Integrand size = 23, antiderivative size = 106 \[ \int \csc ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=-\frac {19 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{4 d}-\frac {9 a^3 \cot (c+d x)}{4 d \sqrt {a+a \sin (c+d x)}}-\frac {a^2 \cot (c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)}}{2 d} \] Output:
-19/4*a^(5/2)*arctanh(a^(1/2)*cos(d*x+c)/(a+a*sin(d*x+c))^(1/2))/d-9/4*a^3 *cot(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-1/2*a^2*cot(d*x+c)*csc(d*x+c)*(a+a*si n(d*x+c))^(1/2)/d
Leaf count is larger than twice the leaf count of optimal. \(252\) vs. \(2(106)=212\).
Time = 6.17 (sec) , antiderivative size = 252, normalized size of antiderivative = 2.38 \[ \int \csc ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\frac {a^2 \csc ^7\left (\frac {1}{2} (c+d x)\right ) \sqrt {a (1+\sin (c+d x))} \left (14 \cos \left (\frac {1}{2} (c+d x)\right )-22 \cos \left (\frac {3}{2} (c+d x)\right )-19 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+19 \cos (2 (c+d x)) \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+19 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-19 \cos (2 (c+d x)) \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-14 \sin \left (\frac {1}{2} (c+d x)\right )-22 \sin \left (\frac {3}{2} (c+d x)\right )\right )}{4 d \left (1+\cot \left (\frac {1}{2} (c+d x)\right )\right ) \left (\csc ^2\left (\frac {1}{4} (c+d x)\right )-\sec ^2\left (\frac {1}{4} (c+d x)\right )\right )^2} \] Input:
Integrate[Csc[c + d*x]^3*(a + a*Sin[c + d*x])^(5/2),x]
Output:
(a^2*Csc[(c + d*x)/2]^7*Sqrt[a*(1 + Sin[c + d*x])]*(14*Cos[(c + d*x)/2] - 22*Cos[(3*(c + d*x))/2] - 19*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 19*Cos[2*(c + d*x)]*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 19*Lo g[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 19*Cos[2*(c + d*x)]*Log[1 - C os[(c + d*x)/2] + Sin[(c + d*x)/2]] - 14*Sin[(c + d*x)/2] - 22*Sin[(3*(c + d*x))/2]))/(4*d*(1 + Cot[(c + d*x)/2])*(Csc[(c + d*x)/4]^2 - Sec[(c + d*x )/4]^2)^2)
Time = 0.58 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3241, 27, 3042, 3459, 3042, 3252, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \csc ^3(c+d x) (a \sin (c+d x)+a)^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (c+d x)+a)^{5/2}}{\sin (c+d x)^3}dx\) |
| \(\Big \downarrow \) 3241 |
| \(\displaystyle -\frac {1}{2} a \int -\frac {1}{2} \csc ^2(c+d x) \sqrt {\sin (c+d x) a+a} (5 \sin (c+d x) a+9 a)dx-\frac {a^2 \cot (c+d x) \csc (c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} a \int \csc ^2(c+d x) \sqrt {\sin (c+d x) a+a} (5 \sin (c+d x) a+9 a)dx-\frac {a^2 \cot (c+d x) \csc (c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} a \int \frac {\sqrt {\sin (c+d x) a+a} (5 \sin (c+d x) a+9 a)}{\sin (c+d x)^2}dx-\frac {a^2 \cot (c+d x) \csc (c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 3459 |
| \(\displaystyle \frac {1}{4} a \left (\frac {19}{2} a \int \csc (c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {9 a^2 \cot (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )-\frac {a^2 \cot (c+d x) \csc (c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} a \left (\frac {19}{2} a \int \frac {\sqrt {\sin (c+d x) a+a}}{\sin (c+d x)}dx-\frac {9 a^2 \cot (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )-\frac {a^2 \cot (c+d x) \csc (c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {1}{4} a \left (-\frac {19 a^2 \int \frac {1}{a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{d}-\frac {9 a^2 \cot (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )-\frac {a^2 \cot (c+d x) \csc (c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{4} a \left (-\frac {19 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}-\frac {9 a^2 \cot (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )-\frac {a^2 \cot (c+d x) \csc (c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\) |
Input:
Int[Csc[c + d*x]^3*(a + a*Sin[c + d*x])^(5/2),x]
Output:
-1/2*(a^2*Cot[c + d*x]*Csc[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/d + (a*((-19 *a^(3/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/d - (9* a^2*Cot[c + d*x])/(d*Sqrt[a + a*Sin[c + d*x]])))/4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*(b*c - a*d)*Cos[e + f*x]*(a + b *Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a* d))), x] + Simp[b^2/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b* c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], x], 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[m, 1] && LtQ[n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)]), x_Symbol] :> Simp[-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]
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] + Simp[(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]
Time = 8.01 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.19
| method | result | size |
| default | \(-\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \sqrt {a}\, \left (19 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) \sin \left (d x +c \right )^{2} a^{2}+13 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {3}{2}}-11 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {3}{2}} \sqrt {a}\right )}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(126\) |
Input:
int(csc(d*x+c)^3*(a+a*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/4*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*a^(1/2)*(19*arctanh((-a*(sin (d*x+c)-1))^(1/2)/a^(1/2))*sin(d*x+c)^2*a^2+13*(-a*(sin(d*x+c)-1))^(1/2)*a ^(3/2)-11*(-a*(sin(d*x+c)-1))^(3/2)*a^(1/2))/sin(d*x+c)^2/cos(d*x+c)/(a+a* sin(d*x+c))^(1/2)/d
Leaf count of result is larger than twice the leaf count of optimal. 359 vs. \(2 (90) = 180\).
Time = 0.11 (sec) , antiderivative size = 359, normalized size of antiderivative = 3.39 \[ \int \csc ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\frac {19 \, {\left (a^{2} \cos \left (d x + c\right )^{3} + a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right ) - a^{2} + {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) + 4 \, {\left (11 \, a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \cos \left (d x + c\right ) - 9 \, a^{2} + {\left (11 \, a^{2} \cos \left (d x + c\right ) + 9 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{16 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right ) - d\right )}} \] Input:
integrate(csc(d*x+c)^3*(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")
Output:
1/16*(19*(a^2*cos(d*x + c)^3 + a^2*cos(d*x + c)^2 - a^2*cos(d*x + c) - a^2 + (a^2*cos(d*x + c)^2 - a^2)*sin(d*x + c))*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*(11*a^2*cos(d*x + c)^2 + 2*a^2*cos(d*x + c) - 9*a^2 + (11*a^2*cos (d*x + c) + 9*a^2)*sin(d*x + c))*sqrt(a*sin(d*x + c) + a))/(d*cos(d*x + c) ^3 + d*cos(d*x + c)^2 - d*cos(d*x + c) + (d*cos(d*x + c)^2 - d)*sin(d*x + c) - d)
Timed out. \[ \int \csc ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:
integrate(csc(d*x+c)**3*(a+a*sin(d*x+c))**(5/2),x)
Output:
Timed out
\[ \int \csc ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \csc \left (d x + c\right )^{3} \,d x } \] Input:
integrate(csc(d*x+c)^3*(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")
Output:
integrate((a*sin(d*x + c) + a)^(5/2)*csc(d*x + c)^3, x)
Time = 0.17 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.55 \[ \int \csc ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=-\frac {\sqrt {2} {\left (19 \, \sqrt {2} a^{2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + \frac {4 \, {\left (22 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 13 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}\right )} \sqrt {a}}{16 \, d} \] Input:
integrate(csc(d*x+c)^3*(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")
Output:
-1/16*sqrt(2)*(19*sqrt(2)*a^2*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)) + 4*(22*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*si n(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 13*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c) )*sin(-1/4*pi + 1/2*d*x + 1/2*c))/(2*sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1) ^2)*sqrt(a)/d
Timed out. \[ \int \csc ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}}{{\sin \left (c+d\,x\right )}^3} \,d x \] Input:
int((a + a*sin(c + d*x))^(5/2)/sin(c + d*x)^3,x)
Output:
int((a + a*sin(c + d*x))^(5/2)/sin(c + d*x)^3, x)
\[ \int \csc ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\sqrt {a}\, a^{2} \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \csc \left (d x +c \right )^{3} \sin \left (d x +c \right )^{2}d x +2 \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \csc \left (d x +c \right )^{3} \sin \left (d x +c \right )d x \right )+\int \sqrt {\sin \left (d x +c \right )+1}\, \csc \left (d x +c \right )^{3}d x \right ) \] Input:
int(csc(d*x+c)^3*(a+a*sin(d*x+c))^(5/2),x)
Output:
sqrt(a)*a**2*(int(sqrt(sin(c + d*x) + 1)*csc(c + d*x)**3*sin(c + d*x)**2,x ) + 2*int(sqrt(sin(c + d*x) + 1)*csc(c + d*x)**3*sin(c + d*x),x) + int(sqr t(sin(c + d*x) + 1)*csc(c + d*x)**3,x))