3.48 \(\int \csc (c+d x) (a+a \sin (c+d x))^{3/2} \, dx\)
Optimal. Leaf size=66 \[ -\frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}-\frac {2 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}} \]
[Out]
-2*a^(3/2)*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))/d-2*a^2*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)
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Rubi [A] time = 0.10, antiderivative size = 66, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used =
{2763, 21, 2773, 206} \[ -\frac {2 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}-\frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d} \]
Antiderivative was successfully verified.
[In]
Int[Csc[c + d*x]*(a + a*Sin[c + d*x])^(3/2),x]
[Out]
(-2*a^(3/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/d - (2*a^2*Cos[c + d*x])/(d*Sqrt[a + a*S
in[c + d*x]])
Rule 21
Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*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 2763
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d*
(m + n)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c*(m - 2) + b^2*d*(n + 1) + a^2*d*(
m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && !LtQ[n, -1] && (IntegersQ[2*m, 2*
n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 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]
Rubi steps
\begin {align*} \int \csc (c+d x) (a+a \sin (c+d x))^{3/2} \, dx &=-\frac {2 a^2 \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}+2 \int \frac {\csc (c+d x) \left (\frac {a^2}{2}+\frac {1}{2} a^2 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=-\frac {2 a^2 \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}+a \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {2 a^2 \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}-\frac {\left (2 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 {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}-\frac {2 a^2 \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 0.15, size = 118, normalized size = 1.79 \[ \frac {(a (\sin (c+d x)+1))^{3/2} \left (2 \sin \left (\frac {1}{2} (c+d x)\right )-2 \cos \left (\frac {1}{2} (c+d x)\right )-\log \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )+1\right )+\log \left (\sin \left (\frac {1}{2} (c+d x)\right )-\cos \left (\frac {1}{2} (c+d x)\right )+1\right )\right )}{d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3} \]
Antiderivative was successfully verified.
[In]
Integrate[Csc[c + d*x]*(a + a*Sin[c + d*x])^(3/2),x]
[Out]
((-2*Cos[(c + d*x)/2] - Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x
)/2]] + 2*Sin[(c + d*x)/2])*(a*(1 + Sin[c + d*x]))^(3/2))/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3)
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fricas [B] time = 0.45, size = 239, normalized size = 3.62 \[ \frac {{\left (a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + a\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) - 4 \, {\left (a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) + a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{2 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(d*x+c)*(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")
[Out]
1/2*((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 + cos(d*x + c)^2 + (cos(d*x +
c)^2 - 1)*sin(d*x + c) - cos(d*x + c) - 1)) - 4*(a*cos(d*x + c) - a*sin(d*x + c) + a)*sqrt(a*sin(d*x + c) + a)
)/(d*cos(d*x + c) + d*sin(d*x + c) + d)
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giac [B] time = 5.43, size = 2251, normalized size = 34.11 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(d*x+c)*(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")
[Out]
1/2*sqrt(2)*sqrt(a)*((sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2*c)^3*tan(1/4*c)^6 - 6*sqrt(2)*a*sg
n(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2*c)^3*tan(1/4*c)^5 + 3*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*
tan(1/2*c)^2*tan(1/4*c)^6 - 15*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2*c)^3*tan(1/4*c)^4 + 18*sq
rt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2*c)^2*tan(1/4*c)^5 - 3*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x
+ 1/2*c))*tan(1/2*c)*tan(1/4*c)^6 + 20*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2*c)^3*tan(1/4*c)^3
- 45*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2*c)^2*tan(1/4*c)^4 + 18*sqrt(2)*a*sgn(cos(-1/4*pi +
1/2*d*x + 1/2*c))*tan(1/2*c)*tan(1/4*c)^5 - sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*c)^6 + 15*s
qrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2*c)^3*tan(1/4*c)^2 - 60*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*
x + 1/2*c))*tan(1/2*c)^2*tan(1/4*c)^3 + 45*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2*c)*tan(1/4*c)
^4 - 6*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*c)^5 - 6*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/
2*c))*tan(1/2*c)^3*tan(1/4*c) + 45*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2*c)^2*tan(1/4*c)^2 - 6
0*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2*c)*tan(1/4*c)^3 + 15*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d
*x + 1/2*c))*tan(1/4*c)^4 - sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2*c)^3 + 18*sqrt(2)*a*sgn(cos(
-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2*c)^2*tan(1/4*c) - 45*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2
*c)*tan(1/4*c)^2 + 20*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*c)^3 - 3*sqrt(2)*a*sgn(cos(-1/4*pi
+ 1/2*d*x + 1/2*c))*tan(1/2*c)^2 + 18*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2*c)*tan(1/4*c) - 1
5*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*c)^2 + 3*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))
*tan(1/2*c) - 6*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*c) + sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x
+ 1/2*c)))*log(abs(-2*tan(1/4*d*x + c)*tan(1/2*c)^3 + 6*tan(1/4*d*x + c)*tan(1/2*c) - 6*tan(1/2*c)^2 - 2*(tan
(1/2*c)^2 + 1)^(3/2) + 2)/abs(-2*tan(1/4*d*x + c)*tan(1/2*c)^3 + 6*tan(1/4*d*x + c)*tan(1/2*c) - 6*tan(1/2*c)^
2 + 2*(tan(1/2*c)^2 + 1)^(3/2) + 2))/((tan(1/4*c)^6 + 3*tan(1/4*c)^4 + 3*tan(1/4*c)^2 + 1)*(tan(1/2*c)^2 + 1)^
(3/2)) + (sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2*c)^3*tan(1/4*c)^6 + 6*sqrt(2)*a*sgn(cos(-1/4*p
i + 1/2*d*x + 1/2*c))*tan(1/2*c)^3*tan(1/4*c)^5 - 3*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2*c)^2
*tan(1/4*c)^6 - 15*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2*c)^3*tan(1/4*c)^4 + 18*sqrt(2)*a*sgn(
cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2*c)^2*tan(1/4*c)^5 - 3*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*ta
n(1/2*c)*tan(1/4*c)^6 - 20*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2*c)^3*tan(1/4*c)^3 + 45*sqrt(2
)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2*c)^2*tan(1/4*c)^4 - 18*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1
/2*c))*tan(1/2*c)*tan(1/4*c)^5 + sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*c)^6 + 15*sqrt(2)*a*sgn
(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2*c)^3*tan(1/4*c)^2 - 60*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*
tan(1/2*c)^2*tan(1/4*c)^3 + 45*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2*c)*tan(1/4*c)^4 - 6*sqrt(
2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*c)^5 + 6*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/
2*c)^3*tan(1/4*c) - 45*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2*c)^2*tan(1/4*c)^2 + 60*sqrt(2)*a*
sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2*c)*tan(1/4*c)^3 - 15*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))
*tan(1/4*c)^4 - sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2*c)^3 + 18*sqrt(2)*a*sgn(cos(-1/4*pi + 1/
2*d*x + 1/2*c))*tan(1/2*c)^2*tan(1/4*c) - 45*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2*c)*tan(1/4*
c)^2 + 20*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*c)^3 + 3*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x +
1/2*c))*tan(1/2*c)^2 - 18*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2*c)*tan(1/4*c) + 15*sqrt(2)*a*
sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*c)^2 + 3*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/2*c)
- 6*sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*c) - sqrt(2)*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))*
log(abs(-6*tan(1/4*d*x + c)*tan(1/2*c)^2 + 2*tan(1/2*c)^3 - 2*(tan(1/2*c)^2 + 1)^(3/2) + 2*tan(1/4*d*x + c) -
6*tan(1/2*c))/abs(-6*tan(1/4*d*x + c)*tan(1/2*c)^2 + 2*tan(1/2*c)^3 + 2*(tan(1/2*c)^2 + 1)^(3/2) + 2*tan(1/4*d
*x + c) - 6*tan(1/2*c)))/((tan(1/4*c)^6 + 3*tan(1/4*c)^4 + 3*tan(1/4*c)^2 + 1)*(tan(1/2*c)^2 + 1)^(3/2)) - 8*(
a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x + c)*tan(1/4*c)^6 + 6*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*
tan(1/4*d*x + c)*tan(1/4*c)^5 - a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*c)^6 - 15*a*sgn(cos(-1/4*pi + 1/
2*d*x + 1/2*c))*tan(1/4*d*x + c)*tan(1/4*c)^4 + 6*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*c)^5 - 20*a*sg
n(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x + c)*tan(1/4*c)^3 + 15*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan
(1/4*c)^4 + 15*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x + c)*tan(1/4*c)^2 - 20*a*sgn(cos(-1/4*pi + 1/
2*d*x + 1/2*c))*tan(1/4*c)^3 + 6*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x + c)*tan(1/4*c) - 15*a*sgn(
cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*c)^2 - a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x + c) + 6*a*sg
n(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*c) + a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))/((sqrt(2)*tan(1/4*c)^6 +
3*sqrt(2)*tan(1/4*c)^4 + 3*sqrt(2)*tan(1/4*c)^2 + sqrt(2))*(tan(1/4*d*x + c)^2 + 1)))/d
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maple [A] time = 0.64, size = 84, normalized size = 1.27 \[ -\frac {2 \left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a \left (\sqrt {a -a \sin \left (d x +c \right )}+\sqrt {a}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right )\right )}{\cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(csc(d*x+c)*(a+a*sin(d*x+c))^(3/2),x)
[Out]
-2*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*a*((a-a*sin(d*x+c))^(1/2)+a^(1/2)*arctanh((a-a*sin(d*x+c))^(1/2)/a
^(1/2)))/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 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \csc \left (d x + c\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(d*x+c)*(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")
[Out]
integrate((a*sin(d*x + c) + a)^(3/2)*csc(d*x + c), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}}{\sin \left (c+d\,x\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + a*sin(c + d*x))^(3/2)/sin(c + d*x),x)
[Out]
int((a + a*sin(c + d*x))^(3/2)/sin(c + d*x), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \csc {\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(d*x+c)*(a+a*sin(d*x+c))**(3/2),x)
[Out]
Integral((a*(sin(c + d*x) + 1))**(3/2)*csc(c + d*x), x)
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