3.4.98 \(\int \frac {(d+e x)^2 (a+b x^2)^p}{x^3} \, dx\) [398]

3.4.98.1 Optimal result

Integrand size = 20, antiderivative size = 127 \[ \int \frac {(d+e x)^2 \left (a+b x^2\right )^p}{x^3} \, dx=-\frac {d^2 \left (a+b x^2\right )^{1+p}}{2 a x^2}-\frac {2 d e \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b x^2}{a}\right )}{x}-\frac {\left (a e^2+b d^2 p\right ) \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^2}{a}\right )}{2 a^2 (1+p)} \]

output

-1/2*d^2*(b*x^2+a)^(p+1)/a/x^2-2*d*e*(b*x^2+a)^p*hypergeom([-1/2, -p],[1/2 
],-b*x^2/a)/x/((1+b*x^2/a)^p)-1/2*(b*d^2*p+a*e^2)*(b*x^2+a)^(p+1)*hypergeo 
m([1, p+1],[2+p],1+b*x^2/a)/a^2/(p+1)
 

3.4.98.2 Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.94 \[ \int \frac {(d+e x)^2 \left (a+b x^2\right )^p}{x^3} \, dx=\frac {1}{2} \left (a+b x^2\right )^p \left (-\frac {4 d e \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b x^2}{a}\right )}{x}-\frac {\left (a+b x^2\right ) \left (a e^2 \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^2}{a}\right )-b d^2 \operatorname {Hypergeometric2F1}\left (2,1+p,2+p,1+\frac {b x^2}{a}\right )\right )}{a^2 (1+p)}\right ) \]

input

Integrate[((d + e*x)^2*(a + b*x^2)^p)/x^3,x]
 

output

((a + b*x^2)^p*((-4*d*e*Hypergeometric2F1[-1/2, -p, 1/2, -((b*x^2)/a)])/(x 
*(1 + (b*x^2)/a)^p) - ((a + b*x^2)*(a*e^2*Hypergeometric2F1[1, 1 + p, 2 + 
p, 1 + (b*x^2)/a] - b*d^2*Hypergeometric2F1[2, 1 + p, 2 + p, 1 + (b*x^2)/a 
]))/(a^2*(1 + p))))/2
 

3.4.98.3 Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {543, 27, 279, 278, 354, 87, 75}

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.

\(\displaystyle \int \frac {(d+e x)^2 \left (a+b x^2\right )^p}{x^3} \, dx\)

\(\Big \downarrow \) 543

\(\displaystyle \int \frac {\left (b x^2+a\right )^p \left (d^2+e^2 x^2\right )}{x^3}dx+\int \frac {2 d e \left (b x^2+a\right )^p}{x^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {\left (b x^2+a\right )^p \left (d^2+e^2 x^2\right )}{x^3}dx+2 d e \int \frac {\left (b x^2+a\right )^p}{x^2}dx\)

\(\Big \downarrow \) 279

\(\displaystyle \int \frac {\left (b x^2+a\right )^p \left (d^2+e^2 x^2\right )}{x^3}dx+2 d e \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \int \frac {\left (\frac {b x^2}{a}+1\right )^p}{x^2}dx\)

\(\Big \downarrow \) 278

\(\displaystyle \int \frac {\left (b x^2+a\right )^p \left (d^2+e^2 x^2\right )}{x^3}dx-\frac {2 d e \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b x^2}{a}\right )}{x}\)

\(\Big \downarrow \) 354

\(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^p \left (d^2+e^2 x^2\right )}{x^4}dx^2-\frac {2 d e \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b x^2}{a}\right )}{x}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {1}{2} \left (\frac {\left (a e^2+b d^2 p\right ) \int \frac {\left (b x^2+a\right )^p}{x^2}dx^2}{a}-\frac {d^2 \left (a+b x^2\right )^{p+1}}{a x^2}\right )-\frac {2 d e \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b x^2}{a}\right )}{x}\)

\(\Big \downarrow \) 75

\(\displaystyle \frac {1}{2} \left (-\frac {\left (a+b x^2\right )^{p+1} \left (a e^2+b d^2 p\right ) \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b x^2}{a}+1\right )}{a^2 (p+1)}-\frac {d^2 \left (a+b x^2\right )^{p+1}}{a x^2}\right )-\frac {2 d e \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b x^2}{a}\right )}{x}\)

input

Int[((d + e*x)^2*(a + b*x^2)^p)/x^3,x]
 

output

(-2*d*e*(a + b*x^2)^p*Hypergeometric2F1[-1/2, -p, 1/2, -((b*x^2)/a)])/(x*( 
1 + (b*x^2)/a)^p) + (-((d^2*(a + b*x^2)^(1 + p))/(a*x^2)) - ((a*e^2 + b*d^ 
2*p)*(a + b*x^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*x^2)/a] 
)/(a^2*(1 + p)))/2
 

3.4.98.3.1 Defintions of rubi rules used

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 75

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x 
)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + 
 d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (IntegerQ[m] 
 || GtQ[-d/(b*c), 0])
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

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 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

rule 543

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Module[{k}, Int[x^m*Sum[Binomial[n, 2*k]*c^(n - 2*k)*d^(2*k)*x^(2*k), 
 {k, 0, n/2}]*(a + b*x^2)^p, x] + Int[x^(m + 1)*Sum[Binomial[n, 2*k + 1]*c^ 
(n - 2*k - 1)*d^(2*k + 1)*x^(2*k), {k, 0, (n - 1)/2}]*(a + b*x^2)^p, x]] /; 
 FreeQ[{a, b, c, d, p}, x] && IGtQ[n, 1] && IntegerQ[m] &&  !IntegerQ[2*p] 
&&  !(EqQ[m, 1] && EqQ[b*c^2 + a*d^2, 0])
 

3.4.98.4 Maple [F]

\[\int \frac {\left (e x +d \right )^{2} \left (b \,x^{2}+a \right )^{p}}{x^{3}}d x\]

input

int((e*x+d)^2*(b*x^2+a)^p/x^3,x)
 

output

int((e*x+d)^2*(b*x^2+a)^p/x^3,x)
 

3.4.98.5 Fricas [F]

\[ \int \frac {(d+e x)^2 \left (a+b x^2\right )^p}{x^3} \, dx=\int { \frac {{\left (e x + d\right )}^{2} {\left (b x^{2} + a\right )}^{p}}{x^{3}} \,d x } \]

input

integrate((e*x+d)^2*(b*x^2+a)^p/x^3,x, algorithm="fricas")
 

output

integral((e^2*x^2 + 2*d*e*x + d^2)*(b*x^2 + a)^p/x^3, x)
 

3.4.98.6 Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 6.65 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.92 \[ \int \frac {(d+e x)^2 \left (a+b x^2\right )^p}{x^3} \, dx=- \frac {2 a^{p} d e {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - p \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{x} - \frac {b^{p} d^{2} x^{2 p - 2} \Gamma \left (1 - p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, 1 - p \\ 2 - p \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 \Gamma \left (2 - p\right )} - \frac {b^{p} e^{2} x^{2 p} \Gamma \left (- p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, - p \\ 1 - p \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 \Gamma \left (1 - p\right )} \]

input

integrate((e*x+d)**2*(b*x**2+a)**p/x**3,x)
 

output

-2*a**p*d*e*hyper((-1/2, -p), (1/2,), b*x**2*exp_polar(I*pi)/a)/x - b**p*d 
**2*x**(2*p - 2)*gamma(1 - p)*hyper((-p, 1 - p), (2 - p,), a*exp_polar(I*p 
i)/(b*x**2))/(2*gamma(2 - p)) - b**p*e**2*x**(2*p)*gamma(-p)*hyper((-p, -p 
), (1 - p,), a*exp_polar(I*pi)/(b*x**2))/(2*gamma(1 - p))
 

3.4.98.7 Maxima [F]

\[ \int \frac {(d+e x)^2 \left (a+b x^2\right )^p}{x^3} \, dx=\int { \frac {{\left (e x + d\right )}^{2} {\left (b x^{2} + a\right )}^{p}}{x^{3}} \,d x } \]

input

integrate((e*x+d)^2*(b*x^2+a)^p/x^3,x, algorithm="maxima")
 

output

integrate((e*x + d)^2*(b*x^2 + a)^p/x^3, x)
 

3.4.98.8 Giac [F]

\[ \int \frac {(d+e x)^2 \left (a+b x^2\right )^p}{x^3} \, dx=\int { \frac {{\left (e x + d\right )}^{2} {\left (b x^{2} + a\right )}^{p}}{x^{3}} \,d x } \]

input

integrate((e*x+d)^2*(b*x^2+a)^p/x^3,x, algorithm="giac")
 

output

integrate((e*x + d)^2*(b*x^2 + a)^p/x^3, x)
 

3.4.98.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^2 \left (a+b x^2\right )^p}{x^3} \, dx=\int \frac {{\left (b\,x^2+a\right )}^p\,{\left (d+e\,x\right )}^2}{x^3} \,d x \]

input

int(((a + b*x^2)^p*(d + e*x)^2)/x^3,x)
 

output

int(((a + b*x^2)^p*(d + e*x)^2)/x^3, x)