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

Optimal. Leaf size=199 \[ -\frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}-\frac {34 a^2 \sin ^3(c+d x) \cos (c+d x)}{63 d \sqrt {a \sin (c+d x)+a}}-\frac {14 a^2 \cos (c+d x)}{45 d \sqrt {a \sin (c+d x)+a}}+\frac {16 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{105 d}+\frac {388 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{315 d} \]

[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

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Rubi [A]  time = 0.71, antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 12, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.414, Rules used = {2881, 2763, 21, 2770, 2759, 2751, 2646, 3046, 2976, 2981, 2773, 206} \[ -\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}-\frac {34 a^2 \sin ^3(c+d x) \cos (c+d x)}{63 d \sqrt {a \sin (c+d x)+a}}-\frac {14 a^2 \cos (c+d x)}{45 d \sqrt {a \sin (c+d x)+a}}-\frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}+\frac {16 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{105 d}+\frac {388 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{315 d} \]

Antiderivative was successfully verified.

[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 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 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 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 2770

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 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 3046

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*Simp
[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*} \int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^{3/2} \, dx &=\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 ) \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 {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} \tanh ^{-1}\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*}

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Mathematica [A]  time = 0.48, size = 219, normalized size = 1.10 \[ \frac {(a (\sin (c+d x)+1))^{3/2} \left (-1260 \sin \left (\frac {1}{2} (c+d x)\right )+1470 \sin \left (\frac {3}{2} (c+d x)\right )+126 \sin \left (\frac {5}{2} (c+d x)\right )+135 \sin \left (\frac {7}{2} (c+d x)\right )+35 \sin \left (\frac {9}{2} (c+d x)\right )+1260 \cos \left (\frac {1}{2} (c+d x)\right )+1470 \cos \left (\frac {3}{2} (c+d x)\right )-126 \cos \left (\frac {5}{2} (c+d x)\right )+135 \cos \left (\frac {7}{2} (c+d x)\right )-35 \cos \left (\frac {9}{2} (c+d x)\right )-2520 \log \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )+1\right )+2520 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )-\cos \left (\frac {1}{2} (c+d x)\right )+1\right )\right )}{2520 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[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)

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fricas [A]  time = 0.47, size = 332, normalized size = 1.67 \[ \frac {315 \, {\left (a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + a\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) - 4 \, {\left (35 \, a \cos \left (d x + c\right )^{5} - 50 \, a \cos \left (d x + c\right )^{4} - 46 \, a \cos \left (d x + c\right )^{3} - 118 \, a \cos \left (d x + c\right )^{2} - 158 \, a \cos \left (d x + c\right ) - {\left (35 \, a \cos \left (d x + c\right )^{4} + 85 \, a \cos \left (d x + c\right )^{3} + 39 \, a \cos \left (d x + c\right )^{2} + 157 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{630 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

[Out]

Timed out

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maple [A]  time = 1.29, size = 159, normalized size = 0.80 \[ -\frac {2 \left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (315 a^{\frac {9}{2}} \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right )+35 \left (a -a \sin \left (d x +c \right )\right )^{\frac {9}{2}}-225 a \left (a -a \sin \left (d x +c \right )\right )^{\frac {7}{2}}+441 \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}} a^{2}-105 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} a^{3}-315 a^{4} \sqrt {a -a \sin \left (d x +c \right )}\right )}{315 a^{3} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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-a*sin(d*x+c))^(7/2)+441*(a-a*sin(d*x+c))^(5/2)*a^2-105*(a-a*sin(d*x+c))^(3/2)*a^3-31
5*a^4*(a-a*sin(d*x+c))^(1/2))/a^3/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 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{4} \csc \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

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),x)

[Out]

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

[Out]

Timed out

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