3.1171 \(\int \frac{\cos ^3(c+d x) \cot (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx\)
Optimal. Leaf size=288 \[ -\frac{2 a \left (8 a^2-23 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{15 b^3 d \sqrt{a+b \sin (c+d x)}}+\frac{2 \left (8 a^2-21 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{15 b^3 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{8 a \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{15 b^2 d}-\frac{2 \sin (c+d x) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{5 b d}+\frac{2 \sqrt{\frac{a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \sqrt{a+b \sin (c+d x)}} \]
[Out]
(8*a*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(15*b^2*d) - (2*Cos[c + d*x]*Sin[c + d*x]*Sqrt[a + b*Sin[c + d*x]]
)/(5*b*d) + (2*(8*a^2 - 21*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(15*b^3
*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) - (2*a*(8*a^2 - 23*b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sq
rt[(a + b*Sin[c + d*x])/(a + b)])/(15*b^3*d*Sqrt[a + b*Sin[c + d*x]]) + (2*EllipticPi[2, (c - Pi/2 + d*x)/2, (
2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(d*Sqrt[a + b*Sin[c + d*x]])
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Rubi [A] time = 0.654878, antiderivative size = 288, 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 =
{2895, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805} \[ -\frac{2 a \left (8 a^2-23 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{15 b^3 d \sqrt{a+b \sin (c+d x)}}+\frac{2 \left (8 a^2-21 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{15 b^3 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{8 a \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{15 b^2 d}-\frac{2 \sin (c+d x) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{5 b d}+\frac{2 \sqrt{\frac{a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \sqrt{a+b \sin (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]^3*Cot[c + d*x])/Sqrt[a + b*Sin[c + d*x]],x]
[Out]
(8*a*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(15*b^2*d) - (2*Cos[c + d*x]*Sin[c + d*x]*Sqrt[a + b*Sin[c + d*x]]
)/(5*b*d) + (2*(8*a^2 - 21*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(15*b^3
*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) - (2*a*(8*a^2 - 23*b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sq
rt[(a + b*Sin[c + d*x])/(a + b)])/(15*b^3*d*Sqrt[a + b*Sin[c + d*x]]) + (2*EllipticPi[2, (c - Pi/2 + d*x)/2, (
2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(d*Sqrt[a + b*Sin[c + d*x]])
Rule 2895
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 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 ^3(c+d x) \cot (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx &=\frac{8 a \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{15 b^2 d}-\frac{2 \cos (c+d x) \sin (c+d x) \sqrt{a+b \sin (c+d x)}}{5 b d}-\frac{4 \int \frac{\csc (c+d x) \left (-\frac{15 b^2}{4}-\frac{1}{2} a b \sin (c+d x)-\frac{1}{4} \left (8 a^2-21 b^2\right ) \sin ^2(c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx}{15 b^2}\\ &=\frac{8 a \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{15 b^2 d}-\frac{2 \cos (c+d x) \sin (c+d x) \sqrt{a+b \sin (c+d x)}}{5 b d}+\frac{4 \int \frac{\csc (c+d x) \left (\frac{15 b^3}{4}-\frac{1}{4} a \left (8 a^2-23 b^2\right ) \sin (c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx}{15 b^3}+\frac{\left (8 a^2-21 b^2\right ) \int \sqrt{a+b \sin (c+d x)} \, dx}{15 b^3}\\ &=\frac{8 a \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{15 b^2 d}-\frac{2 \cos (c+d x) \sin (c+d x) \sqrt{a+b \sin (c+d x)}}{5 b d}-\frac{\left (a \left (8 a^2-23 b^2\right )\right ) \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx}{15 b^3}+\frac{\left (\left (8 a^2-21 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}{15 b^3 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\int \frac{\csc (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx\\ &=\frac{8 a \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{15 b^2 d}-\frac{2 \cos (c+d x) \sin (c+d x) \sqrt{a+b \sin (c+d x)}}{5 b d}+\frac{2 \left (8 a^2-21 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)}}{15 b^3 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\sqrt{\frac{a+b \sin (c+d x)}{a+b}} \int \frac{\csc (c+d x)}{\sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}}} \, dx}{\sqrt{a+b \sin (c+d x)}}-\frac{\left (a \left (8 a^2-23 b^2\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}{15 b^3 \sqrt{a+b \sin (c+d x)}}\\ &=\frac{8 a \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{15 b^2 d}-\frac{2 \cos (c+d x) \sin (c+d x) \sqrt{a+b \sin (c+d x)}}{5 b d}+\frac{2 \left (8 a^2-21 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)}}{15 b^3 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{2 a \left (8 a^2-23 b^2\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}}}{15 b^3 d \sqrt{a+b \sin (c+d x)}}+\frac{2 \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}}}{d \sqrt{a+b \sin (c+d x)}}\\ \end{align*}
Mathematica [C] time = 3.55528, size = 408, normalized size = 1.42 \[ \frac{-\frac{2 \left (8 a^2+9 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )}{\sqrt{a+b \sin (c+d x)}}+\frac{2 i \left (21 b^2-8 a^2\right ) \sec (c+d x) \sqrt{-\frac{b (\sin (c+d x)-1)}{a+b}} \sqrt{\frac{b (\sin (c+d x)+1)}{b-a}} \left (b \left (b \Pi \left (\frac{a+b}{a};i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \sin (c+d x)}\right )|\frac{a+b}{a-b}\right )-2 a F\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \sin (c+d x)}\right )|\frac{a+b}{a-b}\right )\right )-2 a (a-b) E\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \sin (c+d x)}\right )|\frac{a+b}{a-b}\right )\right )}{a b^2 \sqrt{-\frac{1}{a+b}}}+4 \cos (c+d x) (4 a-3 b \sin (c+d x)) \sqrt{a+b \sin (c+d x)}-\frac{8 a b \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )}{\sqrt{a+b \sin (c+d x)}}}{30 b^2 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(Cos[c + d*x]^3*Cot[c + d*x])/Sqrt[a + b*Sin[c + d*x]],x]
[Out]
(((2*I)*(-8*a^2 + 21*b^2)*(-2*a*(a - b)*EllipticE[I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[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*b^2*Sqrt[-(a + b)^(-1)])
+ 4*Cos[c + d*x]*(4*a - 3*b*Sin[c + d*x])*Sqrt[a + b*Sin[c + d*x]] - (8*a*b*EllipticF[(-2*c + Pi - 2*d*x)/4,
(2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/Sqrt[a + b*Sin[c + d*x]] - (2*(8*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)])/Sqrt[a + b*Sin[c + d*x]])/(30*b^2*d
)
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Maple [B] time = 1.494, size = 1018, normalized size = 3.5 \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)^3*cot(d*x+c)/(a+b*sin(d*x+c))^(1/2),x)
[Out]
2/15*(8*((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^4*b-6*((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^2-23*((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
pticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^2*b^3+21*((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^4-8*((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
llipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^5+29*((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^2-21*((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^4-15*((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)*b^4*EllipticPi(((a+b*sin(d*x+c))/(a-b))^(1/2),(a-b)/a
,((a-b)/(a+b))^(1/2))*a+15*((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)*b^5*EllipticPi(((a+b*sin(d*x+c))/(a-b))^(1/2),(a-b)/a,((a-b)/(a+b))^(1/2))+3*a*b^4*sin(d*x+c)^4-a^
2*b^3*sin(d*x+c)^3-4*a^3*b^2*sin(d*x+c)^2-3*a*b^4*sin(d*x+c)^2+a^2*b^3*sin(d*x+c)+4*a^3*b^2)/a/b^4/cos(d*x+c)/
(a+b*sin(d*x+c))^(1/2)/d
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{3} \cot \left (d x + c\right )}{\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)^3*cot(d*x+c)/(a+b*sin(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
integrate(cos(d*x + c)^3*cot(d*x + c)/sqrt(b*sin(d*x + c) + a), x)
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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 )^{3} \cot \left (d x + c\right )}{\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)^3*cot(d*x+c)/(a+b*sin(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
integral(cos(d*x + c)^3*cot(d*x + c)/sqrt(b*sin(d*x + c) + a), x)
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Sympy [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)**3*cot(d*x+c)/(a+b*sin(d*x+c))**(1/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{3} \cot \left (d x + c\right )}{\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)^3*cot(d*x+c)/(a+b*sin(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate(cos(d*x + c)^3*cot(d*x + c)/sqrt(b*sin(d*x + c) + a), x)