3.456 \(\int \cos ^2(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=178 \[ -\frac {3 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}+\frac {171 a^2 \cos (c+d x)}{35 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{5/2}}{7 a d}+\frac {4 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{35 d}+\frac {69 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{35 d}-\frac {\cot (c+d x) (a \sin (c+d x)+a)^{3/2}}{d} \]

[Out]

-3*a^(3/2)*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))/d+4/35*cos(d*x+c)*(a+a*sin(d*x+c))^(3/2)/d-cot(d
*x+c)*(a+a*sin(d*x+c))^(3/2)/d-2/7*cos(d*x+c)*(a+a*sin(d*x+c))^(5/2)/a/d+171/35*a^2*cos(d*x+c)/d/(a+a*sin(d*x+
c))^(1/2)+69/35*a*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.65, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {2881, 2759, 2751, 2647, 2646, 3044, 2976, 2981, 2773, 206} \[ \frac {171 a^2 \cos (c+d x)}{35 d \sqrt {a \sin (c+d x)+a}}-\frac {3 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{5/2}}{7 a d}+\frac {4 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{35 d}+\frac {69 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{35 d}-\frac {\cot (c+d x) (a \sin (c+d x)+a)^{3/2}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a^(3/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/d + (171*a^2*Cos[c + d*x])/(35*d*Sqrt[a
+ a*Sin[c + d*x]]) + (69*a*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(35*d) + (4*Cos[c + d*x]*(a + a*Sin[c + d*x]
)^(3/2))/(35*d) - (Cot[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/d - (2*Cos[c + d*x]*(a + a*Sin[c + d*x])^(5/2))/(7
*a*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 2646

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

Rule 2647

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n -
1))/(d*n), x] + Dist[(a*(2*n - 1))/n, Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && Eq
Q[a^2 - b^2, 0] && IGtQ[n - 1/2, 0]

Rule 2751

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

Rule 2759

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

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 2881

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

Rule 2976

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

Rule 2981

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 \cos ^2(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx &=\int \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx+\int \csc ^2(c+d x) (a+a \sin (c+d x))^{3/2} \left (1-2 \sin ^2(c+d x)\right ) \, dx\\ &=-\frac {\cot (c+d x) (a+a \sin (c+d x))^{3/2}}{d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{7 a d}+\frac {2 \int \left (\frac {5 a}{2}-a \sin (c+d x)\right ) (a+a \sin (c+d x))^{3/2} \, dx}{7 a}+\frac {\int \csc (c+d x) \left (\frac {3 a}{2}-\frac {7}{2} a \sin (c+d x)\right ) (a+a \sin (c+d x))^{3/2} \, dx}{a}\\ &=\frac {7 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}+\frac {4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{35 d}-\frac {\cot (c+d x) (a+a \sin (c+d x))^{3/2}}{d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{7 a d}+\frac {19}{35} \int (a+a \sin (c+d x))^{3/2} \, dx+\frac {2 \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \left (\frac {9 a^2}{4}-\frac {19}{4} a^2 \sin (c+d x)\right ) \, dx}{3 a}\\ &=\frac {19 a^2 \cos (c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}+\frac {69 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{35 d}+\frac {4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{35 d}-\frac {\cot (c+d x) (a+a \sin (c+d x))^{3/2}}{d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{7 a d}+\frac {1}{105} (76 a) \int \sqrt {a+a \sin (c+d x)} \, dx+\frac {1}{2} (3 a) \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=\frac {171 a^2 \cos (c+d x)}{35 d \sqrt {a+a \sin (c+d x)}}+\frac {69 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{35 d}+\frac {4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{35 d}-\frac {\cot (c+d x) (a+a \sin (c+d x))^{3/2}}{d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{7 a d}-\frac {\left (3 a^2\right ) \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 )}{d}\\ &=-\frac {3 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}+\frac {171 a^2 \cos (c+d x)}{35 d \sqrt {a+a \sin (c+d x)}}+\frac {69 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{35 d}+\frac {4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{35 d}-\frac {\cot (c+d x) (a+a \sin (c+d x))^{3/2}}{d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{7 a d}\\ \end {align*}

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Mathematica [A]  time = 1.32, size = 283, normalized size = 1.59 \[ -\frac {a \csc ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\sin (c+d x)+1)} \left (-840 \sin \left (\frac {1}{2} (c+d x)\right )-574 \sin \left (\frac {3}{2} (c+d x)\right )-30 \sin \left (\frac {5}{2} (c+d x)\right )-21 \sin \left (\frac {7}{2} (c+d x)\right )-5 \sin \left (\frac {9}{2} (c+d x)\right )+840 \cos \left (\frac {1}{2} (c+d x)\right )-574 \cos \left (\frac {3}{2} (c+d x)\right )+30 \cos \left (\frac {5}{2} (c+d x)\right )-21 \cos \left (\frac {7}{2} (c+d x)\right )+5 \cos \left (\frac {9}{2} (c+d x)\right )+420 \sin (c+d x) \log \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )+1\right )-420 \sin (c+d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )-\cos \left (\frac {1}{2} (c+d x)\right )+1\right )\right )}{140 d \left (\cot \left (\frac {1}{2} (c+d x)\right )+1\right ) \left (\csc \left (\frac {1}{4} (c+d x)\right )-\sec \left (\frac {1}{4} (c+d x)\right )\right ) \left (\csc \left (\frac {1}{4} (c+d x)\right )+\sec \left (\frac {1}{4} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/140*(a*Csc[(c + d*x)/2]^4*Sqrt[a*(1 + Sin[c + d*x])]*(840*Cos[(c + d*x)/2] - 574*Cos[(3*(c + d*x))/2] + 30*
Cos[(5*(c + d*x))/2] - 21*Cos[(7*(c + d*x))/2] + 5*Cos[(9*(c + d*x))/2] - 840*Sin[(c + d*x)/2] + 420*Log[1 + C
os[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[c + d*x] - 420*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[c + d*
x] - 574*Sin[(3*(c + d*x))/2] - 30*Sin[(5*(c + d*x))/2] - 21*Sin[(7*(c + d*x))/2] - 5*Sin[(9*(c + d*x))/2]))/(
d*(1 + Cot[(c + d*x)/2])*(Csc[(c + d*x)/4] - Sec[(c + d*x)/4])*(Csc[(c + d*x)/4] + Sec[(c + d*x)/4]))

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fricas [B]  time = 0.46, size = 360, normalized size = 2.02 \[ \frac {105 \, {\left (a \cos \left (d x + c\right )^{2} - {\left (a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) - a\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) + 4 \, {\left (10 \, a \cos \left (d x + c\right )^{5} - 16 \, a \cos \left (d x + c\right )^{4} - 8 \, a \cos \left (d x + c\right )^{3} - 120 \, a \cos \left (d x + c\right )^{2} + 33 \, a \cos \left (d x + c\right ) - {\left (10 \, a \cos \left (d x + c\right )^{4} + 26 \, a \cos \left (d x + c\right )^{3} + 18 \, a \cos \left (d x + c\right )^{2} + 138 \, a \cos \left (d x + c\right ) + 171 \, a\right )} \sin \left (d x + c\right ) + 171 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{140 \, {\left (d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right ) + d\right )} \sin \left (d x + c\right ) - d\right )}} \]

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))^(3/2),x, algorithm="fricas")

[Out]

1/140*(105*(a*cos(d*x + c)^2 - (a*cos(d*x + c) + a)*sin(d*x + c) - a)*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 + c
os(d*x + c)^2 + (cos(d*x + c)^2 - 1)*sin(d*x + c) - cos(d*x + c) - 1)) + 4*(10*a*cos(d*x + c)^5 - 16*a*cos(d*x
+ c)^4 - 8*a*cos(d*x + c)^3 - 120*a*cos(d*x + c)^2 + 33*a*cos(d*x + c) - (10*a*cos(d*x + c)^4 + 26*a*cos(d*x
+ c)^3 + 18*a*cos(d*x + c)^2 + 138*a*cos(d*x + c) + 171*a)*sin(d*x + c) + 171*a)*sqrt(a*sin(d*x + c) + a))/(d*
cos(d*x + c)^2 - (d*cos(d*x + c) + d)*sin(d*x + c) - d)

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giac [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))^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 1.28, size = 180, normalized size = 1.01 \[ \frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (\sin \left (d x +c \right ) \left (140 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {7}{2}}+70 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} a^{\frac {5}{2}}-56 \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}} a^{\frac {3}{2}}+10 \sqrt {a}\, \left (a -a \sin \left (d x +c \right )\right )^{\frac {7}{2}}-105 \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right ) a^{4}\right )-35 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {7}{2}}\right )}{35 a^{\frac {5}{2}} \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

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))^(3/2),x)

[Out]

1/35*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)/a^(5/2)*(sin(d*x+c)*(140*(a-a*sin(d*x+c))^(1/2)*a^(7/2)+70*(a-a*
sin(d*x+c))^(3/2)*a^(5/2)-56*(a-a*sin(d*x+c))^(5/2)*a^(3/2)+10*a^(1/2)*(a-a*sin(d*x+c))^(7/2)-105*arctanh((a-a
*sin(d*x+c))^(1/2)/a^(1/2))*a^4)-35*(a-a*sin(d*x+c))^(1/2)*a^(7/2))/sin(d*x+c)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/
2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{4} \csc \left (d x + c\right )^{2}\,{d x} \]

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))^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^4\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}}{{\sin \left (c+d\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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