Integrand size = 16, antiderivative size = 119 \[ \int \left (a+b \csc ^2(c+d x)\right )^{3/2} \, dx=-\frac {a^{3/2} \arctan \left (\frac {\sqrt {a} \cot (c+d x)}{\sqrt {a+b+b \cot ^2(c+d x)}}\right )}{d}-\frac {\sqrt {b} (3 a+b) \text {arctanh}\left (\frac {\sqrt {b} \cot (c+d x)}{\sqrt {a+b+b \cot ^2(c+d x)}}\right )}{2 d}-\frac {b \cot (c+d x) \sqrt {a+b+b \cot ^2(c+d x)}}{2 d} \] Output:
-a^(3/2)*arctan(a^(1/2)*cot(d*x+c)/(a+b+b*cot(d*x+c)^2)^(1/2))/d-1/2*b^(1/ 2)*(3*a+b)*arctanh(b^(1/2)*cot(d*x+c)/(a+b+b*cot(d*x+c)^2)^(1/2))/d-1/2*b* cot(d*x+c)*(a+b+b*cot(d*x+c)^2)^(1/2)/d
Time = 0.80 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.75 \[ \int \left (a+b \csc ^2(c+d x)\right )^{3/2} \, dx=\frac {b \left (a+b \csc ^2(c+d x)\right )^{3/2} \left (\sqrt {2} b (3 a+b) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {-b} \cos (c+d x)}{\sqrt {-a-2 b+a \cos (2 (c+d x))}}\right )+\sqrt {-b} \left (-b \sqrt {-a-2 b+a \cos (2 (c+d x))} \cot (c+d x) \csc (c+d x)+2 \sqrt {2} a^{3/2} \log \left (\sqrt {2} \sqrt {a} \cos (c+d x)+\sqrt {-a-2 b+a \cos (2 (c+d x))}\right )\right )\right ) \sin ^3(c+d x)}{(-b)^{3/2} d (-a-2 b+a \cos (2 (c+d x)))^{3/2}} \] Input:
Integrate[(a + b*Csc[c + d*x]^2)^(3/2),x]
Output:
(b*(a + b*Csc[c + d*x]^2)^(3/2)*(Sqrt[2]*b*(3*a + b)*ArcTanh[(Sqrt[2]*Sqrt [-b]*Cos[c + d*x])/Sqrt[-a - 2*b + a*Cos[2*(c + d*x)]]] + Sqrt[-b]*(-(b*Sq rt[-a - 2*b + a*Cos[2*(c + d*x)]]*Cot[c + d*x]*Csc[c + d*x]) + 2*Sqrt[2]*a ^(3/2)*Log[Sqrt[2]*Sqrt[a]*Cos[c + d*x] + Sqrt[-a - 2*b + a*Cos[2*(c + d*x )]]]))*Sin[c + d*x]^3)/((-b)^(3/2)*d*(-a - 2*b + a*Cos[2*(c + d*x)])^(3/2) )
Time = 0.28 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.98, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4616, 318, 398, 224, 219, 291, 216}
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.
\(\displaystyle \int \left (a+b \csc ^2(c+d x)\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a+b \sec \left (c+d x+\frac {\pi }{2}\right )^2\right )^{3/2}dx\) |
\(\Big \downarrow \) 4616 |
\(\displaystyle -\frac {\int \frac {\left (b \cot ^2(c+d x)+a+b\right )^{3/2}}{\cot ^2(c+d x)+1}d\cot (c+d x)}{d}\) |
\(\Big \downarrow \) 318 |
\(\displaystyle -\frac {\frac {1}{2} \int \frac {b (3 a+b) \cot ^2(c+d x)+(a+b) (2 a+b)}{\left (\cot ^2(c+d x)+1\right ) \sqrt {b \cot ^2(c+d x)+a+b}}d\cot (c+d x)+\frac {1}{2} b \cot (c+d x) \sqrt {a+b \cot ^2(c+d x)+b}}{d}\) |
\(\Big \downarrow \) 398 |
\(\displaystyle -\frac {\frac {1}{2} \left (2 a^2 \int \frac {1}{\left (\cot ^2(c+d x)+1\right ) \sqrt {b \cot ^2(c+d x)+a+b}}d\cot (c+d x)+b (3 a+b) \int \frac {1}{\sqrt {b \cot ^2(c+d x)+a+b}}d\cot (c+d x)\right )+\frac {1}{2} b \cot (c+d x) \sqrt {a+b \cot ^2(c+d x)+b}}{d}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle -\frac {\frac {1}{2} \left (2 a^2 \int \frac {1}{\left (\cot ^2(c+d x)+1\right ) \sqrt {b \cot ^2(c+d x)+a+b}}d\cot (c+d x)+b (3 a+b) \int \frac {1}{1-\frac {b \cot ^2(c+d x)}{b \cot ^2(c+d x)+a+b}}d\frac {\cot (c+d x)}{\sqrt {b \cot ^2(c+d x)+a+b}}\right )+\frac {1}{2} b \cot (c+d x) \sqrt {a+b \cot ^2(c+d x)+b}}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\frac {1}{2} \left (2 a^2 \int \frac {1}{\left (\cot ^2(c+d x)+1\right ) \sqrt {b \cot ^2(c+d x)+a+b}}d\cot (c+d x)+\sqrt {b} (3 a+b) \text {arctanh}\left (\frac {\sqrt {b} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)+b}}\right )\right )+\frac {1}{2} b \cot (c+d x) \sqrt {a+b \cot ^2(c+d x)+b}}{d}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle -\frac {\frac {1}{2} \left (2 a^2 \int \frac {1}{\frac {a \cot ^2(c+d x)}{b \cot ^2(c+d x)+a+b}+1}d\frac {\cot (c+d x)}{\sqrt {b \cot ^2(c+d x)+a+b}}+\sqrt {b} (3 a+b) \text {arctanh}\left (\frac {\sqrt {b} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)+b}}\right )\right )+\frac {1}{2} b \cot (c+d x) \sqrt {a+b \cot ^2(c+d x)+b}}{d}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {\frac {1}{2} \left (2 a^{3/2} \arctan \left (\frac {\sqrt {a} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)+b}}\right )+\sqrt {b} (3 a+b) \text {arctanh}\left (\frac {\sqrt {b} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)+b}}\right )\right )+\frac {1}{2} b \cot (c+d x) \sqrt {a+b \cot ^2(c+d x)+b}}{d}\) |
Input:
Int[(a + b*Csc[c + d*x]^2)^(3/2),x]
Output:
-(((2*a^(3/2)*ArcTan[(Sqrt[a]*Cot[c + d*x])/Sqrt[a + b + b*Cot[c + d*x]^2] ] + Sqrt[b]*(3*a + b)*ArcTanh[(Sqrt[b]*Cot[c + d*x])/Sqrt[a + b + b*Cot[c + d*x]^2]])/2 + (b*Cot[c + d*x]*Sqrt[a + b + b*Cot[c + d*x]^2])/2)/d)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S imp[1/(b*(2*(p + q) + 1)) Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b *c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G tQ[q, 1] && NeQ[2*(p + q) + 1, 0] && !IGtQ[p, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]) , x_Symbol] :> Simp[f/b Int[1/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/ b Int[1/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f} , x]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[(a + b + b*ff^2*x^2)^p /(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && NeQ[a + b, 0] && NeQ[p, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(675\) vs. \(2(101)=202\).
Time = 0.46 (sec) , antiderivative size = 676, normalized size of antiderivative = 5.68
method | result | size |
default | \(-\frac {\sqrt {4}\, \sin \left (d x +c \right ) \left (\left (-3 \cos \left (d x +c \right )+3\right ) \sqrt {b}\, \ln \left (\frac {4 \sqrt {b}\, \sqrt {\frac {a \sin \left (d x +c \right )^{2}+b}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, \sin \left (d x +c \right )^{2}+4 a \cos \left (d x +c \right )^{2}+2 b \cos \left (d x +c \right )^{2}+2 b \sin \left (d x +c \right )^{2}-8 \cos \left (d x +c \right ) a -4 \cos \left (d x +c \right ) b +4 a +2 b}{\left (-1+\cos \left (d x +c \right )\right )^{2}}\right ) a \sqrt {-a}+\left (3 \cos \left (d x +c \right )-3\right ) \sqrt {b}\, a \ln \left (\frac {2 \sqrt {\frac {a \sin \left (d x +c \right )^{2}+b}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, \sqrt {b}\, \cos \left (d x +c \right )+2 \sqrt {b}\, \sqrt {\frac {a \sin \left (d x +c \right )^{2}+b}{\left (1+\cos \left (d x +c \right )\right )^{2}}}+2 \cos \left (d x +c \right ) a +2 a +2 b}{\sqrt {b}\, \left (1+\cos \left (d x +c \right )\right )}\right ) \sqrt {-a}+\left (1-\cos \left (d x +c \right )\right ) b^{\frac {3}{2}} \ln \left (\frac {4 \sqrt {b}\, \sqrt {\frac {a \sin \left (d x +c \right )^{2}+b}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, \sin \left (d x +c \right )^{2}+4 a \cos \left (d x +c \right )^{2}+2 b \cos \left (d x +c \right )^{2}+2 b \sin \left (d x +c \right )^{2}-8 \cos \left (d x +c \right ) a -4 \cos \left (d x +c \right ) b +4 a +2 b}{\left (-1+\cos \left (d x +c \right )\right )^{2}}\right ) \sqrt {-a}+\left (-1+\cos \left (d x +c \right )\right ) b^{\frac {3}{2}} \ln \left (\frac {2 \sqrt {\frac {a \sin \left (d x +c \right )^{2}+b}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, \sqrt {b}\, \cos \left (d x +c \right )+2 \sqrt {b}\, \sqrt {\frac {a \sin \left (d x +c \right )^{2}+b}{\left (1+\cos \left (d x +c \right )\right )^{2}}}+2 \cos \left (d x +c \right ) a +2 a +2 b}{\sqrt {b}\, \left (1+\cos \left (d x +c \right )\right )}\right ) \sqrt {-a}+\left (-4 \cos \left (d x +c \right )+4\right ) a^{2} \ln \left (4 \sqrt {-a}\, \sqrt {\frac {a \sin \left (d x +c \right )^{2}+b}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, \cos \left (d x +c \right )+4 \sqrt {-a}\, \sqrt {\frac {a \sin \left (d x +c \right )^{2}+b}{\left (1+\cos \left (d x +c \right )\right )^{2}}}-4 \cos \left (d x +c \right ) a \right )+2 \sqrt {\frac {a \sin \left (d x +c \right )^{2}+b}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, \cos \left (d x +c \right ) \sqrt {-a}\, b \right ) \left (a +b \csc \left (d x +c \right )^{2}\right )^{\frac {3}{2}}}{8 d \sqrt {-a}\, \left (a \sin \left (d x +c \right )^{2}+b \right ) \sqrt {\frac {a \sin \left (d x +c \right )^{2}+b}{\left (1+\cos \left (d x +c \right )\right )^{2}}}}\) | \(676\) |
Input:
int((a+b*csc(d*x+c)^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/8/d*4^(1/2)/(-a)^(1/2)*sin(d*x+c)*((-3*cos(d*x+c)+3)*b^(1/2)*ln(2*(2*b^ (1/2)*((a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/2)*sin(d*x+c)^2+2*a*cos(d*x +c)^2+b*cos(d*x+c)^2+b*sin(d*x+c)^2-4*cos(d*x+c)*a-2*cos(d*x+c)*b+2*a+b)/( -1+cos(d*x+c))^2)*a*(-a)^(1/2)+(3*cos(d*x+c)-3)*b^(1/2)*a*ln(2/b^(1/2)*((( a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/2)*b^(1/2)*cos(d*x+c)+b^(1/2)*((a*s in(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/2)+cos(d*x+c)*a+a+b)/(1+cos(d*x+c)))*( -a)^(1/2)+(1-cos(d*x+c))*b^(3/2)*ln(2*(2*b^(1/2)*((a*sin(d*x+c)^2+b)/(1+co s(d*x+c))^2)^(1/2)*sin(d*x+c)^2+2*a*cos(d*x+c)^2+b*cos(d*x+c)^2+b*sin(d*x+ c)^2-4*cos(d*x+c)*a-2*cos(d*x+c)*b+2*a+b)/(-1+cos(d*x+c))^2)*(-a)^(1/2)+(- 1+cos(d*x+c))*b^(3/2)*ln(2/b^(1/2)*(((a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2)^ (1/2)*b^(1/2)*cos(d*x+c)+b^(1/2)*((a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/ 2)+cos(d*x+c)*a+a+b)/(1+cos(d*x+c)))*(-a)^(1/2)+(-4*cos(d*x+c)+4)*a^2*ln(4 *(-a)^(1/2)*((a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)+4*(-a)^ (1/2)*((a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/2)-4*cos(d*x+c)*a)+2*((a*si n(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)*(-a)^(1/2)*b)*(a+b*csc(d* x+c)^2)^(3/2)/(a*sin(d*x+c)^2+b)/((a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/ 2)
Leaf count of result is larger than twice the leaf count of optimal. 321 vs. \(2 (101) = 202\).
Time = 0.38 (sec) , antiderivative size = 1607, normalized size of antiderivative = 13.50 \[ \int \left (a+b \csc ^2(c+d x)\right )^{3/2} \, dx=\text {Too large to display} \] Input:
integrate((a+b*csc(d*x+c)^2)^(3/2),x, algorithm="fricas")
Output:
[1/8*(sqrt(-a)*a*log(128*a^4*cos(d*x + c)^8 - 256*(a^4 + a^3*b)*cos(d*x + c)^6 + 160*(a^4 + 2*a^3*b + a^2*b^2)*cos(d*x + c)^4 + a^4 + 4*a^3*b + 6*a^ 2*b^2 + 4*a*b^3 + b^4 - 32*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cos(d*x + c )^2 - 8*(16*a^3*cos(d*x + c)^7 - 24*(a^3 + a^2*b)*cos(d*x + c)^5 + 10*(a^3 + 2*a^2*b + a*b^2)*cos(d*x + c)^3 - (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(d *x + c))*sqrt(-a)*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1))*si n(d*x + c))*sin(d*x + c) + (3*a + b)*sqrt(b)*log(2*((a^2 - 6*a*b + b^2)*co s(d*x + c)^4 - 2*(a^2 - 2*a*b - 3*b^2)*cos(d*x + c)^2 + 4*((a - b)*cos(d*x + c)^3 - (a + b)*cos(d*x + c))*sqrt(b)*sqrt((a*cos(d*x + c)^2 - a - b)/(c os(d*x + c)^2 - 1))*sin(d*x + c) + a^2 + 2*a*b + b^2)/(cos(d*x + c)^4 - 2* cos(d*x + c)^2 + 1))*sin(d*x + c) - 4*b*sqrt((a*cos(d*x + c)^2 - a - b)/(c os(d*x + c)^2 - 1))*cos(d*x + c))/(d*sin(d*x + c)), -1/8*(2*(3*a + b)*sqrt (-b)*arctan(-1/2*((a - b)*cos(d*x + c)^2 - a - b)*sqrt(-b)*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1))*sin(d*x + c)/(a*b*cos(d*x + c)^3 - (a*b + b^2)*cos(d*x + c)))*sin(d*x + c) - sqrt(-a)*a*log(128*a^4*cos(d*x + c)^8 - 256*(a^4 + a^3*b)*cos(d*x + c)^6 + 160*(a^4 + 2*a^3*b + a^2*b^2)*c os(d*x + c)^4 + a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4 - 32*(a^4 + 3*a^ 3*b + 3*a^2*b^2 + a*b^3)*cos(d*x + c)^2 - 8*(16*a^3*cos(d*x + c)^7 - 24*(a ^3 + a^2*b)*cos(d*x + c)^5 + 10*(a^3 + 2*a^2*b + a*b^2)*cos(d*x + c)^3 - ( a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x +...
\[ \int \left (a+b \csc ^2(c+d x)\right )^{3/2} \, dx=\int \left (a + b \csc ^{2}{\left (c + d x \right )}\right )^{\frac {3}{2}}\, dx \] Input:
integrate((a+b*csc(d*x+c)**2)**(3/2),x)
Output:
Integral((a + b*csc(c + d*x)**2)**(3/2), x)
\[ \int \left (a+b \csc ^2(c+d x)\right )^{3/2} \, dx=\int { {\left (b \csc \left (d x + c\right )^{2} + a\right )}^{\frac {3}{2}} \,d x } \] Input:
integrate((a+b*csc(d*x+c)^2)^(3/2),x, algorithm="maxima")
Output:
integrate((b*csc(d*x + c)^2 + a)^(3/2), x)
Exception generated. \[ \int \left (a+b \csc ^2(c+d x)\right )^{3/2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+b*csc(d*x+c)^2)^(3/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
Timed out. \[ \int \left (a+b \csc ^2(c+d x)\right )^{3/2} \, dx=\int {\left (a+\frac {b}{{\sin \left (c+d\,x\right )}^2}\right )}^{3/2} \,d x \] Input:
int((a + b/sin(c + d*x)^2)^(3/2),x)
Output:
int((a + b/sin(c + d*x)^2)^(3/2), x)
\[ \int \left (a+b \csc ^2(c+d x)\right )^{3/2} \, dx=\left (\int \sqrt {\csc \left (d x +c \right )^{2} b +a}d x \right ) a +\left (\int \sqrt {\csc \left (d x +c \right )^{2} b +a}\, \csc \left (d x +c \right )^{2}d x \right ) b \] Input:
int((a+b*csc(d*x+c)^2)^(3/2),x)
Output:
int(sqrt(csc(c + d*x)**2*b + a),x)*a + int(sqrt(csc(c + d*x)**2*b + a)*csc (c + d*x)**2,x)*b