3.1172 \(\int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx\)

Optimal. Leaf size=285 \[ \frac {\left (4 a^2-7 b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 b^2 d \sqrt {a+b \sin (c+d x)}}-\frac {\left (4 a^2+3 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 a b^2 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 b d}-\frac {\cot (c+d x) \sqrt {a+b \sin (c+d x)}}{a d}-\frac {b \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{a d \sqrt {a+b \sin (c+d x)}} \]

[Out]

-2/3*cos(d*x+c)*(a+b*sin(d*x+c))^(1/2)/b/d-cot(d*x+c)*(a+b*sin(d*x+c))^(1/2)/a/d+1/3*(4*a^2+3*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)/a/b^2/d/((a+b*sin(d*x+c))/(a+b))^(1/2)-1/3*(4*a^2-7*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)*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^2/d/(a+b*sin(d*x+c))^(1/2)+b*(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)*E
llipticPi(cos(1/2*c+1/4*Pi+1/2*d*x),2,2^(1/2)*(b/(a+b))^(1/2))*((a+b*sin(d*x+c))/(a+b))^(1/2)/a/d/(a+b*sin(d*x
+c))^(1/2)

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Rubi [A]  time = 0.67, antiderivative size = 285, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {2894, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805} \[ \frac {\left (4 a^2-7 b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 b^2 d \sqrt {a+b \sin (c+d x)}}-\frac {\left (4 a^2+3 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 a b^2 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 b d}-\frac {\cot (c+d x) \sqrt {a+b \sin (c+d x)}}{a d}-\frac {b \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{a d \sqrt {a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(3*b*d) - (Cot[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(a*d) - ((4*a^2 +
3*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(3*a*b^2*d*Sqrt[(a + b*Sin[c +
d*x])/(a + b)]) + ((4*a^2 - 7*b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a +
b)])/(3*b^2*d*Sqrt[a + b*Sin[c + d*x]]) - (b*EllipticPi[2, (c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin
[c + d*x])/(a + b)])/(a*d*Sqrt[a + b*Sin[c + d*x]])

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 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]

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 2894

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*b*d*(n + 1)*(m + n + 4)), Int[(a + b*Sin[e + f*x])^m*(d*Sin[e + f*x])^(n + 1)*Simp[a^2*(n + 1)*(n
+ 2) - b^2*(m + n + 2)*(m + n + 4) + a*b*(m + 3)*Sin[e + f*x] - (a^2*(n + 1)*(n + 3) - b^2*(m + n + 3)*(m + n
+ 4))*Sin[e + f*x]^2, x], x], x] - Simp[(Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 2))/(
b*d^2*f*(m + n + 4)), x]) /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m
, 2*n]) &&  !m < -1 && LtQ[n, -1] && NeQ[m + n + 4, 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 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]

Rubi steps

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

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Mathematica [C]  time = 3.64, size = 416, normalized size = 1.46 \[ \frac {\frac {2 \left (4 a^2+9 b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )}{a b \sqrt {a+b \sin (c+d x)}}+\frac {2 i \left (4 a^2+3 b^2\right ) \sec (c+d x) \sqrt {-\frac {b (\sin (c+d x)-1)}{a+b}} \sqrt {-\frac {b (\sin (c+d x)+1)}{a-b}} \left (b \left (b \Pi \left (\frac {a+b}{a};i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \sin (c+d x)}\right )|\frac {a+b}{a-b}\right )-2 a F\left (i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \sin (c+d x)}\right )|\frac {a+b}{a-b}\right )\right )-2 a (a-b) E\left (i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \sin (c+d x)}\right )|\frac {a+b}{a-b}\right )\right )}{a^2 b^3 \sqrt {-\frac {1}{a+b}}}-\frac {4 \cot (c+d x) (2 a \sin (c+d x)+3 b) \sqrt {a+b \sin (c+d x)}}{a b}+\frac {40 \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )}{\sqrt {a+b \sin (c+d x)}}}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((2*I)*(4*a^2 + 3*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^2*b^3*Sqrt[-(a + b)^(-1)
]) - (4*Cot[c + d*x]*(3*b + 2*a*Sin[c + d*x])*Sqrt[a + b*Sin[c + d*x]])/(a*b) + (40*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*(4*a^2 + 9*b^2)*Ellipt
icPi[2, (-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(a*b*Sqrt[a + b*Sin[c + d*x]
]))/(12*d)

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 3.08, size = 704, normalized size = 2.47 \[ \frac {\sqrt {-\left (-a -b \sin \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right )}\, \left (\left (-2 a^{3} b^{2}-3 a \,b^{4}\right ) \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+\sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, \left (3 \EllipticPi \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \frac {a b -b^{2}}{a b}, \sqrt {\frac {a -b}{a +b}}\right ) a \,b^{4}-3 \EllipticPi \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \frac {a b -b^{2}}{a b}, \sqrt {\frac {a -b}{a +b}}\right ) b^{5}-4 \EllipticF \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{4} b -6 \EllipticF \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{3} b^{2}+7 \EllipticF \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{2} b^{3}+3 \EllipticF \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a \,b^{4}+4 \EllipticE \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{5}-\EllipticE \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{3} b^{2}-3 \EllipticE \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a \,b^{4}\right ) \sin \left (d x +c \right )+2 a^{2} b^{3} \left (\cos ^{4}\left (d x +c \right )\right )-5 a^{2} b^{3} \left (\cos ^{2}\left (d x +c \right )\right )\right )}{3 b^{3} \sqrt {\left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) b +\left (\cos ^{2}\left (d x +c \right )\right ) a}\, a^{2} \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {a +b \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(-(-a-b*sin(d*x+c))*cos(d*x+c)^2)^(1/2)*((-2*a^3*b^2-3*a*b^4)*sin(d*x+c)*cos(d*x+c)^2+(-b/(a+b)*sin(d*x+c)
+b/(a+b))^(1/2)*(b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2)*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*(3*EllipticPi((b/(a-b)*
sin(d*x+c)+a/(a-b))^(1/2),(a*b-b^2)/a/b,((a-b)/(a+b))^(1/2))*a*b^4-3*EllipticPi((b/(a-b)*sin(d*x+c)+a/(a-b))^(
1/2),(a*b-b^2)/a/b,((a-b)/(a+b))^(1/2))*b^5-4*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2)
)*a^4*b-6*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^3*b^2+7*EllipticF((b/(a-b)*sin(d
*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^2*b^3+3*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))
^(1/2))*a*b^4+4*EllipticE((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^5-EllipticE((b/(a-b)*sin(d
*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^3*b^2-3*EllipticE((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))
^(1/2))*a*b^4)*sin(d*x+c)+2*a^2*b^3*cos(d*x+c)^4-5*a^2*b^3*cos(d*x+c)^2)/b^3/(cos(d*x+c)^2*sin(d*x+c)*b+cos(d*
x+c)^2*a)^(1/2)/a^2/sin(d*x+c)/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 \[ \int \frac {\cos \left (d x + c\right )^{2} \cot \left (d x + c\right )^{2}}{\sqrt {b \sin \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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