Integrand size = 23, antiderivative size = 585 \[ \int (f x)^m \left (d+e x^2\right )^3 \left (a+b \csc ^{-1}(c x)\right ) \, dx=\frac {b e \left (e^2 \left (15+8 m+m^2\right )^2+3 c^2 d e (3+m)^2 \left (42+13 m+m^2\right )+3 c^4 d^2 \left (840+638 m+179 m^2+22 m^3+m^4\right )\right ) x (f x)^{1+m} \sqrt {-1+c^2 x^2}}{c^5 f (2+m) (3+m) (4+m) (5+m) (6+m) (7+m) \sqrt {c^2 x^2}}+\frac {b e^2 \left (e (5+m)^2+3 c^2 d \left (42+13 m+m^2\right )\right ) x (f x)^{3+m} \sqrt {-1+c^2 x^2}}{c^3 f^3 (4+m) (5+m) (6+m) (7+m) \sqrt {c^2 x^2}}+\frac {b e^3 x (f x)^{5+m} \sqrt {-1+c^2 x^2}}{c f^5 (6+m) (7+m) \sqrt {c^2 x^2}}+\frac {d^3 (f x)^{1+m} \left (a+b \csc ^{-1}(c x)\right )}{f (1+m)}+\frac {3 d^2 e (f x)^{3+m} \left (a+b \csc ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {3 d e^2 (f x)^{5+m} \left (a+b \csc ^{-1}(c x)\right )}{f^5 (5+m)}+\frac {e^3 (f x)^{7+m} \left (a+b \csc ^{-1}(c x)\right )}{f^7 (7+m)}+\frac {b \left (\frac {c^6 d^3 (2+m) (4+m) (6+m)}{1+m}+\frac {e (1+m) \left (e^2 \left (15+8 m+m^2\right )^2+3 c^2 d e (3+m)^2 \left (42+13 m+m^2\right )+3 c^4 d^2 \left (840+638 m+179 m^2+22 m^3+m^4\right )\right )}{(3+m) (5+m) (7+m)}\right ) x (f x)^{1+m} \sqrt {1-c^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},c^2 x^2\right )}{c^5 f (1+m) (2+m) (4+m) (6+m) \sqrt {c^2 x^2} \sqrt {-1+c^2 x^2}} \]
d^3*(f*x)^(1+m)*(a+b*arccsc(c*x))/f/(1+m)+3*d^2*e*(f*x)^(3+m)*(a+b*arccsc( c*x))/f^3/(3+m)+3*d*e^2*(f*x)^(5+m)*(a+b*arccsc(c*x))/f^5/(5+m)+e^3*(f*x)^ (7+m)*(a+b*arccsc(c*x))/f^7/(7+m)+b*(c^6*d^3*(2+m)*(4+m)*(6+m)/(1+m)+e*(1+ m)*(e^2*(m^2+8*m+15)^2+3*c^2*d*e*(3+m)^2*(m^2+13*m+42)+3*c^4*d^2*(m^4+22*m ^3+179*m^2+638*m+840))/(m^3+15*m^2+71*m+105))*x*(f*x)^(1+m)*hypergeom([1/2 , 1/2+1/2*m],[3/2+1/2*m],c^2*x^2)*(-c^2*x^2+1)^(1/2)/c^5/f/(1+m)/(2+m)/(4+ m)/(6+m)/(c^2*x^2)^(1/2)/(c^2*x^2-1)^(1/2)+b*e*(e^2*(m^2+8*m+15)^2+3*c^2*d *e*(3+m)^2*(m^2+13*m+42)+3*c^4*d^2*(m^4+22*m^3+179*m^2+638*m+840))*x*(f*x) ^(1+m)*(c^2*x^2-1)^(1/2)/c^5/f/(6+m)/(m^2+6*m+8)/(m^3+15*m^2+71*m+105)/(c^ 2*x^2)^(1/2)+b*e^2*(e*(5+m)^2+3*c^2*d*(m^2+13*m+42))*x*(f*x)^(3+m)*(c^2*x^ 2-1)^(1/2)/c^3/f^3/(4+m)/(5+m)/(6+m)/(7+m)/(c^2*x^2)^(1/2)+b*e^3*x*(f*x)^( 5+m)*(c^2*x^2-1)^(1/2)/c/f^5/(6+m)/(7+m)/(c^2*x^2)^(1/2)
Time = 0.88 (sec) , antiderivative size = 402, normalized size of antiderivative = 0.69 \[ \int (f x)^m \left (d+e x^2\right )^3 \left (a+b \csc ^{-1}(c x)\right ) \, dx=x (f x)^m \left (\frac {a d^3}{1+m}+\frac {3 a d^2 e x^2}{3+m}+\frac {3 a d e^2 x^4}{5+m}+\frac {a e^3 x^6}{7+m}+\frac {b d^3 \csc ^{-1}(c x)}{1+m}+\frac {3 b d^2 e x^2 \csc ^{-1}(c x)}{3+m}+\frac {3 b d e^2 x^4 \csc ^{-1}(c x)}{5+m}+\frac {b e^3 x^6 \csc ^{-1}(c x)}{7+m}-\frac {b c d^3 \sqrt {1-\frac {1}{c^2 x^2}} x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},c^2 x^2\right )}{(1+m)^2 \sqrt {1-c^2 x^2}}-\frac {3 b c d^2 e \sqrt {1-\frac {1}{c^2 x^2}} x^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},c^2 x^2\right )}{(3+m)^2 \sqrt {1-c^2 x^2}}-\frac {3 b c d e^2 \sqrt {1-\frac {1}{c^2 x^2}} x^5 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5+m}{2},\frac {7+m}{2},c^2 x^2\right )}{(5+m)^2 \sqrt {1-c^2 x^2}}-\frac {b c e^3 \sqrt {1-\frac {1}{c^2 x^2}} x^7 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7+m}{2},\frac {9+m}{2},c^2 x^2\right )}{(7+m)^2 \sqrt {1-c^2 x^2}}\right ) \]
x*(f*x)^m*((a*d^3)/(1 + m) + (3*a*d^2*e*x^2)/(3 + m) + (3*a*d*e^2*x^4)/(5 + m) + (a*e^3*x^6)/(7 + m) + (b*d^3*ArcCsc[c*x])/(1 + m) + (3*b*d^2*e*x^2* ArcCsc[c*x])/(3 + m) + (3*b*d*e^2*x^4*ArcCsc[c*x])/(5 + m) + (b*e^3*x^6*Ar cCsc[c*x])/(7 + m) - (b*c*d^3*Sqrt[1 - 1/(c^2*x^2)]*x*Hypergeometric2F1[1/ 2, (1 + m)/2, (3 + m)/2, c^2*x^2])/((1 + m)^2*Sqrt[1 - c^2*x^2]) - (3*b*c* d^2*e*Sqrt[1 - 1/(c^2*x^2)]*x^3*Hypergeometric2F1[1/2, (3 + m)/2, (5 + m)/ 2, c^2*x^2])/((3 + m)^2*Sqrt[1 - c^2*x^2]) - (3*b*c*d*e^2*Sqrt[1 - 1/(c^2* x^2)]*x^5*Hypergeometric2F1[1/2, (5 + m)/2, (7 + m)/2, c^2*x^2])/((5 + m)^ 2*Sqrt[1 - c^2*x^2]) - (b*c*e^3*Sqrt[1 - 1/(c^2*x^2)]*x^7*Hypergeometric2F 1[1/2, (7 + m)/2, (9 + m)/2, c^2*x^2])/((7 + m)^2*Sqrt[1 - c^2*x^2]))
Time = 2.15 (sec) , antiderivative size = 541, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {5762, 2340, 1590, 363, 279, 278}
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.
| \(\displaystyle \int \left (d+e x^2\right )^3 (f x)^m \left (a+b \csc ^{-1}(c x)\right ) \, dx\) |
| \(\Big \downarrow \) 5762 |
| \(\displaystyle \frac {b c x \int \frac {(f x)^m \left (\frac {e^3 x^6}{m+7}+\frac {3 d e^2 x^4}{m+5}+\frac {3 d^2 e x^2}{m+3}+\frac {d^3}{m+1}\right )}{\sqrt {c^2 x^2-1}}dx}{\sqrt {c^2 x^2}}+\frac {d^3 (f x)^{m+1} \left (a+b \csc ^{-1}(c x)\right )}{f (m+1)}+\frac {3 d^2 e (f x)^{m+3} \left (a+b \csc ^{-1}(c x)\right )}{f^3 (m+3)}+\frac {3 d e^2 (f x)^{m+5} \left (a+b \csc ^{-1}(c x)\right )}{f^5 (m+5)}+\frac {e^3 (f x)^{m+7} \left (a+b \csc ^{-1}(c x)\right )}{f^7 (m+7)}\) |
| \(\Big \downarrow \) 2340 |
| \(\displaystyle \frac {b c x \left (\frac {\int \frac {(f x)^m \left (\frac {e^2 \left (3 d \left (m^2+13 m+42\right ) c^2+e (m+5)^2\right ) x^4}{(m+5) (m+7)}+\frac {3 c^2 d^2 e (m+6) x^2}{m+3}+\frac {c^2 d^3 (m+6)}{m+1}\right )}{\sqrt {c^2 x^2-1}}dx}{c^2 (m+6)}+\frac {e^3 \sqrt {c^2 x^2-1} (f x)^{m+5}}{c^2 f^5 (m+6) (m+7)}\right )}{\sqrt {c^2 x^2}}+\frac {d^3 (f x)^{m+1} \left (a+b \csc ^{-1}(c x)\right )}{f (m+1)}+\frac {3 d^2 e (f x)^{m+3} \left (a+b \csc ^{-1}(c x)\right )}{f^3 (m+3)}+\frac {3 d e^2 (f x)^{m+5} \left (a+b \csc ^{-1}(c x)\right )}{f^5 (m+5)}+\frac {e^3 (f x)^{m+7} \left (a+b \csc ^{-1}(c x)\right )}{f^7 (m+7)}\) |
| \(\Big \downarrow \) 1590 |
| \(\displaystyle \frac {b c x \left (\frac {\frac {\int \frac {(f x)^m \left (\frac {d^3 (m+4) (m+6) c^4}{m+1}+\frac {e \left (3 d^2 \left (m^4+22 m^3+179 m^2+638 m+840\right ) c^4+3 d e (m+3)^2 \left (m^2+13 m+42\right ) c^2+e^2 \left (m^2+8 m+15\right )^2\right ) x^2}{(m+3) (m+5) (m+7)}\right )}{\sqrt {c^2 x^2-1}}dx}{c^2 (m+4)}+\frac {e^2 \sqrt {c^2 x^2-1} (f x)^{m+3} \left (3 c^2 d \left (m^2+13 m+42\right )+e (m+5)^2\right )}{c^2 f^3 (m+4) (m+5) (m+7)}}{c^2 (m+6)}+\frac {e^3 \sqrt {c^2 x^2-1} (f x)^{m+5}}{c^2 f^5 (m+6) (m+7)}\right )}{\sqrt {c^2 x^2}}+\frac {d^3 (f x)^{m+1} \left (a+b \csc ^{-1}(c x)\right )}{f (m+1)}+\frac {3 d^2 e (f x)^{m+3} \left (a+b \csc ^{-1}(c x)\right )}{f^3 (m+3)}+\frac {3 d e^2 (f x)^{m+5} \left (a+b \csc ^{-1}(c x)\right )}{f^5 (m+5)}+\frac {e^3 (f x)^{m+7} \left (a+b \csc ^{-1}(c x)\right )}{f^7 (m+7)}\) |
| \(\Big \downarrow \) 363 |
| \(\displaystyle \frac {b c x \left (\frac {\frac {\left (\frac {c^4 d^3 (m+4) (m+6)}{m+1}+\frac {e (m+1) \left (3 c^4 d^2 \left (m^4+22 m^3+179 m^2+638 m+840\right )+3 c^2 d e (m+3)^2 \left (m^2+13 m+42\right )+e^2 \left (m^2+8 m+15\right )^2\right )}{c^2 (m+2) (m+3) (m+5) (m+7)}\right ) \int \frac {(f x)^m}{\sqrt {c^2 x^2-1}}dx+\frac {e \sqrt {c^2 x^2-1} (f x)^{m+1} \left (3 c^4 d^2 \left (m^4+22 m^3+179 m^2+638 m+840\right )+3 c^2 d e (m+3)^2 \left (m^2+13 m+42\right )+e^2 \left (m^2+8 m+15\right )^2\right )}{c^2 f (m+2) (m+3) (m+5) (m+7)}}{c^2 (m+4)}+\frac {e^2 \sqrt {c^2 x^2-1} (f x)^{m+3} \left (3 c^2 d \left (m^2+13 m+42\right )+e (m+5)^2\right )}{c^2 f^3 (m+4) (m+5) (m+7)}}{c^2 (m+6)}+\frac {e^3 \sqrt {c^2 x^2-1} (f x)^{m+5}}{c^2 f^5 (m+6) (m+7)}\right )}{\sqrt {c^2 x^2}}+\frac {d^3 (f x)^{m+1} \left (a+b \csc ^{-1}(c x)\right )}{f (m+1)}+\frac {3 d^2 e (f x)^{m+3} \left (a+b \csc ^{-1}(c x)\right )}{f^3 (m+3)}+\frac {3 d e^2 (f x)^{m+5} \left (a+b \csc ^{-1}(c x)\right )}{f^5 (m+5)}+\frac {e^3 (f x)^{m+7} \left (a+b \csc ^{-1}(c x)\right )}{f^7 (m+7)}\) |
| \(\Big \downarrow \) 279 |
| \(\displaystyle \frac {b c x \left (\frac {\frac {\frac {\sqrt {1-c^2 x^2} \left (\frac {c^4 d^3 (m+4) (m+6)}{m+1}+\frac {e (m+1) \left (3 c^4 d^2 \left (m^4+22 m^3+179 m^2+638 m+840\right )+3 c^2 d e (m+3)^2 \left (m^2+13 m+42\right )+e^2 \left (m^2+8 m+15\right )^2\right )}{c^2 (m+2) (m+3) (m+5) (m+7)}\right ) \int \frac {(f x)^m}{\sqrt {1-c^2 x^2}}dx}{\sqrt {c^2 x^2-1}}+\frac {e \sqrt {c^2 x^2-1} (f x)^{m+1} \left (3 c^4 d^2 \left (m^4+22 m^3+179 m^2+638 m+840\right )+3 c^2 d e (m+3)^2 \left (m^2+13 m+42\right )+e^2 \left (m^2+8 m+15\right )^2\right )}{c^2 f (m+2) (m+3) (m+5) (m+7)}}{c^2 (m+4)}+\frac {e^2 \sqrt {c^2 x^2-1} (f x)^{m+3} \left (3 c^2 d \left (m^2+13 m+42\right )+e (m+5)^2\right )}{c^2 f^3 (m+4) (m+5) (m+7)}}{c^2 (m+6)}+\frac {e^3 \sqrt {c^2 x^2-1} (f x)^{m+5}}{c^2 f^5 (m+6) (m+7)}\right )}{\sqrt {c^2 x^2}}+\frac {d^3 (f x)^{m+1} \left (a+b \csc ^{-1}(c x)\right )}{f (m+1)}+\frac {3 d^2 e (f x)^{m+3} \left (a+b \csc ^{-1}(c x)\right )}{f^3 (m+3)}+\frac {3 d e^2 (f x)^{m+5} \left (a+b \csc ^{-1}(c x)\right )}{f^5 (m+5)}+\frac {e^3 (f x)^{m+7} \left (a+b \csc ^{-1}(c x)\right )}{f^7 (m+7)}\) |
| \(\Big \downarrow \) 278 |
| \(\displaystyle \frac {d^3 (f x)^{m+1} \left (a+b \csc ^{-1}(c x)\right )}{f (m+1)}+\frac {3 d^2 e (f x)^{m+3} \left (a+b \csc ^{-1}(c x)\right )}{f^3 (m+3)}+\frac {3 d e^2 (f x)^{m+5} \left (a+b \csc ^{-1}(c x)\right )}{f^5 (m+5)}+\frac {e^3 (f x)^{m+7} \left (a+b \csc ^{-1}(c x)\right )}{f^7 (m+7)}+\frac {b c x \left (\frac {e^3 \sqrt {c^2 x^2-1} (f x)^{m+5}}{c^2 f^5 (m+6) (m+7)}+\frac {\frac {e^2 \sqrt {c^2 x^2-1} (f x)^{m+3} \left (3 c^2 d \left (m^2+13 m+42\right )+e (m+5)^2\right )}{c^2 f^3 (m+4) (m+5) (m+7)}+\frac {\frac {e \sqrt {c^2 x^2-1} (f x)^{m+1} \left (3 c^4 d^2 \left (m^4+22 m^3+179 m^2+638 m+840\right )+3 c^2 d e (m+3)^2 \left (m^2+13 m+42\right )+e^2 \left (m^2+8 m+15\right )^2\right )}{c^2 f (m+2) (m+3) (m+5) (m+7)}+\frac {\sqrt {1-c^2 x^2} (f x)^{m+1} \left (\frac {c^4 d^3 (m+4) (m+6)}{m+1}+\frac {e (m+1) \left (3 c^4 d^2 \left (m^4+22 m^3+179 m^2+638 m+840\right )+3 c^2 d e (m+3)^2 \left (m^2+13 m+42\right )+e^2 \left (m^2+8 m+15\right )^2\right )}{c^2 (m+2) (m+3) (m+5) (m+7)}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},c^2 x^2\right )}{f (m+1) \sqrt {c^2 x^2-1}}}{c^2 (m+4)}}{c^2 (m+6)}\right )}{\sqrt {c^2 x^2}}\) |
(d^3*(f*x)^(1 + m)*(a + b*ArcCsc[c*x]))/(f*(1 + m)) + (3*d^2*e*(f*x)^(3 + m)*(a + b*ArcCsc[c*x]))/(f^3*(3 + m)) + (3*d*e^2*(f*x)^(5 + m)*(a + b*ArcC sc[c*x]))/(f^5*(5 + m)) + (e^3*(f*x)^(7 + m)*(a + b*ArcCsc[c*x]))/(f^7*(7 + m)) + (b*c*x*((e^3*(f*x)^(5 + m)*Sqrt[-1 + c^2*x^2])/(c^2*f^5*(6 + m)*(7 + m)) + ((e^2*(e*(5 + m)^2 + 3*c^2*d*(42 + 13*m + m^2))*(f*x)^(3 + m)*Sqr t[-1 + c^2*x^2])/(c^2*f^3*(4 + m)*(5 + m)*(7 + m)) + ((e*(e^2*(15 + 8*m + m^2)^2 + 3*c^2*d*e*(3 + m)^2*(42 + 13*m + m^2) + 3*c^4*d^2*(840 + 638*m + 179*m^2 + 22*m^3 + m^4))*(f*x)^(1 + m)*Sqrt[-1 + c^2*x^2])/(c^2*f*(2 + m)* (3 + m)*(5 + m)*(7 + m)) + (((c^4*d^3*(4 + m)*(6 + m))/(1 + m) + (e*(1 + m )*(e^2*(15 + 8*m + m^2)^2 + 3*c^2*d*e*(3 + m)^2*(42 + 13*m + m^2) + 3*c^4* d^2*(840 + 638*m + 179*m^2 + 22*m^3 + m^4)))/(c^2*(2 + m)*(3 + m)*(5 + m)* (7 + m)))*(f*x)^(1 + m)*Sqrt[1 - c^2*x^2]*Hypergeometric2F1[1/2, (1 + m)/2 , (3 + m)/2, c^2*x^2])/(f*(1 + m)*Sqrt[-1 + c^2*x^2]))/(c^2*(4 + m)))/(c^2 *(6 + m))))/Sqrt[c^2*x^2]
3.2.65.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[(c*x)^m* (1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && !(ILtQ[p, 0] || GtQ[a, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3)) Int[(e*x)^ m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d , 0] && NeQ[m + 2*p + 3, 0]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[c^p*(f*x)^(m + 4*p - 1)*((d + e*x^2)^ (q + 1)/(e*f^(4*p - 1)*(m + 4*p + 2*q + 1))), x] + Simp[1/(e*(m + 4*p + 2*q + 1)) Int[(f*x)^m*(d + e*x^2)^q*ExpandToSum[e*(m + 4*p + 2*q + 1)*((a + b*x^2 + c*x^4)^p - c^p*x^(4*p)) - d*c^p*(m + 4*p - 1)*x^(4*p - 2), x], x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && !IntegerQ[q] && NeQ[m + 4*p + 2*q + 1, 0]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ {q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(c*x)^(m + q - 1 )*((a + b*x^2)^(p + 1)/(b*c^(q - 1)*(m + q + 2*p + 1))), x] + Simp[1/(b*(m + q + 2*p + 1)) Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1) *Pq - b*f*(m + q + 2*p + 1)*x^q - a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ [Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])
Int[((a_.) + ArcCsc[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x _)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Sim p[(a + b*ArcCsc[c*x]) u, x] + Simp[b*c*(x/Sqrt[c^2*x^2]) Int[SimplifyIn tegrand[u/(x*Sqrt[c^2*x^2 - 1]), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && ((IGtQ[p, 0] && !(ILtQ[(m - 1)/2, 0] && GtQ[m + 2*p + 3, 0])) | | (IGtQ[(m + 1)/2, 0] && !(ILtQ[p, 0] && GtQ[m + 2*p + 3, 0])) || (ILtQ[(m + 2*p + 1)/2, 0] && !ILtQ[(m - 1)/2, 0]))
\[\int \left (f x \right )^{m} \left (e \,x^{2}+d \right )^{3} \left (a +b \,\operatorname {arccsc}\left (c x \right )\right )d x\]
\[ \int (f x)^m \left (d+e x^2\right )^3 \left (a+b \csc ^{-1}(c x)\right ) \, dx=\int { {\left (e x^{2} + d\right )}^{3} {\left (b \operatorname {arccsc}\left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \]
integral((a*e^3*x^6 + 3*a*d*e^2*x^4 + 3*a*d^2*e*x^2 + a*d^3 + (b*e^3*x^6 + 3*b*d*e^2*x^4 + 3*b*d^2*e*x^2 + b*d^3)*arccsc(c*x))*(f*x)^m, x)
Timed out. \[ \int (f x)^m \left (d+e x^2\right )^3 \left (a+b \csc ^{-1}(c x)\right ) \, dx=\text {Timed out} \]
\[ \int (f x)^m \left (d+e x^2\right )^3 \left (a+b \csc ^{-1}(c x)\right ) \, dx=\int { {\left (e x^{2} + d\right )}^{3} {\left (b \operatorname {arccsc}\left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \]
a*e^3*f^m*x^7*x^m/(m + 7) + 3*a*d*e^2*f^m*x^5*x^m/(m + 5) + 3*a*d^2*e*f^m* x^3*x^m/(m + 3) + (f*x)^(m + 1)*a*d^3/(f*(m + 1)) + (((b*e^3*f^m*m^3*arcta n2(1, sqrt(c*x + 1)*sqrt(c*x - 1)) + 9*b*e^3*f^m*m^2*arctan2(1, sqrt(c*x + 1)*sqrt(c*x - 1)) + 23*b*e^3*f^m*m*arctan2(1, sqrt(c*x + 1)*sqrt(c*x - 1) ) + 15*b*e^3*f^m*arctan2(1, sqrt(c*x + 1)*sqrt(c*x - 1)))*x^7 + 3*(b*d*e^2 *f^m*m^3*arctan2(1, sqrt(c*x + 1)*sqrt(c*x - 1)) + 11*b*d*e^2*f^m*m^2*arct an2(1, sqrt(c*x + 1)*sqrt(c*x - 1)) + 31*b*d*e^2*f^m*m*arctan2(1, sqrt(c*x + 1)*sqrt(c*x - 1)) + 21*b*d*e^2*f^m*arctan2(1, sqrt(c*x + 1)*sqrt(c*x - 1)))*x^5 + 3*(b*d^2*e*f^m*m^3*arctan2(1, sqrt(c*x + 1)*sqrt(c*x - 1)) + 13 *b*d^2*e*f^m*m^2*arctan2(1, sqrt(c*x + 1)*sqrt(c*x - 1)) + 47*b*d^2*e*f^m* m*arctan2(1, sqrt(c*x + 1)*sqrt(c*x - 1)) + 35*b*d^2*e*f^m*arctan2(1, sqrt (c*x + 1)*sqrt(c*x - 1)))*x^3 + (b*d^3*f^m*m^3*arctan2(1, sqrt(c*x + 1)*sq rt(c*x - 1)) + 15*b*d^3*f^m*m^2*arctan2(1, sqrt(c*x + 1)*sqrt(c*x - 1)) + 71*b*d^3*f^m*m*arctan2(1, sqrt(c*x + 1)*sqrt(c*x - 1)) + 105*b*d^3*f^m*arc tan2(1, sqrt(c*x + 1)*sqrt(c*x - 1)))*x)*x^m + (m^4 + 16*m^3 + 86*m^2 + 17 6*m + 105)*integrate(-(b*d^3*f^m*m^3 + 15*b*d^3*f^m*m^2 + (b*e^3*f^m*m^3 + 9*b*e^3*f^m*m^2 + 23*b*e^3*f^m*m + 15*b*e^3*f^m)*x^6 + 71*b*d^3*f^m*m + 1 05*b*d^3*f^m + 3*(b*d*e^2*f^m*m^3 + 11*b*d*e^2*f^m*m^2 + 31*b*d*e^2*f^m*m + 21*b*d*e^2*f^m)*x^4 + 3*(b*d^2*e*f^m*m^3 + 13*b*d^2*e*f^m*m^2 + 47*b*d^2 *e*f^m*m + 35*b*d^2*e*f^m)*x^2)*sqrt(c*x + 1)*sqrt(c*x - 1)*x^m/(m^4 + ...
\[ \int (f x)^m \left (d+e x^2\right )^3 \left (a+b \csc ^{-1}(c x)\right ) \, dx=\int { {\left (e x^{2} + d\right )}^{3} {\left (b \operatorname {arccsc}\left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \]
Timed out. \[ \int (f x)^m \left (d+e x^2\right )^3 \left (a+b \csc ^{-1}(c x)\right ) \, dx=\int {\left (f\,x\right )}^m\,{\left (e\,x^2+d\right )}^3\,\left (a+b\,\mathrm {asin}\left (\frac {1}{c\,x}\right )\right ) \,d x \]