + 6*a*b^3*cos(d*x + c)*sin(d*x + c) - 2*(4*a^2*b^2 - 5*b^4)*cos(d*x + c
))*sqrt(b*sin(d*x + c) + a))/(b^5*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cos ^{4}{\left (c + d x \right )}}{\sqrt {a + b \sin {\left (c + d x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\cos \left (c+d\,x\right )}^4}{\sqrt {a+b\,\sin \left (c+d\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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