3.1173 \(\int \frac{\cos (c+d x) \cot ^3(c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx\)

Optimal. Leaf size=307 \[ -\frac{\left (8 a^2+b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{4 a b d \sqrt{a+b \sin (c+d x)}}+\frac{\left (8 a^2+3 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{4 a^2 b d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{3 \left (4 a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{4 a^2 d \sqrt{a+b \sin (c+d x)}}+\frac{3 b \cot (c+d x) \sqrt{a+b \sin (c+d x)}}{4 a^2 d}-\frac{\cot (c+d x) \csc (c+d x) \sqrt{a+b \sin (c+d x)}}{2 a d} \]

[Out]

(3*b*Cot[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(4*a^2*d) - (Cot[c + d*x]*Csc[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(
2*a*d) + ((8*a^2 + 3*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(4*a^2*b*d*Sq
rt[(a + b*Sin[c + d*x])/(a + b)]) - ((8*a^2 + b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Si
n[c + d*x])/(a + b)])/(4*a*b*d*Sqrt[a + b*Sin[c + d*x]]) - (3*(4*a^2 - b^2)*EllipticPi[2, (c - Pi/2 + d*x)/2,
(2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(4*a^2*d*Sqrt[a + b*Sin[c + d*x]])

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Rubi [A]  time = 0.665193, antiderivative size = 307, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.31, Rules used = {2893, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805} \[ -\frac{\left (8 a^2+b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{4 a b d \sqrt{a+b \sin (c+d x)}}+\frac{\left (8 a^2+3 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{4 a^2 b d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{3 \left (4 a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{4 a^2 d \sqrt{a+b \sin (c+d x)}}+\frac{3 b \cot (c+d x) \sqrt{a+b \sin (c+d x)}}{4 a^2 d}-\frac{\cot (c+d x) \csc (c+d x) \sqrt{a+b \sin (c+d x)}}{2 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]*Cot[c + d*x]^3)/Sqrt[a + b*Sin[c + d*x]],x]

[Out]

(3*b*Cot[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(4*a^2*d) - (Cot[c + d*x]*Csc[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(
2*a*d) + ((8*a^2 + 3*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(4*a^2*b*d*Sq
rt[(a + b*Sin[c + d*x])/(a + b)]) - ((8*a^2 + b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Si
n[c + d*x])/(a + b)])/(4*a*b*d*Sqrt[a + b*Sin[c + d*x]]) - (3*(4*a^2 - b^2)*EllipticPi[2, (c - Pi/2 + d*x)/2,
(2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(4*a^2*d*Sqrt[a + b*Sin[c + d*x]])

Rule 2893

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

Rule 3059

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

Rule 3002

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^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 2807

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d*
Sin[e + f*x])/(c + d)]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && N
eQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rubi steps

\begin{align*} \int \frac{\cos (c+d x) \cot ^3(c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx &=\frac{3 b \cot (c+d x) \sqrt{a+b \sin (c+d x)}}{4 a^2 d}-\frac{\cot (c+d x) \csc (c+d x) \sqrt{a+b \sin (c+d x)}}{2 a d}-\frac{\int \frac{\csc (c+d x) \left (\frac{3}{4} \left (4 a^2-b^2\right )-\frac{1}{2} a b \sin (c+d x)-\frac{1}{4} \left (8 a^2+3 b^2\right ) \sin ^2(c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx}{2 a^2}\\ &=\frac{3 b \cot (c+d x) \sqrt{a+b \sin (c+d x)}}{4 a^2 d}-\frac{\cot (c+d x) \csc (c+d x) \sqrt{a+b \sin (c+d x)}}{2 a d}+\frac{\int \frac{\csc (c+d x) \left (-\frac{3}{4} b \left (4 a^2-b^2\right )-\frac{1}{4} a \left (8 a^2+b^2\right ) \sin (c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx}{2 a^2 b}-\frac{\left (-8 a^2-3 b^2\right ) \int \sqrt{a+b \sin (c+d x)} \, dx}{8 a^2 b}\\ &=\frac{3 b \cot (c+d x) \sqrt{a+b \sin (c+d x)}}{4 a^2 d}-\frac{\cot (c+d x) \csc (c+d x) \sqrt{a+b \sin (c+d x)}}{2 a d}-\frac{1}{8} \left (\frac{8 a}{b}+\frac{b}{a}\right ) \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx-\frac{1}{8} \left (3 \left (4-\frac{b^2}{a^2}\right )\right ) \int \frac{\csc (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx-\frac{\left (\left (-8 a^2-3 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}{8 a^2 b \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}\\ &=\frac{3 b \cot (c+d x) \sqrt{a+b \sin (c+d x)}}{4 a^2 d}-\frac{\cot (c+d x) \csc (c+d x) \sqrt{a+b \sin (c+d x)}}{2 a d}+\frac{\left (8 a^2+3 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)}}{4 a^2 b d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{\left (\left (\frac{8 a}{b}+\frac{b}{a}\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}{8 \sqrt{a+b \sin (c+d x)}}-\frac{\left (3 \left (4-\frac{b^2}{a^2}\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}\right ) \int \frac{\csc (c+d x)}{\sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}}} \, dx}{8 \sqrt{a+b \sin (c+d x)}}\\ &=\frac{3 b \cot (c+d x) \sqrt{a+b \sin (c+d x)}}{4 a^2 d}-\frac{\cot (c+d x) \csc (c+d x) \sqrt{a+b \sin (c+d x)}}{2 a d}+\frac{\left (8 a^2+3 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)}}{4 a^2 b d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{\left (\frac{8 a}{b}+\frac{b}{a}\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}}}{4 d \sqrt{a+b \sin (c+d x)}}-\frac{3 \left (4-\frac{b^2}{a^2}\right ) \Pi \left (2;\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}}}{4 d \sqrt{a+b \sin (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 3.36653, size = 443, normalized size = 1.44 \[ \frac{\frac{2 \left (16 a^2-9 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )}{a^2 \sqrt{a+b \sin (c+d x)}}+\frac{2 i \left (8 a^2+3 b^2\right ) \cos (2 (c+d x)) \csc ^2(c+d x) \sec (c+d x) \sqrt{-\frac{b (\sin (c+d x)-1)}{a+b}} \sqrt{-\frac{b (\sin (c+d x)+1)}{a-b}} \left (2 a (a-b) E\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \sin (c+d x)}\right )|\frac{a+b}{a-b}\right )+b \left (2 a F\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \sin (c+d x)}\right )|\frac{a+b}{a-b}\right )-b \Pi \left (\frac{a+b}{a};i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \sin (c+d x)}\right )|\frac{a+b}{a-b}\right )\right )\right )}{a^3 b^2 \sqrt{-\frac{1}{a+b}} \left (\csc ^2(c+d x)-2\right )}-\frac{4 \cot (c+d x) \sqrt{a+b \sin (c+d x)} (2 a \csc (c+d x)-3 b)}{a^2}-\frac{8 b \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )}{a \sqrt{a+b \sin (c+d x)}}}{16 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]*Cot[c + d*x]^3)/Sqrt[a + b*Sin[c + d*x]],x]

[Out]

(((2*I)*(8*a^2 + 3*b^2)*Cos[2*(c + d*x)]*Csc[c + d*x]^2*(2*a*(a - b)*EllipticE[I*ArcSinh[Sqrt[-(a + b)^(-1)]*S
qrt[a + b*Sin[c + d*x]]], (a + b)/(a - b)] + b*(2*a*EllipticF[I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Sin[c +
d*x]]], (a + b)/(a - b)] - b*EllipticPi[(a + b)/a, I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Sin[c + d*x]]], (
a + b)/(a - b)]))*Sec[c + d*x]*Sqrt[-((b*(-1 + Sin[c + d*x]))/(a + b))]*Sqrt[-((b*(1 + Sin[c + d*x]))/(a - b))
])/(a^3*b^2*Sqrt[-(a + b)^(-1)]*(-2 + Csc[c + d*x]^2)) - (4*Cot[c + d*x]*(-3*b + 2*a*Csc[c + d*x])*Sqrt[a + b*
Sin[c + d*x]])/a^2 - (8*b*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/
(a*Sqrt[a + b*Sin[c + d*x]]) + (2*(16*a^2 - 9*b^2)*EllipticPi[2, (-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a
+ b*Sin[c + d*x])/(a + b)])/(a^2*Sqrt[a + b*Sin[c + d*x]]))/(16*d)

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Maple [B]  time = 3.119, size = 913, normalized size = 3. \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)*cot(d*x+c)^3/(a+b*sin(d*x+c))^(1/2),x)

[Out]

(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)*(2*(a/b-1)*((a+b*sin(d*x+c))/(a-b))^(1/2)*(b*(1-sin(d*x+c))/(a+b))^(1/
2)*((-sin(d*x+c)-1)*b/(a-b))^(1/2)/(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)*((-a/b-1)*EllipticE(((a+b*sin(d*x+c
))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))+EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2)))+4*(a/b-1)*
((a+b*sin(d*x+c))/(a-b))^(1/2)*(b*(1-sin(d*x+c))/(a+b))^(1/2)*((-sin(d*x+c)-1)*b/(a-b))^(1/2)/(-(-b*sin(d*x+c)
-a)*cos(d*x+c)^2)^(1/2)/a*b*EllipticPi(((a+b*sin(d*x+c))/(a-b))^(1/2),-(-a/b+1)/a*b,((a-b)/(a+b))^(1/2))-1/2/a
*(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)/sin(d*x+c)^2+3/4/a^2*b*(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)/sin(d*
x+c)+1/2/a*b*(a/b-1)*((a+b*sin(d*x+c))/(a-b))^(1/2)*(b*(1-sin(d*x+c))/(a+b))^(1/2)*((-sin(d*x+c)-1)*b/(a-b))^(
1/2)/(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))+3/4
/a^2*b^2*(a/b-1)*((a+b*sin(d*x+c))/(a-b))^(1/2)*(b*(1-sin(d*x+c))/(a+b))^(1/2)*((-sin(d*x+c)-1)*b/(a-b))^(1/2)
/(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)*((-a/b-1)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2
))+EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2)))-1/4*(4*a^2+3*b^2)/a^3*(a/b-1)*((a+b*sin(d*x+
c))/(a-b))^(1/2)*(b*(1-sin(d*x+c))/(a+b))^(1/2)*((-sin(d*x+c)-1)*b/(a-b))^(1/2)/(-(-b*sin(d*x+c)-a)*cos(d*x+c)
^2)^(1/2)*b*EllipticPi(((a+b*sin(d*x+c))/(a-b))^(1/2),-(-a/b+1)/a*b,((a-b)/(a+b))^(1/2)))/cos(d*x+c)/(a+b*sin(
d*x+c))^(1/2)/d

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Maxima [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)*cot(d*x+c)^3/(a+b*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\cos \left (d x + c\right ) \cot \left (d x + c\right )^{3}}{\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)*cot(d*x+c)^3/(a+b*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral(cos(d*x + c)*cot(d*x + c)^3/sqrt(b*sin(d*x + c) + a), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos{\left (c + d x \right )} \cot ^{3}{\left (c + d x \right )}}{\sqrt{a + b \sin{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*cot(d*x+c)**3/(a+b*sin(d*x+c))**(1/2),x)

[Out]

Integral(cos(c + d*x)*cot(c + d*x)**3/sqrt(a + b*sin(c + d*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right ) \cot \left (d x + c\right )^{3}}{\sqrt{b \sin \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*cot(d*x+c)^3/(a+b*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)*cot(d*x + c)^3/sqrt(b*sin(d*x + c) + a), x)