Integrand size = 34, antiderivative size = 294 \[ \int \frac {(c x)^m \left (A+B x^2+C x^4+D x^6\right )}{\sqrt {a+b x^2}} \, dx=\frac {\left (a^2 D \left (15+8 m+m^2\right )-a b C \left (18+9 m+m^2\right )+b^2 B \left (24+10 m+m^2\right )\right ) (c x)^{1+m} \sqrt {a+b x^2}}{b^3 c (2+m) (4+m) (6+m)}-\frac {(a D (5+m)-b C (6+m)) (c x)^{3+m} \sqrt {a+b x^2}}{b^2 c^3 (4+m) (6+m)}+\frac {D (c x)^{5+m} \sqrt {a+b x^2}}{b c^5 (6+m)}+\frac {\left (\frac {A}{1+m}-\frac {a \left (a^2 D \left (15+8 m+m^2\right )-a b C \left (18+9 m+m^2\right )+b^2 B \left (24+10 m+m^2\right )\right )}{b^3 (2+m) (4+m) (6+m)}\right ) (c x)^{1+m} \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{c \sqrt {a+b x^2}} \] Output:
(a^2*D*(m^2+8*m+15)-a*b*C*(m^2+9*m+18)+b^2*B*(m^2+10*m+24))*(c*x)^(1+m)*(b *x^2+a)^(1/2)/b^3/c/(2+m)/(4+m)/(6+m)-(a*D*(5+m)-b*C*(6+m))*(c*x)^(3+m)*(b *x^2+a)^(1/2)/b^2/c^3/(4+m)/(6+m)+D*(c*x)^(5+m)*(b*x^2+a)^(1/2)/b/c^5/(6+m )+(A/(1+m)-a*(a^2*D*(m^2+8*m+15)-a*b*C*(m^2+9*m+18)+b^2*B*(m^2+10*m+24))/b ^3/(2+m)/(4+m)/(6+m))*(c*x)^(1+m)*(1+b*x^2/a)^(1/2)*hypergeom([1/2, 1/2+1/ 2*m],[3/2+1/2*m],-b*x^2/a)/c/(b*x^2+a)^(1/2)
Time = 2.54 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.61 \[ \int \frac {(c x)^m \left (A+B x^2+C x^4+D x^6\right )}{\sqrt {a+b x^2}} \, dx=\frac {x (c x)^m \sqrt {1+\frac {b x^2}{a}} \left (\frac {A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{1+m}+\frac {B x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},-\frac {b x^2}{a}\right )}{3+m}+\frac {C x^4 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5+m}{2},\frac {7+m}{2},-\frac {b x^2}{a}\right )}{5+m}+\frac {D x^6 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7+m}{2},\frac {9+m}{2},-\frac {b x^2}{a}\right )}{7+m}\right )}{\sqrt {a+b x^2}} \] Input:
Integrate[((c*x)^m*(A + B*x^2 + C*x^4 + D*x^6))/Sqrt[a + b*x^2],x]
Output:
(x*(c*x)^m*Sqrt[1 + (b*x^2)/a]*((A*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(1 + m) + (B*x^2*Hypergeometric2F1[1/2, (3 + m)/2, (5 + m)/2, -((b*x^2)/a)])/(3 + m) + (C*x^4*Hypergeometric2F1[1/2, (5 + m)/2, (7 + m)/2, -((b*x^2)/a)])/(5 + m) + (D*x^6*Hypergeometric2F1[1/2, (7 + m) /2, (9 + m)/2, -((b*x^2)/a)])/(7 + m)))/Sqrt[a + b*x^2]
Time = 0.58 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {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.
\(\displaystyle \int \frac {(c x)^m \left (A+B x^2+C x^4+D x^6\right )}{\sqrt {a+b x^2}} \, dx\) |
\(\Big \downarrow \) 2340 |
\(\displaystyle \frac {\int \frac {(c x)^m \left (-\left ((a D (m+5)-b C (m+6)) x^4\right )+b B (m+6) x^2+A b (m+6)\right )}{\sqrt {b x^2+a}}dx}{b (m+6)}+\frac {D \sqrt {a+b x^2} (c x)^{m+5}}{b c^5 (m+6)}\) |
\(\Big \downarrow \) 1590 |
\(\displaystyle \frac {\frac {\int \frac {(c x)^m \left (A (m+4) (m+6) b^2+\left (D \left (m^2+8 m+15\right ) a^2-b C \left (m^2+9 m+18\right ) a+b^2 B \left (m^2+10 m+24\right )\right ) x^2\right )}{\sqrt {b x^2+a}}dx}{b (m+4)}-\frac {\sqrt {a+b x^2} (c x)^{m+3} (a D (m+5)-b C (m+6))}{b c^3 (m+4)}}{b (m+6)}+\frac {D \sqrt {a+b x^2} (c x)^{m+5}}{b c^5 (m+6)}\) |
\(\Big \downarrow \) 363 |
\(\displaystyle \frac {\frac {\frac {\left (A b^3 (m+4) (m+6)-\frac {a (m+1) \left (a^2 D \left (m^2+8 m+15\right )-a b C \left (m^2+9 m+18\right )+b^2 B \left (m^2+10 m+24\right )\right )}{m+2}\right ) \int \frac {(c x)^m}{\sqrt {b x^2+a}}dx}{b}+\frac {\sqrt {a+b x^2} (c x)^{m+1} \left (a^2 D \left (m^2+8 m+15\right )-a b C \left (m^2+9 m+18\right )+b^2 B \left (m^2+10 m+24\right )\right )}{b c (m+2)}}{b (m+4)}-\frac {\sqrt {a+b x^2} (c x)^{m+3} (a D (m+5)-b C (m+6))}{b c^3 (m+4)}}{b (m+6)}+\frac {D \sqrt {a+b x^2} (c x)^{m+5}}{b c^5 (m+6)}\) |
\(\Big \downarrow \) 279 |
\(\displaystyle \frac {\frac {\frac {\sqrt {\frac {b x^2}{a}+1} \left (A b^3 (m+4) (m+6)-\frac {a (m+1) \left (a^2 D \left (m^2+8 m+15\right )-a b C \left (m^2+9 m+18\right )+b^2 B \left (m^2+10 m+24\right )\right )}{m+2}\right ) \int \frac {(c x)^m}{\sqrt {\frac {b x^2}{a}+1}}dx}{b \sqrt {a+b x^2}}+\frac {\sqrt {a+b x^2} (c x)^{m+1} \left (a^2 D \left (m^2+8 m+15\right )-a b C \left (m^2+9 m+18\right )+b^2 B \left (m^2+10 m+24\right )\right )}{b c (m+2)}}{b (m+4)}-\frac {\sqrt {a+b x^2} (c x)^{m+3} (a D (m+5)-b C (m+6))}{b c^3 (m+4)}}{b (m+6)}+\frac {D \sqrt {a+b x^2} (c x)^{m+5}}{b c^5 (m+6)}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {\frac {\frac {\sqrt {\frac {b x^2}{a}+1} (c x)^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right ) \left (A b^3 (m+4) (m+6)-\frac {a (m+1) \left (a^2 D \left (m^2+8 m+15\right )-a b C \left (m^2+9 m+18\right )+b^2 B \left (m^2+10 m+24\right )\right )}{m+2}\right )}{b c (m+1) \sqrt {a+b x^2}}+\frac {\sqrt {a+b x^2} (c x)^{m+1} \left (a^2 D \left (m^2+8 m+15\right )-a b C \left (m^2+9 m+18\right )+b^2 B \left (m^2+10 m+24\right )\right )}{b c (m+2)}}{b (m+4)}-\frac {\sqrt {a+b x^2} (c x)^{m+3} (a D (m+5)-b C (m+6))}{b c^3 (m+4)}}{b (m+6)}+\frac {D \sqrt {a+b x^2} (c x)^{m+5}}{b c^5 (m+6)}\) |
Input:
Int[((c*x)^m*(A + B*x^2 + C*x^4 + D*x^6))/Sqrt[a + b*x^2],x]
Output:
(D*(c*x)^(5 + m)*Sqrt[a + b*x^2])/(b*c^5*(6 + m)) + (-(((a*D*(5 + m) - b*C *(6 + m))*(c*x)^(3 + m)*Sqrt[a + b*x^2])/(b*c^3*(4 + m))) + (((a^2*D*(15 + 8*m + m^2) - a*b*C*(18 + 9*m + m^2) + b^2*B*(24 + 10*m + m^2))*(c*x)^(1 + m)*Sqrt[a + b*x^2])/(b*c*(2 + m)) + ((A*b^3*(4 + m)*(6 + m) - (a*(1 + m)* (a^2*D*(15 + 8*m + m^2) - a*b*C*(18 + 9*m + m^2) + b^2*B*(24 + 10*m + m^2) ))/(2 + m))*(c*x)^(1 + m)*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(b*c*(1 + m)*Sqrt[a + b*x^2]))/(b*(4 + m)) )/(b*(6 + m))
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])
\[\int \frac {\left (c x \right )^{m} \left (D x^{6}+C \,x^{4}+x^{2} B +A \right )}{\sqrt {b \,x^{2}+a}}d x\]
Input:
int((c*x)^m*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(1/2),x)
Output:
int((c*x)^m*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(1/2),x)
\[ \int \frac {(c x)^m \left (A+B x^2+C x^4+D x^6\right )}{\sqrt {a+b x^2}} \, dx=\int { \frac {{\left (D x^{6} + C x^{4} + B x^{2} + A\right )} \left (c x\right )^{m}}{\sqrt {b x^{2} + a}} \,d x } \] Input:
integrate((c*x)^m*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(1/2),x, algorithm="fric as")
Output:
integral((D*x^6 + C*x^4 + B*x^2 + A)*(c*x)^m/sqrt(b*x^2 + a), x)
Result contains complex when optimal does not.
Time = 4.58 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.80 \[ \int \frac {(c x)^m \left (A+B x^2+C x^4+D x^6\right )}{\sqrt {a+b x^2}} \, dx=\frac {A c^{m} x^{m + 1} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} + \frac {1}{2} \\ \frac {m}{2} + \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {B c^{m} x^{m + 3} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} + \frac {3}{2} \\ \frac {m}{2} + \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {C c^{m} x^{m + 5} \Gamma \left (\frac {m}{2} + \frac {5}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} + \frac {5}{2} \\ \frac {m}{2} + \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )} + \frac {D c^{m} x^{m + 7} \Gamma \left (\frac {m}{2} + \frac {7}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} + \frac {7}{2} \\ \frac {m}{2} + \frac {9}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {m}{2} + \frac {9}{2}\right )} \] Input:
integrate((c*x)**m*(D*x**6+C*x**4+B*x**2+A)/(b*x**2+a)**(1/2),x)
Output:
A*c**m*x**(m + 1)*gamma(m/2 + 1/2)*hyper((1/2, m/2 + 1/2), (m/2 + 3/2,), b *x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*gamma(m/2 + 3/2)) + B*c**m*x**(m + 3)* gamma(m/2 + 3/2)*hyper((1/2, m/2 + 3/2), (m/2 + 5/2,), b*x**2*exp_polar(I* pi)/a)/(2*sqrt(a)*gamma(m/2 + 5/2)) + C*c**m*x**(m + 5)*gamma(m/2 + 5/2)*h yper((1/2, m/2 + 5/2), (m/2 + 7/2,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)* gamma(m/2 + 7/2)) + D*c**m*x**(m + 7)*gamma(m/2 + 7/2)*hyper((1/2, m/2 + 7 /2), (m/2 + 9/2,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*gamma(m/2 + 9/2))
\[ \int \frac {(c x)^m \left (A+B x^2+C x^4+D x^6\right )}{\sqrt {a+b x^2}} \, dx=\int { \frac {{\left (D x^{6} + C x^{4} + B x^{2} + A\right )} \left (c x\right )^{m}}{\sqrt {b x^{2} + a}} \,d x } \] Input:
integrate((c*x)^m*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(1/2),x, algorithm="maxi ma")
Output:
integrate((D*x^6 + C*x^4 + B*x^2 + A)*(c*x)^m/sqrt(b*x^2 + a), x)
\[ \int \frac {(c x)^m \left (A+B x^2+C x^4+D x^6\right )}{\sqrt {a+b x^2}} \, dx=\int { \frac {{\left (D x^{6} + C x^{4} + B x^{2} + A\right )} \left (c x\right )^{m}}{\sqrt {b x^{2} + a}} \,d x } \] Input:
integrate((c*x)^m*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(1/2),x, algorithm="giac ")
Output:
integrate((D*x^6 + C*x^4 + B*x^2 + A)*(c*x)^m/sqrt(b*x^2 + a), x)
Timed out. \[ \int \frac {(c x)^m \left (A+B x^2+C x^4+D x^6\right )}{\sqrt {a+b x^2}} \, dx=\int \frac {{\left (c\,x\right )}^m\,\left (A+B\,x^2+C\,x^4+x^6\,D\right )}{\sqrt {b\,x^2+a}} \,d x \] Input:
int(((c*x)^m*(A + B*x^2 + C*x^4 + x^6*D))/(a + b*x^2)^(1/2),x)
Output:
int(((c*x)^m*(A + B*x^2 + C*x^4 + x^6*D))/(a + b*x^2)^(1/2), x)
\[ \int \frac {(c x)^m \left (A+B x^2+C x^4+D x^6\right )}{\sqrt {a+b x^2}} \, dx=c^{m} \left (\left (\int \frac {x^{m}}{\sqrt {b \,x^{2}+a}}d x \right ) a +\left (\int \frac {x^{m} x^{6}}{\sqrt {b \,x^{2}+a}}d x \right ) d +\left (\int \frac {x^{m} x^{4}}{\sqrt {b \,x^{2}+a}}d x \right ) c +\left (\int \frac {x^{m} x^{2}}{\sqrt {b \,x^{2}+a}}d x \right ) b \right ) \] Input:
int((c*x)^m*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(1/2),x)
Output:
c**m*(int(x**m/sqrt(a + b*x**2),x)*a + int((x**m*x**6)/sqrt(a + b*x**2),x) *d + int((x**m*x**4)/sqrt(a + b*x**2),x)*c + int((x**m*x**2)/sqrt(a + b*x* *2),x)*b)