\(\int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^{3/2} \, dx\) [455]

Optimal result

Integrand size = 29, antiderivative size = 199 \[ \int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}-\frac {14 a^2 \cos (c+d x)}{45 d \sqrt {a+a \sin (c+d x)}}-\frac {34 a^2 \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt {a+a \sin (c+d x)}}+\frac {388 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{315 d}+\frac {16 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 d} \]

[Out]

-2*a^(3/2)*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))/d+16/105*cos(d*x+c)*(a+a*sin(d*x+c))^(3/2)/d-14/
45*a^2*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-34/63*a^2*cos(d*x+c)*sin(d*x+c)^3/d/(a+a*sin(d*x+c))^(1/2)-2/9*a^2*
cos(d*x+c)*sin(d*x+c)^4/d/(a+a*sin(d*x+c))^(1/2)+388/315*a*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.414, Rules used = {2960, 2842, 21, 2849, 2838, 2830, 2725, 3125, 3055, 3060, 2852, 212} \[ \int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}-\frac {34 a^2 \sin ^3(c+d x) \cos (c+d x)}{63 d \sqrt {a \sin (c+d x)+a}}-\frac {14 a^2 \cos (c+d x)}{45 d \sqrt {a \sin (c+d x)+a}}+\frac {16 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{105 d}+\frac {388 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{315 d} \]

[In]

Int[Cos[c + d*x]^3*Cot[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 - (14*a^2*Cos[c + d*x])/(45*d*Sqrt[a +
a*Sin[c + d*x]]) - (34*a^2*Cos[c + d*x]*Sin[c + d*x]^3)/(63*d*Sqrt[a + a*Sin[c + d*x]]) - (2*a^2*Cos[c + d*x]
*Sin[c + d*x]^4)/(9*d*Sqrt[a + a*Sin[c + d*x]]) + (388*a*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(315*d) + (16*
Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(105*d)

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 212

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

Rule 2725

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 2830

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*S
in[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 2838

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 2842

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-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 2849

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

Rule 2852

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 2960

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 3055

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)))
*Sin[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 3060

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)*Sqrt
[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 3125

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

Rubi steps \begin{align*} \text {integral}& = \int \sin ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx+\int \csc (c+d x) (a+a \sin (c+d x))^{3/2} \left (1-2 \sin ^2(c+d x)\right ) \, dx \\ & = -\frac {2 a^2 \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt {a+a \sin (c+d x)}}+\frac {4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}+\frac {2}{9} \int \frac {\sin ^3(c+d x) \left (\frac {17 a^2}{2}+\frac {17}{2} a^2 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx+\frac {2 \int \csc (c+d x) \left (\frac {5 a}{2}-3 a \sin (c+d x)\right ) (a+a \sin (c+d x))^{3/2} \, dx}{5 a} \\ & = -\frac {2 a^2 \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt {a+a \sin (c+d x)}}+\frac {4 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}+\frac {4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}+\frac {4 \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \left (\frac {15 a^2}{4}-\frac {9}{4} a^2 \sin (c+d x)\right ) \, dx}{15 a}+\frac {1}{9} (17 a) \int \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx \\ & = \frac {6 a^2 \cos (c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}-\frac {34 a^2 \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt {a+a \sin (c+d x)}}+\frac {4 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}+\frac {4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}+a \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx+\frac {1}{21} (34 a) \int \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx \\ & = \frac {6 a^2 \cos (c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}-\frac {34 a^2 \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt {a+a \sin (c+d x)}}+\frac {4 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}+\frac {16 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 d}+\frac {68}{105} \int \left (\frac {3 a}{2}-a \sin (c+d x)\right ) \sqrt {a+a \sin (c+d x)} \, dx-\frac {\left (2 a^2\right ) \text {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} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}+\frac {6 a^2 \cos (c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}-\frac {34 a^2 \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt {a+a \sin (c+d x)}}+\frac {388 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{315 d}+\frac {16 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 d}+\frac {1}{45} (34 a) \int \sqrt {a+a \sin (c+d x)} \, dx \\ & = -\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}-\frac {14 a^2 \cos (c+d x)}{45 d \sqrt {a+a \sin (c+d x)}}-\frac {34 a^2 \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt {a+a \sin (c+d x)}}+\frac {388 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{315 d}+\frac {16 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.76 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.10 \[ \int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {(a (1+\sin (c+d x)))^{3/2} \left (1260 \cos \left (\frac {1}{2} (c+d x)\right )+1470 \cos \left (\frac {3}{2} (c+d x)\right )-126 \cos \left (\frac {5}{2} (c+d x)\right )+135 \cos \left (\frac {7}{2} (c+d x)\right )-35 \cos \left (\frac {9}{2} (c+d x)\right )-2520 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2520 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-1260 \sin \left (\frac {1}{2} (c+d x)\right )+1470 \sin \left (\frac {3}{2} (c+d x)\right )+126 \sin \left (\frac {5}{2} (c+d x)\right )+135 \sin \left (\frac {7}{2} (c+d x)\right )+35 \sin \left (\frac {9}{2} (c+d x)\right )\right )}{2520 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3} \]

[In]

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

[Out]

((a*(1 + Sin[c + d*x]))^(3/2)*(1260*Cos[(c + d*x)/2] + 1470*Cos[(3*(c + d*x))/2] - 126*Cos[(5*(c + d*x))/2] +
135*Cos[(7*(c + d*x))/2] - 35*Cos[(9*(c + d*x))/2] - 2520*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2520*
Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 1260*Sin[(c + d*x)/2] + 1470*Sin[(3*(c + d*x))/2] + 126*Sin[(5*
(c + d*x))/2] + 135*Sin[(7*(c + d*x))/2] + 35*Sin[(9*(c + d*x))/2]))/(2520*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)
/2])^3)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.80

method	result	size
default	\(-\frac {2 \left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (315 a^{\frac {9}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right )+35 \left (a -a \sin \left (d x +c \right )\right )^{\frac {9}{2}}-225 \left (a -a \sin \left (d x +c \right )\right )^{\frac {7}{2}} a +441 a^{2} \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}}-105 a^{3} \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}}-315 \sqrt {a -a \sin \left (d x +c \right )}\, a^{4}\right )}{315 a^{3} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(159\)

[In]

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

[Out]

-2/315*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(315*a^(9/2)*arctanh((a-a*sin(d*x+c))^(1/2)/a^(1/2))+35*(a-a*s
in(d*x+c))^(9/2)-225*(a-a*sin(d*x+c))^(7/2)*a+441*a^2*(a-a*sin(d*x+c))^(5/2)-105*a^3*(a-a*sin(d*x+c))^(3/2)-31
5*(a-a*sin(d*x+c))^(1/2)*a^4)/a^3/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.67 \[ \int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {315 \, {\left (a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + a\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) - 4 \, {\left (35 \, a \cos \left (d x + c\right )^{5} - 50 \, a \cos \left (d x + c\right )^{4} - 46 \, a \cos \left (d x + c\right )^{3} - 118 \, a \cos \left (d x + c\right )^{2} - 158 \, a \cos \left (d x + c\right ) - {\left (35 \, a \cos \left (d x + c\right )^{4} + 85 \, a \cos \left (d x + c\right )^{3} + 39 \, a \cos \left (d x + c\right )^{2} + 157 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{630 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

[In]

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

[Out]

1/630*(315*(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*(35*a*cos(d*x + c)^5 - 50*a*cos(d*x + c)^4 - 46*a*cos(d*
x + c)^3 - 118*a*cos(d*x + c)^2 - 158*a*cos(d*x + c) - (35*a*cos(d*x + c)^4 + 85*a*cos(d*x + c)^3 + 39*a*cos(d
*x + c)^2 + 157*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)

Sympy [F(-1)]

Timed out. \[ \int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{4} \csc \left (d x + c\right ) \,d x } \]

[In]

integrate(cos(d*x+c)^4*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)*cos(d*x + c)^4*csc(d*x + c), x)

Giac [A] (verification not implemented)

none

Time = 0.46 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.13 \[ \int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {\sqrt {2} {\left (2240 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 7200 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 7056 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 840 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 315 \, \sqrt {2} a \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 1260 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{630 \, d} \]

[In]

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

[Out]

1/630*sqrt(2)*(2240*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^9 - 7200*a*sgn(cos(-1
/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^7 + 7056*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1
/4*pi + 1/2*d*x + 1/2*c)^5 - 840*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 315*
sqrt(2)*a*log(abs(-2*sqrt(2) + 4*sin(-1/4*pi + 1/2*d*x + 1/2*c))/abs(2*sqrt(2) + 4*sin(-1/4*pi + 1/2*d*x + 1/2
*c)))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) - 1260*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x +
1/2*c))*sqrt(a)/d

Mupad [F(-1)]

Timed out. \[ \int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^4\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}}{\sin \left (c+d\,x\right )} \,d x \]

[In]

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

[Out]

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