Integrand size = 23, antiderivative size = 121 \[ \int \cot ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {3 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}+\frac {11 a^2 \cos (c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}+\frac {5 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}-\frac {\cot (c+d x) (a+a \sin (c+d x))^{3/2}}{d} \] Output:
-3*a^(3/2)*arctanh(a^(1/2)*cos(d*x+c)/(a+a*sin(d*x+c))^(1/2))/d+11/3*a^2*c os(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)+5/3*a*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2) /d-cot(d*x+c)*(a+a*sin(d*x+c))^(3/2)/d
Time = 5.86 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.93 \[ \int \cot ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {a \csc ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {a (1+\sin (c+d x))} \left (14 \cos \left (\frac {1}{2} (c+d x)\right )-9 \cos \left (\frac {3}{2} (c+d x)\right )+\cos \left (\frac {5}{2} (c+d x)\right )-14 \sin \left (\frac {1}{2} (c+d x)\right )+9 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)-9 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)-9 \sin \left (\frac {3}{2} (c+d x)\right )-\sin \left (\frac {5}{2} (c+d x)\right )\right )}{3 d \left (1+\cot \left (\frac {1}{2} (c+d x)\right )\right ) \left (\csc \left (\frac {1}{4} (c+d x)\right )-\sec \left (\frac {1}{4} (c+d x)\right )\right ) \left (\csc \left (\frac {1}{4} (c+d x)\right )+\sec \left (\frac {1}{4} (c+d x)\right )\right )} \] Input:
Integrate[Cot[c + d*x]^2*(a + a*Sin[c + d*x])^(3/2),x]
Output:
-1/3*(a*Csc[(c + d*x)/2]^4*Sqrt[a*(1 + Sin[c + d*x])]*(14*Cos[(c + d*x)/2] - 9*Cos[(3*(c + d*x))/2] + Cos[(5*(c + d*x))/2] - 14*Sin[(c + d*x)/2] + 9 *Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[c + d*x] - 9*Log[1 - Cos [(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[c + d*x] - 9*Sin[(3*(c + d*x))/2] - Sin[(5*(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]))
Time = 0.80 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.11, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 3195, 27, 3042, 3455, 27, 3042, 3460, 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.
\(\displaystyle \int \cot ^2(c+d x) (a \sin (c+d x)+a)^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (c+d x)+a)^{3/2}}{\tan (c+d x)^2}dx\) |
\(\Big \downarrow \) 3195 |
\(\displaystyle \frac {\int \frac {1}{2} \csc (c+d x) (3 a-5 a \sin (c+d x)) (\sin (c+d x) a+a)^{3/2}dx}{a}-\frac {\cot (c+d x) (a \sin (c+d x)+a)^{3/2}}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \csc (c+d x) (3 a-5 a \sin (c+d x)) (\sin (c+d x) a+a)^{3/2}dx}{2 a}-\frac {\cot (c+d x) (a \sin (c+d x)+a)^{3/2}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {(3 a-5 a \sin (c+d x)) (\sin (c+d x) a+a)^{3/2}}{\sin (c+d x)}dx}{2 a}-\frac {\cot (c+d x) (a \sin (c+d x)+a)^{3/2}}{d}\) |
\(\Big \downarrow \) 3455 |
\(\displaystyle \frac {\frac {2}{3} \int \frac {1}{2} \csc (c+d x) \sqrt {\sin (c+d x) a+a} \left (9 a^2-11 a^2 \sin (c+d x)\right )dx+\frac {10 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{2 a}-\frac {\cot (c+d x) (a \sin (c+d x)+a)^{3/2}}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{3} \int \csc (c+d x) \sqrt {\sin (c+d x) a+a} \left (9 a^2-11 a^2 \sin (c+d x)\right )dx+\frac {10 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{2 a}-\frac {\cot (c+d x) (a \sin (c+d x)+a)^{3/2}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{3} \int \frac {\sqrt {\sin (c+d x) a+a} \left (9 a^2-11 a^2 \sin (c+d x)\right )}{\sin (c+d x)}dx+\frac {10 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{2 a}-\frac {\cot (c+d x) (a \sin (c+d x)+a)^{3/2}}{d}\) |
\(\Big \downarrow \) 3460 |
\(\displaystyle \frac {\frac {1}{3} \left (9 a^2 \int \csc (c+d x) \sqrt {\sin (c+d x) a+a}dx+\frac {22 a^3 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )+\frac {10 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{2 a}-\frac {\cot (c+d x) (a \sin (c+d x)+a)^{3/2}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{3} \left (9 a^2 \int \frac {\sqrt {\sin (c+d x) a+a}}{\sin (c+d x)}dx+\frac {22 a^3 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )+\frac {10 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{2 a}-\frac {\cot (c+d x) (a \sin (c+d x)+a)^{3/2}}{d}\) |
\(\Big \downarrow \) 3252 |
\(\displaystyle \frac {\frac {1}{3} \left (\frac {22 a^3 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}-\frac {18 a^3 \int \frac {1}{a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{d}\right )+\frac {10 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{2 a}-\frac {\cot (c+d x) (a \sin (c+d x)+a)^{3/2}}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {10 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}+\frac {1}{3} \left (\frac {22 a^3 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}-\frac {18 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}\right )}{2 a}-\frac {\cot (c+d x) (a \sin (c+d x)+a)^{3/2}}{d}\) |
Input:
Int[Cot[c + d*x]^2*(a + a*Sin[c + d*x])^(3/2),x]
Output:
-((Cot[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/d) + ((10*a^2*Cos[c + d*x]*Sqr t[a + a*Sin[c + d*x]])/(3*d) + ((-18*a^(5/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x] )/Sqrt[a + a*Sin[c + d*x]]])/d + (22*a^3*Cos[c + d*x])/(d*Sqrt[a + a*Sin[c + d*x]]))/3)/(2*a)
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_.)/tan[(e_.) + (f_.)*(x_)]^2, x_Symbol] :> Simp[-(a + b*Sin[e + f*x])^m/(f*Tan[e + f*x]), x] + Simp[1/a Int[(a + b*Sin[e + f*x])^m*((b*m - a*(m + 1)*Sin[e + f*x])/Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m - 1 /2] && !LtQ[m, -1]
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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-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] + Simp[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])
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] + Simp[(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]
Time = 0.33 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.27
method | result | size |
default | \(\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (12 \sqrt {a -a \sin \left (d x +c \right )}\, \sin \left (d x +c \right ) a^{\frac {3}{2}}-2 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sin \left (d x +c \right ) \sqrt {a}-9 \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right ) a^{2} \sin \left (d x +c \right )-3 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {3}{2}}\right )}{3 \sin \left (d x +c \right ) \sqrt {a}\, \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(154\) |
Input:
int(cot(d*x+c)^2*(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
Output:
1/3*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(12*(a-a*sin(d*x+c))^(1/2)*si n(d*x+c)*a^(3/2)-2*(a-a*sin(d*x+c))^(3/2)*sin(d*x+c)*a^(1/2)-9*arctanh((a- a*sin(d*x+c))^(1/2)/a^(1/2))*a^2*sin(d*x+c)-3*(a-a*sin(d*x+c))^(1/2)*a^(3/ 2))/sin(d*x+c)/a^(1/2)/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. 315 vs. \(2 (105) = 210\).
Time = 0.09 (sec) , antiderivative size = 315, normalized size of antiderivative = 2.60 \[ \int \cot ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {9 \, {\left (a \cos \left (d x + c\right )^{2} - {\left (a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) - a\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) + 4 \, {\left (2 \, a \cos \left (d x + c\right )^{3} - 8 \, a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right ) - {\left (2 \, a \cos \left (d x + c\right )^{2} + 10 \, a \cos \left (d x + c\right ) + 11 \, a\right )} \sin \left (d x + c\right ) + 11 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{12 \, {\left (d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right ) + d\right )} \sin \left (d x + c\right ) - d\right )}} \] Input:
integrate(cot(d*x+c)^2*(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")
Output:
1/12*(9*(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)*sqr t(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*(2*a*cos(d*x + c)^3 - 8*a*cos(d*x + c)^2 + a*cos(d*x + c) - (2*a*cos(d*x + c)^2 + 10*a*cos(d*x + c) + 11*a)*sin(d*x + c) + 11*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)
\[ \int \cot ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \cot ^{2}{\left (c + d x \right )}\, dx \] Input:
integrate(cot(d*x+c)**2*(a+a*sin(d*x+c))**(3/2),x)
Output:
Integral((a*(sin(c + d*x) + 1))**(3/2)*cot(c + d*x)**2, x)
\[ \int \cot ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cot \left (d x + c\right )^{2} \,d x } \] Input:
integrate(cot(d*x+c)^2*(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")
Output:
integrate((a*sin(d*x + c) + a)^(3/2)*cot(d*x + c)^2, x)
Time = 0.18 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.51 \[ \int \cot ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {\sqrt {2} {\left (16 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, \sqrt {2} a \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 48 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {12 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}\right )} \sqrt {a}}{12 \, d} \] Input:
integrate(cot(d*x+c)^2*(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")
Output:
1/12*sqrt(2)*(16*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 - 9*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*p i + 1/2*d*x + 1/2*c)) - 48*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4* pi + 1/2*d*x + 1/2*c) - 12*a*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))*sqrt(a)/d
Timed out. \[ \int \cot ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^2\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2} \,d x \] Input:
int(cot(c + d*x)^2*(a + a*sin(c + d*x))^(3/2),x)
Output:
int(cot(c + d*x)^2*(a + a*sin(c + d*x))^(3/2), x)
\[ \int \cot ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\sqrt {a}\, a \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \cot \left (d x +c \right )^{2} \sin \left (d x +c \right )d x +\int \sqrt {\sin \left (d x +c \right )+1}\, \cot \left (d x +c \right )^{2}d x \right ) \] Input:
int(cot(d*x+c)^2*(a+a*sin(d*x+c))^(3/2),x)
Output:
sqrt(a)*a*(int(sqrt(sin(c + d*x) + 1)*cot(c + d*x)**2*sin(c + d*x),x) + in t(sqrt(sin(c + d*x) + 1)*cot(c + d*x)**2,x))