\(\int \frac {(c x)^m (A+B x^2+C x^4+D x^6)}{(a+b x^2)^{7/2}} \, dx\) [264]

Optimal result

Integrand size = 34, antiderivative size = 278 \[ \int \frac {(c x)^m \left (A+B x^2+C x^4+D x^6\right )}{\left (a+b x^2\right )^{7/2}} \, dx=\frac {\left (\frac {A}{a}-\frac {b^2 B-a b C+a^2 D}{b^3}\right ) (c x)^{1+m}}{5 c \left (a+b x^2\right )^{5/2}}+\frac {\left (\frac {A (4-m)}{a}+\frac {b^2 B (1+m)-a b C (6+m)+a^2 D (11+m)}{b^3}\right ) (c x)^{1+m}}{15 a c \left (a+b x^2\right )^{3/2}}+\frac {D (c x)^{1+m}}{b^3 c m \sqrt {a+b x^2}}+\frac {\left (a b^2 B (2-m)+a^2 b C (3+m)+\frac {A b^3 \left (8-6 m+m^2\right )}{1+m}-\frac {a^3 D \left (15+8 m+m^2\right )}{m}\right ) (c x)^{1+m} \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{15 a^3 b^3 c \sqrt {a+b x^2}} \] Output:

1/5*(A/a-(B*b^2-C*a*b+D*a^2)/b^3)*(c*x)^(1+m)/c/(b*x^2+a)^(5/2)+1/15*(A*(4 
-m)/a+(b^2*B*(1+m)-a*b*C*(6+m)+a^2*D*(11+m))/b^3)*(c*x)^(1+m)/a/c/(b*x^2+a 
)^(3/2)+D*(c*x)^(1+m)/b^3/c/m/(b*x^2+a)^(1/2)+1/15*(a*b^2*B*(2-m)+a^2*b*C* 
(3+m)+A*b^3*(m^2-6*m+8)/(1+m)-a^3*D*(m^2+8*m+15)/m)*(c*x)^(1+m)*(1+b*x^2/a 
)^(1/2)*hypergeom([3/2, 1/2+1/2*m],[3/2+1/2*m],-b*x^2/a)/a^3/b^3/c/(b*x^2+ 
a)^(1/2)
 

Mathematica [A] (verified)

Time = 2.99 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.65 \[ \int \frac {(c x)^m \left (A+B x^2+C x^4+D x^6\right )}{\left (a+b x^2\right )^{7/2}} \, dx=\frac {x (c x)^m \sqrt {1+\frac {b x^2}{a}} \left (\frac {A \operatorname {Hypergeometric2F1}\left (\frac {7}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{1+m}+\frac {B x^2 \operatorname {Hypergeometric2F1}\left (\frac {7}{2},\frac {3+m}{2},\frac {5+m}{2},-\frac {b x^2}{a}\right )}{3+m}+\frac {C x^4 \operatorname {Hypergeometric2F1}\left (\frac {7}{2},\frac {5+m}{2},\frac {7+m}{2},-\frac {b x^2}{a}\right )}{5+m}+\frac {D x^6 \operatorname {Hypergeometric2F1}\left (\frac {7}{2},\frac {7+m}{2},\frac {9+m}{2},-\frac {b x^2}{a}\right )}{7+m}\right )}{a^3 \sqrt {a+b x^2}} \] Input:

Integrate[((c*x)^m*(A + B*x^2 + C*x^4 + D*x^6))/(a + b*x^2)^(7/2),x]
 

Output:

(x*(c*x)^m*Sqrt[1 + (b*x^2)/a]*((A*Hypergeometric2F1[7/2, (1 + m)/2, (3 + 
m)/2, -((b*x^2)/a)])/(1 + m) + (B*x^2*Hypergeometric2F1[7/2, (3 + m)/2, (5 
 + m)/2, -((b*x^2)/a)])/(3 + m) + (C*x^4*Hypergeometric2F1[7/2, (5 + m)/2, 
 (7 + m)/2, -((b*x^2)/a)])/(5 + m) + (D*x^6*Hypergeometric2F1[7/2, (7 + m) 
/2, (9 + m)/2, -((b*x^2)/a)])/(7 + m)))/(a^3*Sqrt[a + b*x^2])
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.14, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {2337, 25, 1590, 27, 362, 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.

\(\displaystyle \int \frac {(c x)^m \left (A+B x^2+C x^4+D x^6\right )}{\left (a+b x^2\right )^{7/2}} \, dx\)

\(\Big \downarrow \) 2337

\(\displaystyle \frac {(c x)^{m+1} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{5 a c \left (a+b x^2\right )^{5/2}}-\frac {\int -\frac {(c x)^m \left (\frac {5 a D x^4}{b}+\frac {5 a (b C-a D) x^2}{b^2}+A (4-m)+\frac {a \left (D a^2-b C a+b^2 B\right ) (m+1)}{b^3}\right )}{\left (b x^2+a\right )^{5/2}}dx}{5 a}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {(c x)^m \left (\frac {5 a D x^4}{b}+\frac {5 a (b C-a D) x^2}{b^2}+A (4-m)+\frac {a \left (D a^2-b C a+b^2 B\right ) (m+1)}{b^3}\right )}{\left (b x^2+a\right )^{5/2}}dx}{5 a}+\frac {(c x)^{m+1} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{5 a c \left (a+b x^2\right )^{5/2}}\)

\(\Big \downarrow \) 1590

\(\displaystyle \frac {\frac {\int \frac {(c x)^m \left (5 a (b C m-a D (2 m+3)) x^2+b m \left (A b (4-m)+\frac {a \left (D a^2-b C a+b^2 B\right ) (m+1)}{b^2}\right )\right )}{b \left (b x^2+a\right )^{5/2}}dx}{b m}+\frac {5 a D (c x)^{m+3}}{b^2 c^3 m \left (a+b x^2\right )^{3/2}}}{5 a}+\frac {(c x)^{m+1} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{5 a c \left (a+b x^2\right )^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {(c x)^m \left (5 a (b C m-a D (2 m+3)) x^2+b m \left (A b (4-m)+\frac {a \left (D a^2-b C a+b^2 B\right ) (m+1)}{b^2}\right )\right )}{\left (b x^2+a\right )^{5/2}}dx}{b^2 m}+\frac {5 a D (c x)^{m+3}}{b^2 c^3 m \left (a+b x^2\right )^{3/2}}}{5 a}+\frac {(c x)^{m+1} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{5 a c \left (a+b x^2\right )^{5/2}}\)

\(\Big \downarrow \) 362

\(\displaystyle \frac {\frac {\frac {\left ((2-m) m \left (a (m+1) \left (a^2 D-a b C+b^2 B\right )+A b^3 (4-m)\right )+5 a^2 (m+1) (b C m-a D (2 m+3))\right ) \int \frac {(c x)^m}{\left (b x^2+a\right )^{3/2}}dx}{3 a b}+\frac {(c x)^{m+1} \left (a^3 D \left (m^2+11 m+15\right )-a^2 b C m (m+6)+a b^2 B m (m+1)+A b^3 (4-m) m\right )}{3 a b c \left (a+b x^2\right )^{3/2}}}{b^2 m}+\frac {5 a D (c x)^{m+3}}{b^2 c^3 m \left (a+b x^2\right )^{3/2}}}{5 a}+\frac {(c x)^{m+1} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{5 a c \left (a+b x^2\right )^{5/2}}\)

\(\Big \downarrow \) 279

\(\displaystyle \frac {\frac {\frac {\sqrt {\frac {b x^2}{a}+1} \left ((2-m) m \left (a (m+1) \left (a^2 D-a b C+b^2 B\right )+A b^3 (4-m)\right )+5 a^2 (m+1) (b C m-a D (2 m+3))\right ) \int \frac {(c x)^m}{\left (\frac {b x^2}{a}+1\right )^{3/2}}dx}{3 a^2 b \sqrt {a+b x^2}}+\frac {(c x)^{m+1} \left (a^3 D \left (m^2+11 m+15\right )-a^2 b C m (m+6)+a b^2 B m (m+1)+A b^3 (4-m) m\right )}{3 a b c \left (a+b x^2\right )^{3/2}}}{b^2 m}+\frac {5 a D (c x)^{m+3}}{b^2 c^3 m \left (a+b x^2\right )^{3/2}}}{5 a}+\frac {(c x)^{m+1} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{5 a c \left (a+b x^2\right )^{5/2}}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {(c x)^{m+1} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{5 a c \left (a+b x^2\right )^{5/2}}+\frac {\frac {\frac {\sqrt {\frac {b x^2}{a}+1} (c x)^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right ) \left ((2-m) m \left (a (m+1) \left (a^2 D-a b C+b^2 B\right )+A b^3 (4-m)\right )+5 a^2 (m+1) (b C m-a D (2 m+3))\right )}{3 a^2 b c (m+1) \sqrt {a+b x^2}}+\frac {(c x)^{m+1} \left (a^3 D \left (m^2+11 m+15\right )-a^2 b C m (m+6)+a b^2 B m (m+1)+A b^3 (4-m) m\right )}{3 a b c \left (a+b x^2\right )^{3/2}}}{b^2 m}+\frac {5 a D (c x)^{m+3}}{b^2 c^3 m \left (a+b x^2\right )^{3/2}}}{5 a}\)

Input:

Int[((c*x)^m*(A + B*x^2 + C*x^4 + D*x^6))/(a + b*x^2)^(7/2),x]
 

Output:

((A - (a*(b^2*B - a*b*C + a^2*D))/b^3)*(c*x)^(1 + m))/(5*a*c*(a + b*x^2)^( 
5/2)) + ((5*a*D*(c*x)^(3 + m))/(b^2*c^3*m*(a + b*x^2)^(3/2)) + (((A*b^3*(4 
 - m)*m + a*b^2*B*m*(1 + m) - a^2*b*C*m*(6 + m) + a^3*D*(15 + 11*m + m^2)) 
*(c*x)^(1 + m))/(3*a*b*c*(a + b*x^2)^(3/2)) + (((2 - m)*m*(A*b^3*(4 - m) + 
 a*(b^2*B - a*b*C + a^2*D)*(1 + m)) + 5*a^2*(1 + m)*(b*C*m - a*D*(3 + 2*m) 
))*(c*x)^(1 + m)*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[3/2, (1 + m)/2, (3 
+ m)/2, -((b*x^2)/a)])/(3*a^2*b*c*(1 + m)*Sqrt[a + b*x^2]))/(b^2*m))/(5*a)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 278

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])
 

rule 279

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])
 

rule 362

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*b*e 
*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(2*a*b*(p + 1))   I 
nt[(e*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && N 
eQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) || 
  !RationalQ[m] || (ILtQ[p + 1/2, 0] && LeQ[-1, m, -2*(p + 1)]))
 

rule 1590

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]
 

rule 2337

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq 
, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 
 1]}, Simp[(-(c*x)^(m + 1))*(f + g*x)*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))) 
, x] + Simp[1/(2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[2 
*a*(p + 1)*Q + f*(m + 2*p + 3) + g*(m + 2*p + 4)*x, x], x], x]] /; FreeQ[{a 
, b, c, m}, x] && PolyQ[Pq, x] && LtQ[p, -1] &&  !GtQ[m, 0]
 

Maple [F]

\[\int \frac {\left (c x \right )^{m} \left (D x^{6}+C \,x^{4}+x^{2} B +A \right )}{\left (b \,x^{2}+a \right )^{\frac {7}{2}}}d x\]

Input:

int((c*x)^m*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(7/2),x)
 

Output:

int((c*x)^m*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(7/2),x)
 

Fricas [F]

\[ \int \frac {(c x)^m \left (A+B x^2+C x^4+D x^6\right )}{\left (a+b x^2\right )^{7/2}} \, dx=\int { \frac {{\left (D x^{6} + C x^{4} + B x^{2} + A\right )} \left (c x\right )^{m}}{{\left (b x^{2} + a\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate((c*x)^m*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(7/2),x, algorithm="fric 
as")
 

Output:

integral((D*x^6 + C*x^4 + B*x^2 + A)*sqrt(b*x^2 + a)*(c*x)^m/(b^4*x^8 + 4* 
a*b^3*x^6 + 6*a^2*b^2*x^4 + 4*a^3*b*x^2 + a^4), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(c x)^m \left (A+B x^2+C x^4+D x^6\right )}{\left (a+b x^2\right )^{7/2}} \, dx=\text {Timed out} \] Input:

integrate((c*x)**m*(D*x**6+C*x**4+B*x**2+A)/(b*x**2+a)**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(c x)^m \left (A+B x^2+C x^4+D x^6\right )}{\left (a+b x^2\right )^{7/2}} \, dx=\int { \frac {{\left (D x^{6} + C x^{4} + B x^{2} + A\right )} \left (c x\right )^{m}}{{\left (b x^{2} + a\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate((c*x)^m*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(7/2),x, algorithm="maxi 
ma")
 

Output:

integrate((D*x^6 + C*x^4 + B*x^2 + A)*(c*x)^m/(b*x^2 + a)^(7/2), x)
 

Giac [F]

\[ \int \frac {(c x)^m \left (A+B x^2+C x^4+D x^6\right )}{\left (a+b x^2\right )^{7/2}} \, dx=\int { \frac {{\left (D x^{6} + C x^{4} + B x^{2} + A\right )} \left (c x\right )^{m}}{{\left (b x^{2} + a\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate((c*x)^m*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(7/2),x, algorithm="giac 
")
 

Output:

integrate((D*x^6 + C*x^4 + B*x^2 + A)*(c*x)^m/(b*x^2 + a)^(7/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c x)^m \left (A+B x^2+C x^4+D x^6\right )}{\left (a+b x^2\right )^{7/2}} \, dx=\int \frac {{\left (c\,x\right )}^m\,\left (A+B\,x^2+C\,x^4+x^6\,D\right )}{{\left (b\,x^2+a\right )}^{7/2}} \,d x \] Input:

int(((c*x)^m*(A + B*x^2 + C*x^4 + x^6*D))/(a + b*x^2)^(7/2),x)
 

Output:

int(((c*x)^m*(A + B*x^2 + C*x^4 + x^6*D))/(a + b*x^2)^(7/2), x)
 

Reduce [F]

\[ \int \frac {(c x)^m \left (A+B x^2+C x^4+D x^6\right )}{\left (a+b x^2\right )^{7/2}} \, dx=c^{m} \left (\left (\int \frac {x^{m}}{\sqrt {b \,x^{2}+a}\, a^{3}+3 \sqrt {b \,x^{2}+a}\, a^{2} b \,x^{2}+3 \sqrt {b \,x^{2}+a}\, a \,b^{2} x^{4}+\sqrt {b \,x^{2}+a}\, b^{3} x^{6}}d x \right ) a +\left (\int \frac {x^{m} x^{6}}{\sqrt {b \,x^{2}+a}\, a^{3}+3 \sqrt {b \,x^{2}+a}\, a^{2} b \,x^{2}+3 \sqrt {b \,x^{2}+a}\, a \,b^{2} x^{4}+\sqrt {b \,x^{2}+a}\, b^{3} x^{6}}d x \right ) d +\left (\int \frac {x^{m} x^{4}}{\sqrt {b \,x^{2}+a}\, a^{3}+3 \sqrt {b \,x^{2}+a}\, a^{2} b \,x^{2}+3 \sqrt {b \,x^{2}+a}\, a \,b^{2} x^{4}+\sqrt {b \,x^{2}+a}\, b^{3} x^{6}}d x \right ) c +\left (\int \frac {x^{m} x^{2}}{\sqrt {b \,x^{2}+a}\, a^{3}+3 \sqrt {b \,x^{2}+a}\, a^{2} b \,x^{2}+3 \sqrt {b \,x^{2}+a}\, a \,b^{2} x^{4}+\sqrt {b \,x^{2}+a}\, b^{3} x^{6}}d x \right ) b \right ) \] Input:

int((c*x)^m*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(7/2),x)
 

Output:

c**m*(int(x**m/(sqrt(a + b*x**2)*a**3 + 3*sqrt(a + b*x**2)*a**2*b*x**2 + 3 
*sqrt(a + b*x**2)*a*b**2*x**4 + sqrt(a + b*x**2)*b**3*x**6),x)*a + int((x* 
*m*x**6)/(sqrt(a + b*x**2)*a**3 + 3*sqrt(a + b*x**2)*a**2*b*x**2 + 3*sqrt( 
a + b*x**2)*a*b**2*x**4 + sqrt(a + b*x**2)*b**3*x**6),x)*d + int((x**m*x** 
4)/(sqrt(a + b*x**2)*a**3 + 3*sqrt(a + b*x**2)*a**2*b*x**2 + 3*sqrt(a + b* 
x**2)*a*b**2*x**4 + sqrt(a + b*x**2)*b**3*x**6),x)*c + int((x**m*x**2)/(sq 
rt(a + b*x**2)*a**3 + 3*sqrt(a + b*x**2)*a**2*b*x**2 + 3*sqrt(a + b*x**2)* 
a*b**2*x**4 + sqrt(a + b*x**2)*b**3*x**6),x)*b)