3.1174 \(\int \frac {\cot ^4(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx\)
Optimal. Leaf size=353 \[ \frac {\left (16 a^2+5 b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{24 a^2 d \sqrt {a+b \sin (c+d x)}}+\frac {5 b \cot (c+d x) \csc (c+d x) \sqrt {a+b \sin (c+d x)}}{12 a^2 d}+\frac {\left (32 a^2-15 b^2\right ) \cot (c+d x) \sqrt {a+b \sin (c+d x)}}{24 a^3 d}+\frac {\left (32 a^2-15 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{24 a^3 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {b \left (12 a^2-5 b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{8 a^3 d \sqrt {a+b \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^2(c+d x) \sqrt {a+b \sin (c+d x)}}{3 a d} \]
[Out]
1/24*(32*a^2-15*b^2)*cot(d*x+c)*(a+b*sin(d*x+c))^(1/2)/a^3/d+5/12*b*cot(d*x+c)*csc(d*x+c)*(a+b*sin(d*x+c))^(1/
2)/a^2/d-1/3*cot(d*x+c)*csc(d*x+c)^2*(a+b*sin(d*x+c))^(1/2)/a/d-1/24*(32*a^2-15*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^3/d/((a+b*sin(d*x+c))/(a+b))^(1/2)-1/24*(16*a^2+5*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
)/a^2/d/(a+b*sin(d*x+c))^(1/2)-1/8*b*(12*a^2-5*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)*EllipticPi(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^3/d/(a+b*
sin(d*x+c))^(1/2)
________________________________________________________________________________________
Rubi [A] time = 0.88, antiderivative size = 353, normalized size of antiderivative =
1.00, number of steps used = 10, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\)
= 0.435, Rules used = {2725, 3055, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805}
\[ \frac {\left (32 a^2-15 b^2\right ) \cot (c+d x) \sqrt {a+b \sin (c+d x)}}{24 a^3 d}+\frac {\left (16 a^2+5 b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{24 a^2 d \sqrt {a+b \sin (c+d x)}}+\frac {\left (32 a^2-15 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{24 a^3 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {b \left (12 a^2-5 b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{8 a^3 d \sqrt {a+b \sin (c+d x)}}+\frac {5 b \cot (c+d x) \csc (c+d x) \sqrt {a+b \sin (c+d x)}}{12 a^2 d}-\frac {\cot (c+d x) \csc ^2(c+d x) \sqrt {a+b \sin (c+d x)}}{3 a d} \]
Antiderivative was successfully verified.
[In]
Int[Cot[c + d*x]^4/Sqrt[a + b*Sin[c + d*x]],x]
[Out]
((32*a^2 - 15*b^2)*Cot[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(24*a^3*d) + (5*b*Cot[c + d*x]*Csc[c + d*x]*Sqrt[a +
b*Sin[c + d*x]])/(12*a^2*d) - (Cot[c + d*x]*Csc[c + d*x]^2*Sqrt[a + b*Sin[c + d*x]])/(3*a*d) + ((32*a^2 - 15*
b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(24*a^3*d*Sqrt[(a + b*Sin[c + d*x]
)/(a + b)]) + ((16*a^2 + 5*b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)
])/(24*a^2*d*Sqrt[a + b*Sin[c + d*x]]) + (b*(12*a^2 - 5*b^2)*EllipticPi[2, (c - Pi/2 + d*x)/2, (2*b)/(a + b)]*
Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(8*a^3*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 2725
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)/tan[(e_.) + (f_.)*(x_)]^4, x_Symbol] :> -Simp[(Cos[e + f*x]*(a
+ b*Sin[e + f*x])^(m + 1))/(3*a*f*Sin[e + f*x]^3), x] + (-Dist[1/(6*a^2), Int[((a + b*Sin[e + f*x])^m*Simp[8*
a^2 - b^2*(m - 1)*(m - 2) + a*b*m*Sin[e + f*x] - (6*a^2 - b^2*m*(m - 2))*Sin[e + f*x]^2, x])/Sin[e + f*x]^2, x
], x] - Simp[(b*(m - 2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(6*a^2*f*Sin[e + f*x]^2), x]) /; FreeQ[{a,
b, e, f, m}, x] && NeQ[a^2 - b^2, 0] && !LtQ[m, -1] && IntegerQ[2*m]
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 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 3055
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*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] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 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 {\cot ^4(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx &=\frac {5 b \cot (c+d x) \csc (c+d x) \sqrt {a+b \sin (c+d x)}}{12 a^2 d}-\frac {\cot (c+d x) \csc ^2(c+d x) \sqrt {a+b \sin (c+d x)}}{3 a d}-\frac {\int \frac {\csc ^2(c+d x) \left (\frac {1}{4} \left (32 a^2-15 b^2\right )-\frac {1}{2} a b \sin (c+d x)-\frac {1}{4} \left (24 a^2-5 b^2\right ) \sin ^2(c+d x)\right )}{\sqrt {a+b \sin (c+d x)}} \, dx}{6 a^2}\\ &=\frac {\left (32 a^2-15 b^2\right ) \cot (c+d x) \sqrt {a+b \sin (c+d x)}}{24 a^3 d}+\frac {5 b \cot (c+d x) \csc (c+d x) \sqrt {a+b \sin (c+d x)}}{12 a^2 d}-\frac {\cot (c+d x) \csc ^2(c+d x) \sqrt {a+b \sin (c+d x)}}{3 a d}-\frac {\int \frac {\csc (c+d x) \left (-\frac {3}{8} b \left (12 a^2-5 b^2\right )-\frac {1}{4} a \left (24 a^2-5 b^2\right ) \sin (c+d x)-\frac {1}{8} b \left (32 a^2-15 b^2\right ) \sin ^2(c+d x)\right )}{\sqrt {a+b \sin (c+d x)}} \, dx}{6 a^3}\\ &=\frac {\left (32 a^2-15 b^2\right ) \cot (c+d x) \sqrt {a+b \sin (c+d x)}}{24 a^3 d}+\frac {5 b \cot (c+d x) \csc (c+d x) \sqrt {a+b \sin (c+d x)}}{12 a^2 d}-\frac {\cot (c+d x) \csc ^2(c+d x) \sqrt {a+b \sin (c+d x)}}{3 a d}+\frac {\int \frac {\csc (c+d x) \left (\frac {3}{8} b^2 \left (12 a^2-5 b^2\right )+\frac {1}{8} a b \left (16 a^2+5 b^2\right ) \sin (c+d x)\right )}{\sqrt {a+b \sin (c+d x)}} \, dx}{6 a^3 b}-\frac {\left (-32 a^2+15 b^2\right ) \int \sqrt {a+b \sin (c+d x)} \, dx}{48 a^3}\\ &=\frac {\left (32 a^2-15 b^2\right ) \cot (c+d x) \sqrt {a+b \sin (c+d x)}}{24 a^3 d}+\frac {5 b \cot (c+d x) \csc (c+d x) \sqrt {a+b \sin (c+d x)}}{12 a^2 d}-\frac {\cot (c+d x) \csc ^2(c+d x) \sqrt {a+b \sin (c+d x)}}{3 a d}+\frac {\left (b \left (12 a^2-5 b^2\right )\right ) \int \frac {\csc (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx}{16 a^3}+\frac {1}{48} \left (16+\frac {5 b^2}{a^2}\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}} \, dx-\frac {\left (\left (-32 a^2+15 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}{48 a^3 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}\\ &=\frac {\left (32 a^2-15 b^2\right ) \cot (c+d x) \sqrt {a+b \sin (c+d x)}}{24 a^3 d}+\frac {5 b \cot (c+d x) \csc (c+d x) \sqrt {a+b \sin (c+d x)}}{12 a^2 d}-\frac {\cot (c+d x) \csc ^2(c+d x) \sqrt {a+b \sin (c+d x)}}{3 a d}+\frac {\left (32 a^2-15 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)}}{24 a^3 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\left (b \left (12 a^2-5 b^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}{16 a^3 \sqrt {a+b \sin (c+d x)}}+\frac {\left (\left (16+\frac {5 b^2}{a^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}{48 \sqrt {a+b \sin (c+d x)}}\\ &=\frac {\left (32 a^2-15 b^2\right ) \cot (c+d x) \sqrt {a+b \sin (c+d x)}}{24 a^3 d}+\frac {5 b \cot (c+d x) \csc (c+d x) \sqrt {a+b \sin (c+d x)}}{12 a^2 d}-\frac {\cot (c+d x) \csc ^2(c+d x) \sqrt {a+b \sin (c+d x)}}{3 a d}+\frac {\left (32 a^2-15 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)}}{24 a^3 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\left (16+\frac {5 b^2}{a^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}}}{24 d \sqrt {a+b \sin (c+d x)}}+\frac {b \left (12 a^2-5 b^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}}}{8 a^3 d \sqrt {a+b \sin (c+d x)}}\\ \end {align*}
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Mathematica [C] time = 5.72, size = 475, normalized size = 1.35 \[ \frac {-\frac {4 \cot (c+d x) \sqrt {a+b \sin (c+d x)} \left (8 a^2 \csc ^2(c+d x)-32 a^2-10 a b \csc (c+d x)+15 b^2\right )}{a^3}+\frac {-\frac {8 a \left (24 a^2-5 b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )}{\sqrt {a+b \sin (c+d x)}}+\frac {2 b \left (45 b^2-104 a^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \Pi \left (2;\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )}{\sqrt {a+b \sin (c+d x)}}+\frac {2 i \left (32 a^2-15 b^2\right ) \cos (2 (c+d x)) \csc ^2(c+d x) \sec (c+d x) \sqrt {-\frac {b (\sin (c+d x)-1)}{a+b}} \sqrt {-\frac {b (\sin (c+d x)+1)}{a-b}} \left (2 a (a-b) E\left (i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \sin (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (2 a F\left (i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \sin (c+d x)}\right )|\frac {a+b}{a-b}\right )-b \Pi \left (\frac {a+b}{a};i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \sin (c+d x)}\right )|\frac {a+b}{a-b}\right )\right )\right )}{a b \sqrt {-\frac {1}{a+b}} \left (\csc ^2(c+d x)-2\right )}}{a^3}}{96 d} \]
Antiderivative was successfully verified.
[In]
Integrate[Cot[c + d*x]^4/Sqrt[a + b*Sin[c + d*x]],x]
[Out]
((-4*Cot[c + d*x]*(-32*a^2 + 15*b^2 - 10*a*b*Csc[c + d*x] + 8*a^2*Csc[c + d*x]^2)*Sqrt[a + b*Sin[c + d*x]])/a^
3 + (((2*I)*(32*a^2 - 15*b^2)*Cos[2*(c + d*x)]*Csc[c + d*x]^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*S
in[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*Sqrt[-(a + b)^(-1)]*(-2 + Csc[c + d*x]^2)) - (8*a*(24*a^2 - 5*b^2)*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*b*(-104*a^2 + 45*b^2)*Ell
ipticPi[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]])
/a^3)/(96*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(cot(d*x+c)^4/(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 \[ \int \frac {\cot \left (d x + c\right )^{4}}{\sqrt {b \sin \left (d x + c\right ) + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^4/(a+b*sin(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate(cot(d*x + c)^4/sqrt(b*sin(d*x + c) + a), x)
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maple [B] time = 2.13, size = 1496, normalized size = 4.24 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(d*x+c)^4/(a+b*sin(d*x+c))^(1/2),x)
[Out]
1/24*(48*a^5*((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)*El
lipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*sin(d*x+c)^3-16*((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*sin(d*x+c)^3-42*b^2*((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*sin(d*x+c)^3-5*b^3*(
(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*sin(d*x+c)^3+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)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2
))*a*b^4*sin(d*x+c)^3-32*((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^5*sin(d*x+c)^3+47*((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*sin(d*x+c)^3-15*((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)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a*b^4*sin(
d*x+c)^3-36*((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)*Ell
ipticPi(((a+b*sin(d*x+c))/(a-b))^(1/2),(a-b)/a,((a-b)/(a+b))^(1/2))*a^3*b^2*sin(d*x+c)^3+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)*EllipticPi(((a+b*sin(d*x+c))/(a-b
))^(1/2),(a-b)/a,((a-b)/(a+b))^(1/2))*a^2*b^3*sin(d*x+c)^3+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)*EllipticPi(((a+b*sin(d*x+c))/(a-b))^(1/2),(a-b)/a,((a-b)/(a+b))
^(1/2))*a*b^4*sin(d*x+c)^3-15*((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)*EllipticPi(((a+b*sin(d*x+c))/(a-b))^(1/2),(a-b)/a,((a-b)/(a+b))^(1/2))*b^5*sin(d*x+c)^3-32*a^3*
b^2*sin(d*x+c)^5+15*a*b^4*sin(d*x+c)^5-32*a^4*b*sin(d*x+c)^4+5*a^2*b^3*sin(d*x+c)^4+30*a^3*b^2*sin(d*x+c)^3-15
*a*b^4*sin(d*x+c)^3+40*a^4*b*sin(d*x+c)^2-5*a^2*b^3*sin(d*x+c)^2+2*a^3*b^2*sin(d*x+c)-8*a^4*b)/a^4/sin(d*x+c)^
3/b/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 \[ \int \frac {\cot \left (d x + c\right )^{4}}{\sqrt {b \sin \left (d x + c\right ) + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^4/(a+b*sin(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
integrate(cot(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 \[ \int \frac {{\mathrm {cot}\left (c+d\,x\right )}^4}{\sqrt {a+b\,\sin \left (c+d\,x\right )}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(c + d*x)^4/(a + b*sin(c + d*x))^(1/2),x)
[Out]
int(cot(c + d*x)^4/(a + b*sin(c + d*x))^(1/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{4}{\left (c + d x \right )}}{\sqrt {a + b \sin {\left (c + d x \right )}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)**4/(a+b*sin(d*x+c))**(1/2),x)
[Out]
Integral(cot(c + d*x)**4/sqrt(a + b*sin(c + d*x)), x)
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