Integrand size = 16, antiderivative size = 167 \[ \int \left (a+b \csc ^2(c+d x)\right )^{5/2} \, dx=-\frac {a^{5/2} \arctan \left (\frac {\sqrt {a} \cot (c+d x)}{\sqrt {a+b+b \cot ^2(c+d x)}}\right )}{d}-\frac {\sqrt {b} \left (15 a^2+10 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \cot (c+d x)}{\sqrt {a+b+b \cot ^2(c+d x)}}\right )}{8 d}-\frac {b (7 a+3 b) \cot (c+d x) \sqrt {a+b+b \cot ^2(c+d x)}}{8 d}-\frac {b \cot (c+d x) \left (a+b+b \cot ^2(c+d x)\right )^{3/2}}{4 d} \] Output:
-a^(5/2)*arctan(a^(1/2)*cot(d*x+c)/(a+b+b*cot(d*x+c)^2)^(1/2))/d-1/8*b^(1/ 2)*(15*a^2+10*a*b+3*b^2)*arctanh(b^(1/2)*cot(d*x+c)/(a+b+b*cot(d*x+c)^2)^( 1/2))/d-1/8*b*(7*a+3*b)*cot(d*x+c)*(a+b+b*cot(d*x+c)^2)^(1/2)/d-1/4*b*cot( d*x+c)*(a+b+b*cot(d*x+c)^2)^(3/2)/d
Time = 2.75 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.47 \[ \int \left (a+b \csc ^2(c+d x)\right )^{5/2} \, dx=\frac {\left (a+b \csc ^2(c+d x)\right )^{5/2} \left (\sqrt {2} b \left (15 a^2+10 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {-b} \cos (c+d x)}{\sqrt {-a-2 b+a \cos (2 (c+d x))}}\right )+\frac {1}{2} \sqrt {-b} \left (b \sqrt {-a-2 b+a \cos (2 (c+d x))} (-9 a-7 b+3 (3 a+b) \cos (2 (c+d x))) \cot (c+d x) \csc ^3(c+d x)+16 \sqrt {2} a^{5/2} \log \left (\sqrt {2} \sqrt {a} \cos (c+d x)+\sqrt {-a-2 b+a \cos (2 (c+d x))}\right )\right )\right ) \sin ^5(c+d x)}{2 \sqrt {-b} d (-a-2 b+a \cos (2 (c+d x)))^{5/2}} \] Input:
Integrate[(a + b*Csc[c + d*x]^2)^(5/2),x]
Output:
((a + b*Csc[c + d*x]^2)^(5/2)*(Sqrt[2]*b*(15*a^2 + 10*a*b + 3*b^2)*ArcTanh [(Sqrt[2]*Sqrt[-b]*Cos[c + d*x])/Sqrt[-a - 2*b + a*Cos[2*(c + d*x)]]] + (S qrt[-b]*(b*Sqrt[-a - 2*b + a*Cos[2*(c + d*x)]]*(-9*a - 7*b + 3*(3*a + b)*C os[2*(c + d*x)])*Cot[c + d*x]*Csc[c + d*x]^3 + 16*Sqrt[2]*a^(5/2)*Log[Sqrt [2]*Sqrt[a]*Cos[c + d*x] + Sqrt[-a - 2*b + a*Cos[2*(c + d*x)]]]))/2)*Sin[c + d*x]^5)/(2*Sqrt[-b]*d*(-a - 2*b + a*Cos[2*(c + d*x)])^(5/2))
Time = 0.35 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {3042, 4616, 318, 403, 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.
\(\displaystyle \int \left (a+b \csc ^2(c+d x)\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a+b \sec \left (c+d x+\frac {\pi }{2}\right )^2\right )^{5/2}dx\) |
\(\Big \downarrow \) 4616 |
\(\displaystyle -\frac {\int \frac {\left (b \cot ^2(c+d x)+a+b\right )^{5/2}}{\cot ^2(c+d x)+1}d\cot (c+d x)}{d}\) |
\(\Big \downarrow \) 318 |
\(\displaystyle -\frac {\frac {1}{4} \int \frac {\sqrt {b \cot ^2(c+d x)+a+b} \left (b (7 a+3 b) \cot ^2(c+d x)+(a+b) (4 a+3 b)\right )}{\cot ^2(c+d x)+1}d\cot (c+d x)+\frac {1}{4} b \cot (c+d x) \left (a+b \cot ^2(c+d x)+b\right )^{3/2}}{d}\) |
\(\Big \downarrow \) 403 |
\(\displaystyle -\frac {\frac {1}{4} \left (\frac {1}{2} \int \frac {b \left (15 a^2+10 b a+3 b^2\right ) \cot ^2(c+d x)+(a+b) \left (8 a^2+7 b a+3 b^2\right )}{\left (\cot ^2(c+d x)+1\right ) \sqrt {b \cot ^2(c+d x)+a+b}}d\cot (c+d x)+\frac {1}{2} b (7 a+3 b) \cot (c+d x) \sqrt {a+b \cot ^2(c+d x)+b}\right )+\frac {1}{4} b \cot (c+d x) \left (a+b \cot ^2(c+d x)+b\right )^{3/2}}{d}\) |
\(\Big \downarrow \) 398 |
\(\displaystyle -\frac {\frac {1}{4} \left (\frac {1}{2} \left (8 a^3 \int \frac {1}{\left (\cot ^2(c+d x)+1\right ) \sqrt {b \cot ^2(c+d x)+a+b}}d\cot (c+d x)+b \left (15 a^2+10 a b+3 b^2\right ) \int \frac {1}{\sqrt {b \cot ^2(c+d x)+a+b}}d\cot (c+d x)\right )+\frac {1}{2} b (7 a+3 b) \cot (c+d x) \sqrt {a+b \cot ^2(c+d x)+b}\right )+\frac {1}{4} b \cot (c+d x) \left (a+b \cot ^2(c+d x)+b\right )^{3/2}}{d}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle -\frac {\frac {1}{4} \left (\frac {1}{2} \left (8 a^3 \int \frac {1}{\left (\cot ^2(c+d x)+1\right ) \sqrt {b \cot ^2(c+d x)+a+b}}d\cot (c+d x)+b \left (15 a^2+10 a b+3 b^2\right ) \int \frac {1}{1-\frac {b \cot ^2(c+d x)}{b \cot ^2(c+d x)+a+b}}d\frac {\cot (c+d x)}{\sqrt {b \cot ^2(c+d x)+a+b}}\right )+\frac {1}{2} b (7 a+3 b) \cot (c+d x) \sqrt {a+b \cot ^2(c+d x)+b}\right )+\frac {1}{4} b \cot (c+d x) \left (a+b \cot ^2(c+d x)+b\right )^{3/2}}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\frac {1}{4} \left (\frac {1}{2} \left (8 a^3 \int \frac {1}{\left (\cot ^2(c+d x)+1\right ) \sqrt {b \cot ^2(c+d x)+a+b}}d\cot (c+d x)+\sqrt {b} \left (15 a^2+10 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)+b}}\right )\right )+\frac {1}{2} b (7 a+3 b) \cot (c+d x) \sqrt {a+b \cot ^2(c+d x)+b}\right )+\frac {1}{4} b \cot (c+d x) \left (a+b \cot ^2(c+d x)+b\right )^{3/2}}{d}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle -\frac {\frac {1}{4} \left (\frac {1}{2} \left (8 a^3 \int \frac {1}{\frac {a \cot ^2(c+d x)}{b \cot ^2(c+d x)+a+b}+1}d\frac {\cot (c+d x)}{\sqrt {b \cot ^2(c+d x)+a+b}}+\sqrt {b} \left (15 a^2+10 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)+b}}\right )\right )+\frac {1}{2} b (7 a+3 b) \cot (c+d x) \sqrt {a+b \cot ^2(c+d x)+b}\right )+\frac {1}{4} b \cot (c+d x) \left (a+b \cot ^2(c+d x)+b\right )^{3/2}}{d}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {\frac {1}{4} \left (\frac {1}{2} \left (8 a^{5/2} \arctan \left (\frac {\sqrt {a} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)+b}}\right )+\sqrt {b} \left (15 a^2+10 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)+b}}\right )\right )+\frac {1}{2} b (7 a+3 b) \cot (c+d x) \sqrt {a+b \cot ^2(c+d x)+b}\right )+\frac {1}{4} b \cot (c+d x) \left (a+b \cot ^2(c+d x)+b\right )^{3/2}}{d}\) |
Input:
Int[(a + b*Csc[c + d*x]^2)^(5/2),x]
Output:
-(((b*Cot[c + d*x]*(a + b + b*Cot[c + d*x]^2)^(3/2))/4 + ((8*a^(5/2)*ArcTa n[(Sqrt[a]*Cot[c + d*x])/Sqrt[a + b + b*Cot[c + d*x]^2]] + Sqrt[b]*(15*a^2 + 10*a*b + 3*b^2)*ArcTanh[(Sqrt[b]*Cot[c + d*x])/Sqrt[a + b + b*Cot[c + d *x]^2]])/2 + (b*(7*a + 3*b)*Cot[c + d*x]*Sqrt[a + b + b*Cot[c + d*x]^2])/2 )/4)/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_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*( x_)^2), x_Symbol] :> Simp[f*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(b*(2*(p + q + 1) + 1))), x] + Simp[1/(b*(2*(p + q + 1) + 1)) Int[(a + b*x^2)^p*(c + d*x^2)^(q - 1)*Simp[c*(b*e - a*f + b*e*2*(p + q + 1)) + (d*(b*e - a*f) + f*2*q*(b*c - a*d) + b*d*e*2*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && GtQ[q, 0] && NeQ[2*(p + q + 1) + 1, 0]
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. \(1050\) vs. \(2(145)=290\).
Time = 0.80 (sec) , antiderivative size = 1051, normalized size of antiderivative = 6.29
Input:
int((a+b*csc(d*x+c)^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/32/d*4^(1/2)/(-a)^(1/2)*(a+b*csc(d*x+c)^2)^(5/2)/(a^2*sin(d*x+c)^4+2*b*a *sin(d*x+c)^2+b^2)/((a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/2)*(-18*cos(d* x+c)*sin(d*x+c)^3*(-a)^(1/2)*((a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/2)*a *b+cos(d*x+c)*(6*cos(d*x+c)^2-10)*b^2*((a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2 )^(1/2)*(-a)^(1/2)*sin(d*x+c)+(3*cos(d*x+c)-3)*sin(d*x+c)^3*(-a)^(1/2)*b^( 5/2)*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)+(-3*cos(d*x+c)+3)*sin(d*x+c)^3*(-a)^(1/2) *b^(5/2)*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)))+(10*cos(d*x+c)-10)*sin(d*x+c)^3*(-a)^(1/2)*b^(3/2) *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+(-10*cos(d*x+c)+10)*sin(d*x+c)^3*(-a)^(1/2) *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+(15*cos(d*x+c)-15)*sin(d*x+c)^3*(-a)^(1/2)*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^2+(-15*cos(d*x+c)+15)*sin(d*x+c)^3*(-a...
Leaf count of result is larger than twice the leaf count of optimal. 417 vs. \(2 (145) = 290\).
Time = 0.99 (sec) , antiderivative size = 1986, normalized size of antiderivative = 11.89 \[ \int \left (a+b \csc ^2(c+d x)\right )^{5/2} \, dx=\text {Too large to display} \] Input:
integrate((a+b*csc(d*x+c)^2)^(5/2),x, algorithm="fricas")
Output:
[1/32*(4*(a^2*cos(d*x + c)^2 - a^2)*sqrt(-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))*sin(d*x + c))*sin(d*x + c) + ((15*a^2 + 10*a*b + 3*b^2)*cos(d*x + c)^2 - 15*a^2 - 10*a*b - 3*b^2)*sqrt(b)*log(2*((a^2 - 6* a*b + b^2)*cos(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)/(cos(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*(3*(3*a*b + b^2)*cos(d* x + c)^3 - (9*a*b + 5*b^2)*cos(d*x + c))*sqrt((a*cos(d*x + c)^2 - a - b)/( cos(d*x + c)^2 - 1)))/((d*cos(d*x + c)^2 - d)*sin(d*x + c)), -1/16*(((15*a ^2 + 10*a*b + 3*b^2)*cos(d*x + c)^2 - 15*a^2 - 10*a*b - 3*b^2)*sqrt(-b)*ar ctan(-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) - 2*(a^2*cos(d*x + c)^2 - a^2)*sqrt(-a)*l og(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...
\[ \int \left (a+b \csc ^2(c+d x)\right )^{5/2} \, dx=\int \left (a + b \csc ^{2}{\left (c + d x \right )}\right )^{\frac {5}{2}}\, dx \] Input:
integrate((a+b*csc(d*x+c)**2)**(5/2),x)
Output:
Integral((a + b*csc(c + d*x)**2)**(5/2), x)
\[ \int \left (a+b \csc ^2(c+d x)\right )^{5/2} \, dx=\int { {\left (b \csc \left (d x + c\right )^{2} + a\right )}^{\frac {5}{2}} \,d x } \] Input:
integrate((a+b*csc(d*x+c)^2)^(5/2),x, algorithm="maxima")
Output:
integrate((b*csc(d*x + c)^2 + a)^(5/2), x)
Exception generated. \[ \int \left (a+b \csc ^2(c+d x)\right )^{5/2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+b*csc(d*x+c)^2)^(5/2),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. \[ \int \left (a+b \csc ^2(c+d x)\right )^{5/2} \, dx=\int {\left (a+\frac {b}{{\sin \left (c+d\,x\right )}^2}\right )}^{5/2} \,d x \] Input:
int((a + b/sin(c + d*x)^2)^(5/2),x)
Output:
int((a + b/sin(c + d*x)^2)^(5/2), x)
\[ \int \left (a+b \csc ^2(c+d x)\right )^{5/2} \, dx=\left (\int \sqrt {\csc \left (d x +c \right )^{2} b +a}d x \right ) a^{2}+\left (\int \sqrt {\csc \left (d x +c \right )^{2} b +a}\, \csc \left (d x +c \right )^{4}d x \right ) b^{2}+2 \left (\int \sqrt {\csc \left (d x +c \right )^{2} b +a}\, \csc \left (d x +c \right )^{2}d x \right ) a b \] Input:
int((a+b*csc(d*x+c)^2)^(5/2),x)
Output:
int(sqrt(csc(c + d*x)**2*b + a),x)*a**2 + int(sqrt(csc(c + d*x)**2*b + a)* csc(c + d*x)**4,x)*b**2 + 2*int(sqrt(csc(c + d*x)**2*b + a)*csc(c + d*x)** 2,x)*a*b