3.480 \(\int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=220 \[ -\frac {3 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{128 a^{3/2} d}-\frac {\cot (c+d x) \csc ^4(c+d x) \sqrt {a \sin (c+d x)+a}}{5 a^2 d}-\frac {3 \cot (c+d x)}{128 a d \sqrt {a \sin (c+d x)+a}}+\frac {19 \cot (c+d x) \csc ^3(c+d x)}{40 a d \sqrt {a \sin (c+d x)+a}}-\frac {\cot (c+d x) \csc ^2(c+d x)}{80 a d \sqrt {a \sin (c+d x)+a}}-\frac {\cot (c+d x) \csc (c+d x)}{64 a d \sqrt {a \sin (c+d x)+a}} \]

[Out]

-3/128*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))/a^(3/2)/d-3/128*cot(d*x+c)/a/d/(a+a*sin(d*x+c))^(1/2
)-1/64*cot(d*x+c)*csc(d*x+c)/a/d/(a+a*sin(d*x+c))^(1/2)-1/80*cot(d*x+c)*csc(d*x+c)^2/a/d/(a+a*sin(d*x+c))^(1/2
)+19/40*cot(d*x+c)*csc(d*x+c)^3/a/d/(a+a*sin(d*x+c))^(1/2)-1/5*cot(d*x+c)*csc(d*x+c)^4*(a+a*sin(d*x+c))^(1/2)/
a^2/d

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Rubi [A]  time = 0.88, antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {2880, 2772, 2773, 206, 3044, 2980} \[ -\frac {3 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{128 a^{3/2} d}-\frac {\cot (c+d x) \csc ^4(c+d x) \sqrt {a \sin (c+d x)+a}}{5 a^2 d}-\frac {3 \cot (c+d x)}{128 a d \sqrt {a \sin (c+d x)+a}}+\frac {19 \cot (c+d x) \csc ^3(c+d x)}{40 a d \sqrt {a \sin (c+d x)+a}}-\frac {\cot (c+d x) \csc ^2(c+d x)}{80 a d \sqrt {a \sin (c+d x)+a}}-\frac {\cot (c+d x) \csc (c+d x)}{64 a d \sqrt {a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(Cot[c + d*x]^4*Csc[c + d*x]^2)/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(-3*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(128*a^(3/2)*d) - (3*Cot[c + d*x])/(128*a*d*Sqrt
[a + a*Sin[c + d*x]]) - (Cot[c + d*x]*Csc[c + d*x])/(64*a*d*Sqrt[a + a*Sin[c + d*x]]) - (Cot[c + d*x]*Csc[c +
d*x]^2)/(80*a*d*Sqrt[a + a*Sin[c + d*x]]) + (19*Cot[c + d*x]*Csc[c + d*x]^3)/(40*a*d*Sqrt[a + a*Sin[c + d*x]])
- (Cot[c + d*x]*Csc[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]])/(5*a^2*d)

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 2772

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[((b*c - a*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]]), x]
+ Dist[((2*n + 3)*(b*c - a*d))/(2*b*(n + 1)*(c^2 - d^2)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
+ 1), 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] &
& LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]

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 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 2980

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n + 1)
*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), 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] && LtQ[n, -1]

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 {\cot ^4(c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx &=\frac {\int \csc ^6(c+d x) \sqrt {a+a \sin (c+d x)} \left (1+\sin ^2(c+d x)\right ) \, dx}{a^2}-\frac {2 \int \csc ^5(c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{a^2}\\ &=\frac {\cot (c+d x) \csc ^3(c+d x)}{2 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{5 a^2 d}+\frac {\int \csc ^5(c+d x) \sqrt {a+a \sin (c+d x)} \left (\frac {a}{2}+\frac {17}{2} a \sin (c+d x)\right ) \, dx}{5 a^3}-\frac {7 \int \csc ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{4 a^2}\\ &=\frac {7 \cot (c+d x) \csc ^2(c+d x)}{12 a d \sqrt {a+a \sin (c+d x)}}+\frac {19 \cot (c+d x) \csc ^3(c+d x)}{40 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{5 a^2 d}-\frac {35 \int \csc ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{24 a^2}+\frac {143 \int \csc ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{80 a^2}\\ &=\frac {35 \cot (c+d x) \csc (c+d x)}{48 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^2(c+d x)}{80 a d \sqrt {a+a \sin (c+d x)}}+\frac {19 \cot (c+d x) \csc ^3(c+d x)}{40 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{5 a^2 d}-\frac {35 \int \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{32 a^2}+\frac {143 \int \csc ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{96 a^2}\\ &=\frac {35 \cot (c+d x)}{32 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x)}{64 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^2(c+d x)}{80 a d \sqrt {a+a \sin (c+d x)}}+\frac {19 \cot (c+d x) \csc ^3(c+d x)}{40 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{5 a^2 d}-\frac {35 \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{64 a^2}+\frac {143 \int \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{128 a^2}\\ &=-\frac {3 \cot (c+d x)}{128 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x)}{64 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^2(c+d x)}{80 a d \sqrt {a+a \sin (c+d x)}}+\frac {19 \cot (c+d x) \csc ^3(c+d x)}{40 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{5 a^2 d}+\frac {143 \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{256 a^2}+\frac {35 \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 )}{32 a d}\\ &=\frac {35 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{32 a^{3/2} d}-\frac {3 \cot (c+d x)}{128 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x)}{64 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^2(c+d x)}{80 a d \sqrt {a+a \sin (c+d x)}}+\frac {19 \cot (c+d x) \csc ^3(c+d x)}{40 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{5 a^2 d}-\frac {143 \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 )}{128 a d}\\ &=-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{128 a^{3/2} d}-\frac {3 \cot (c+d x)}{128 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x)}{64 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^2(c+d x)}{80 a d \sqrt {a+a \sin (c+d x)}}+\frac {19 \cot (c+d x) \csc ^3(c+d x)}{40 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{5 a^2 d}\\ \end {align*}

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Mathematica [A]  time = 1.38, size = 412, normalized size = 1.87 \[ -\frac {\csc ^{15}\left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (-7100 \sin \left (\frac {1}{2} (c+d x)\right )-2880 \sin \left (\frac {3}{2} (c+d x)\right )+144 \sin \left (\frac {5}{2} (c+d x)\right )-10 \sin \left (\frac {7}{2} (c+d x)\right )-30 \sin \left (\frac {9}{2} (c+d x)\right )+7100 \cos \left (\frac {1}{2} (c+d x)\right )-2880 \cos \left (\frac {3}{2} (c+d x)\right )-144 \cos \left (\frac {5}{2} (c+d x)\right )-10 \cos \left (\frac {7}{2} (c+d x)\right )+30 \cos \left (\frac {9}{2} (c+d x)\right )+150 \sin (c+d x) \log \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )+1\right )-150 \sin (c+d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )-\cos \left (\frac {1}{2} (c+d x)\right )+1\right )-75 \sin (3 (c+d x)) \log \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )+1\right )+75 \sin (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )-\cos \left (\frac {1}{2} (c+d x)\right )+1\right )+15 \sin (5 (c+d x)) \log \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )+1\right )-15 \sin (5 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )-\cos \left (\frac {1}{2} (c+d x)\right )+1\right )\right )}{640 d (a (\sin (c+d x)+1))^{3/2} \left (\csc ^2\left (\frac {1}{4} (c+d x)\right )-\sec ^2\left (\frac {1}{4} (c+d x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cot[c + d*x]^4*Csc[c + d*x]^2)/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

-1/640*(Csc[(c + d*x)/2]^15*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(7100*Cos[(c + d*x)/2] - 2880*Cos[(3*(c +
d*x))/2] - 144*Cos[(5*(c + d*x))/2] - 10*Cos[(7*(c + d*x))/2] + 30*Cos[(9*(c + d*x))/2] - 7100*Sin[(c + d*x)/2
] + 150*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[c + d*x] - 150*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*
x)/2]]*Sin[c + d*x] - 2880*Sin[(3*(c + d*x))/2] + 144*Sin[(5*(c + d*x))/2] - 75*Log[1 + Cos[(c + d*x)/2] - Sin
[(c + d*x)/2]]*Sin[3*(c + d*x)] + 75*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 10*Sin[(7
*(c + d*x))/2] - 30*Sin[(9*(c + d*x))/2] + 15*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[5*(c + d*x)] -
15*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[5*(c + d*x)]))/(d*(Csc[(c + d*x)/4]^2 - Sec[(c + d*x)/4]^2
)^5*(a*(1 + Sin[c + d*x]))^(3/2))

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fricas [B]  time = 0.49, size = 492, normalized size = 2.24 \[ \frac {15 \, {\left (\cos \left (d x + c\right )^{6} - 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{5} + \cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right ) - 1\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) + 4 \, {\left (15 \, \cos \left (d x + c\right )^{5} + 5 \, \cos \left (d x + c\right )^{4} - 38 \, \cos \left (d x + c\right )^{3} - 194 \, \cos \left (d x + c\right )^{2} - {\left (15 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{3} - 28 \, \cos \left (d x + c\right )^{2} + 166 \, \cos \left (d x + c\right ) + 317\right )} \sin \left (d x + c\right ) + 151 \, \cos \left (d x + c\right ) + 317\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{2560 \, {\left (a^{2} d \cos \left (d x + c\right )^{6} - 3 \, a^{2} d \cos \left (d x + c\right )^{4} + 3 \, a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d - {\left (a^{2} d \cos \left (d x + c\right )^{5} + a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{3} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^6/(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/2560*(15*(cos(d*x + c)^6 - 3*cos(d*x + c)^4 + 3*cos(d*x + c)^2 - (cos(d*x + c)^5 + cos(d*x + c)^4 - 2*cos(d*
x + c)^3 - 2*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)*sq
rt(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)) + 4*(15*cos(d*x + c)^5 + 5*cos(d*x + c)^4 -
38*cos(d*x + c)^3 - 194*cos(d*x + c)^2 - (15*cos(d*x + c)^4 + 10*cos(d*x + c)^3 - 28*cos(d*x + c)^2 + 166*cos
(d*x + c) + 317)*sin(d*x + c) + 151*cos(d*x + c) + 317)*sqrt(a*sin(d*x + c) + a))/(a^2*d*cos(d*x + c)^6 - 3*a^
2*d*cos(d*x + c)^4 + 3*a^2*d*cos(d*x + c)^2 - a^2*d - (a^2*d*cos(d*x + c)^5 + a^2*d*cos(d*x + c)^4 - 2*a^2*d*c
os(d*x + c)^3 - 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c) + a^2*d)*sin(d*x + c))

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giac [B]  time = 1.16, size = 808, normalized size = 3.67 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^6/(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

1/1280*(sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)*((2*((4*tan(1/2*d*x + 1/2*c)/(a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)) -
15/(a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)))*tan(1/2*d*x + 1/2*c) + 28/(a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)))*tan(1/2
*d*x + 1/2*c) - 95/(a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)))*tan(1/2*d*x + 1/2*c) + 128/(a^2*sgn(tan(1/2*d*x + 1/2*
c) + 1))) - (870*sqrt(2)*sqrt(a)*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - 435*sqrt(2)*sqrt(-a)*log(sqrt(
2)*sqrt(a) + sqrt(a)) + 1230*sqrt(a)*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - 615*sqrt(-a)*log(sqrt(2)*s
qrt(a) + sqrt(a)) + 22282*sqrt(2)*sqrt(-a) + 31524*sqrt(-a))*sgn(tan(1/2*d*x + 1/2*c) + 1)/(29*sqrt(2)*sqrt(-a
)*a^(3/2) + 41*sqrt(-a)*a^(3/2)) + 30*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*sgn(tan(1/2*d*x + 1/2*c) + 1)) - 15*log(abs(-sqrt(a)*tan(1/2*d*x + 1/2*c) + sqrt(a*t
an(1/2*d*x + 1/2*c)^2 + a)))/(a^(3/2)*sgn(tan(1/2*d*x + 1/2*c) + 1)) - 2*(95*(sqrt(a)*tan(1/2*d*x + 1/2*c) - s
qrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^9 - 240*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))
^8*sqrt(a) - 70*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^7*a + 560*(sqrt(a)*tan(1/2
*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*a^(3/2) - 720*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan
(1/2*d*x + 1/2*c)^2 + a))^4*a^(5/2) + 70*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^3
*a^3 + 400*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a^(7/2) - 95*(sqrt(a)*tan(1/2
*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))*a^4 - 128*a^(9/2))/(((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(
a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a)^5*a*sgn(tan(1/2*d*x + 1/2*c) + 1)))/d

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maple [A]  time = 1.36, size = 180, normalized size = 0.82 \[ -\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (15 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {9}{2}} a^{\frac {7}{2}}-70 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {7}{2}} a^{\frac {9}{2}}+128 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {5}{2}} a^{\frac {11}{2}}+15 \arctanh \left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) a^{8} \left (\sin ^{5}\left (d x +c \right )\right )+70 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {3}{2}} a^{\frac {13}{2}}-15 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {15}{2}}\right )}{640 a^{\frac {19}{2}} \sin \left (d x +c \right )^{5} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^6/(a+a*sin(d*x+c))^(3/2),x)

[Out]

-1/640/a^(19/2)*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(15*(-a*(sin(d*x+c)-1))^(9/2)*a^(7/2)-70*(-a*(sin(d*x
+c)-1))^(7/2)*a^(9/2)+128*(-a*(sin(d*x+c)-1))^(5/2)*a^(11/2)+15*arctanh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*a^8
*sin(d*x+c)^5+70*(-a*(sin(d*x+c)-1))^(3/2)*a^(13/2)-15*(-a*(sin(d*x+c)-1))^(1/2)*a^(15/2))/sin(d*x+c)^5/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)^6/(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^4}{{\sin \left (c+d\,x\right )}^6\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4/(sin(c + d*x)^6*(a + a*sin(c + d*x))^(3/2)),x)

[Out]

int(cos(c + d*x)^4/(sin(c + d*x)^6*(a + a*sin(c + d*x))^(3/2)), 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)**6/(a+a*sin(d*x+c))**(3/2),x)

[Out]

Timed out

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