3.486 \(\int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=113 \[ \frac {5 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{a^{5/2} d}-\frac {4 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{a^{5/2} d}-\frac {\cot (c+d x)}{a^2 d \sqrt {a \sin (c+d x)+a}} \]

[Out]

5*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))/a^(5/2)/d-4*arctanh(1/2*cos(d*x+c)*a^(1/2)*2^(1/2)/(a+a*s
in(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)-cot(d*x+c)/a^2/d/(a+a*sin(d*x+c))^(1/2)

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Rubi [A]  time = 0.52, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {2880, 2780, 2649, 206, 2773, 3044, 2985} \[ -\frac {\cot (c+d x)}{a^2 d \sqrt {a \sin (c+d x)+a}}+\frac {5 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{a^{5/2} d}-\frac {4 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{a^{5/2} d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^2*Cot[c + d*x]^2)/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(5*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(a^(5/2)*d) - (4*Sqrt[2]*ArcTanh[(Sqrt[a]*Cos[c +
d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(a^(5/2)*d) - Cot[c + d*x]/(a^2*d*Sqrt[a + a*Sin[c + d*x]])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2773

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

Rule 2780

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

Rule 2880

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

Rule 2985

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

Rule 3044

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + A*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^
m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n +
2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ[b
*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0
])

Rubi steps

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

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Mathematica [C]  time = 2.70, size = 170, normalized size = 1.50 \[ \frac {\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^5 \left (-\tan \left (\frac {1}{4} (c+d x)\right )-\cot \left (\frac {1}{4} (c+d x)\right )+2 \sec \left (\frac {1}{2} (c+d x)\right )+(32+32 i) (-1)^{3/4} \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac {1}{4} (c+d x)\right )-1\right )\right )+10 \log \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )+1\right )-10 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )-\cos \left (\frac {1}{2} (c+d x)\right )+1\right )\right )}{4 d (a (\sin (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^2*Cot[c + d*x]^2)/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5*((32 + 32*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(c +
d*x)/4])] - Cot[(c + d*x)/4] + 10*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 10*Log[1 - Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2]] + 2*Sec[(c + d*x)/2] - Tan[(c + d*x)/4]))/(4*d*(a*(1 + Sin[c + d*x]))^(5/2))

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fricas [B]  time = 0.50, size = 421, normalized size = 3.73 \[ \frac {5 \, {\left (\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right ) - 1\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) + \frac {8 \, \sqrt {2} {\left (a \cos \left (d x + c\right )^{2} - {\left (a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) - a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) - \frac {2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (d x + c\right ) + 2}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right )}{\sqrt {a}} + 4 \, \sqrt {a \sin \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{4 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d - {\left (a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^2/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.98, size = 472, normalized size = 4.18 \[ \frac {\frac {{\left (10 \, \sqrt {2} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a} + \sqrt {a}}{\sqrt {-a}}\right ) - 32 \, \sqrt {2} \sqrt {a} \arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right ) - 5 \, \sqrt {2} \sqrt {-a} \log \left (\sqrt {2} \sqrt {a} + \sqrt {a}\right ) + 20 \, \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a} + \sqrt {a}}{\sqrt {-a}}\right ) - 32 \, \sqrt {a} \arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right ) - 10 \, \sqrt {-a} \log \left (\sqrt {2} \sqrt {a} + \sqrt {a}\right ) - 3 \, \sqrt {2} \sqrt {-a} - 2 \, \sqrt {-a}\right )} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{\sqrt {2} \sqrt {-a} a^{\frac {5}{2}} + 2 \, \sqrt {-a} a^{\frac {5}{2}}} + \frac {16 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} + \sqrt {a}\right )}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - \frac {10 \, \arctan \left (-\frac {\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + \frac {5 \, \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + \frac {\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{a^{3} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + \frac {2}{{\left ({\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a\right )} a^{\frac {3}{2}} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^2/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

1/2*((10*sqrt(2)*sqrt(a)*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - 32*sqrt(2)*sqrt(a)*arctan(sqrt(a)/sqrt
(-a)) - 5*sqrt(2)*sqrt(-a)*log(sqrt(2)*sqrt(a) + sqrt(a)) + 20*sqrt(a)*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt
(-a)) - 32*sqrt(a)*arctan(sqrt(a)/sqrt(-a)) - 10*sqrt(-a)*log(sqrt(2)*sqrt(a) + sqrt(a)) - 3*sqrt(2)*sqrt(-a)
- 2*sqrt(-a))*sgn(tan(1/2*d*x + 1/2*c) + 1)/(sqrt(2)*sqrt(-a)*a^(5/2) + 2*sqrt(-a)*a^(5/2)) + 16*sqrt(2)*arcta
n(-1/2*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a) + sqrt(a))/sqrt(-a))/(sqrt(-
a)*a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)) - 10*arctan(-(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)
^2 + a))/sqrt(-a))/(sqrt(-a)*a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 5*log(abs(-sqrt(a)*tan(1/2*d*x + 1/2*c) + sq
rt(a*tan(1/2*d*x + 1/2*c)^2 + a)))/(a^(5/2)*sgn(tan(1/2*d*x + 1/2*c) + 1)) + sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a
)/(a^3*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 2/(((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)
)^2 - a)*a^(3/2)*sgn(tan(1/2*d*x + 1/2*c) + 1)))/d

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maple [A]  time = 1.07, size = 132, normalized size = 1.17 \[ -\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (\sin \left (d x +c \right ) a \left (4 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-5 \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right )\right )+\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {a}\right )}{a^{\frac {7}{2}} \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^2/(a+a*sin(d*x+c))^(5/2),x)

[Out]

-1/a^(7/2)*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(sin(d*x+c)*a*(4*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2
)*2^(1/2)/a^(1/2))-5*arctanh((a-a*sin(d*x+c))^(1/2)/a^(1/2)))+(a-a*sin(d*x+c))^(1/2)*a^(1/2))/sin(d*x+c)/cos(d
*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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maxima [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)^4*csc(d*x+c)^2/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^4}{{\sin \left (c+d\,x\right )}^2\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4/(sin(c + d*x)^2*(a + a*sin(c + d*x))^(5/2)),x)

[Out]

int(cos(c + d*x)^4/(sin(c + d*x)^2*(a + a*sin(c + d*x))^(5/2)), x)

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sympy [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)**4*csc(d*x+c)**2/(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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