3.5.19 \(\int \frac {x (a+b x^2)^p}{(d+e x)^2} \, dx\) [419]

3.5.19.1 Optimal result

Integrand size = 18, antiderivative size = 273 \[ \int \frac {x \left (a+b x^2\right )^p}{(d+e x)^2} \, dx=\frac {d \left (a+b x^2\right )^{1+p}}{\left (b d^2+a e^2\right ) (d+e x)}+\frac {\left (a e^2+b d^2 (1+2 p)\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d e \left (b d^2+a e^2\right )}-\frac {b d (1+2 p) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{e \left (b d^2+a e^2\right )}-\frac {\left (a e^2+b d^2 (1+2 p)\right ) \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{2 \left (b d^2+a e^2\right )^2 (1+p)} \]

output

d*(b*x^2+a)^(p+1)/(a*e^2+b*d^2)/(e*x+d)+(a*e^2+b*d^2*(1+2*p))*x*(b*x^2+a)^ 
p*AppellF1(1/2,1,-p,3/2,e^2*x^2/d^2,-b*x^2/a)/d/e/(a*e^2+b*d^2)/((1+b*x^2/ 
a)^p)-b*d*(1+2*p)*x*(b*x^2+a)^p*hypergeom([1/2, -p],[3/2],-b*x^2/a)/e/(a*e 
^2+b*d^2)/((1+b*x^2/a)^p)-1/2*(a*e^2+b*d^2*(1+2*p))*(b*x^2+a)^(p+1)*hyperg 
eom([1, p+1],[2+p],e^2*(b*x^2+a)/(a*e^2+b*d^2))/(a*e^2+b*d^2)^2/(p+1)
 

3.5.19.2 Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.82 \[ \int \frac {x \left (a+b x^2\right )^p}{(d+e x)^2} \, dx=\frac {\left (\frac {e \left (-\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \left (a+b x^2\right )^p \left (-2 d p \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )+(-1+2 p) (d+e x) \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )\right )}{2 e^2 p (-1+2 p) (d+e x)} \]

input

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

output

((a + b*x^2)^p*(-2*d*p*AppellF1[1 - 2*p, -p, -p, 2 - 2*p, (d - Sqrt[-(a/b) 
]*e)/(d + e*x), (d + Sqrt[-(a/b)]*e)/(d + e*x)] + (-1 + 2*p)*(d + e*x)*App 
ellF1[-2*p, -p, -p, 1 - 2*p, (d - Sqrt[-(a/b)]*e)/(d + e*x), (d + Sqrt[-(a 
/b)]*e)/(d + e*x)]))/(2*e^2*p*(-1 + 2*p)*((e*(-Sqrt[-(a/b)] + x))/(d + e*x 
))^p*((e*(Sqrt[-(a/b)] + x))/(d + e*x))^p*(d + e*x))
 

3.5.19.3 Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.91, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {594, 25, 719, 238, 237, 504, 334, 333, 353, 78}

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.

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

\(\Big \downarrow \) 594

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 719

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

\(\Big \downarrow \) 238

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

\(\Big \downarrow \) 237

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

\(\Big \downarrow \) 504

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

\(\Big \downarrow \) 334

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

\(\Big \downarrow \) 333

\(\displaystyle \frac {\frac {\left (a e^2+b d^2 (2 p+1)\right ) \left (\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d}-e \int \frac {x \left (b x^2+a\right )^p}{d^2-e^2 x^2}dx\right )}{e}-\frac {b d (2 p+1) x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{e}}{a e^2+b d^2}+\frac {d \left (a+b x^2\right )^{p+1}}{(d+e x) \left (a e^2+b d^2\right )}\)

\(\Big \downarrow \) 353

\(\displaystyle \frac {\frac {\left (a e^2+b d^2 (2 p+1)\right ) \left (\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d}-\frac {1}{2} e \int \frac {\left (b x^2+a\right )^p}{d^2-e^2 x^2}dx^2\right )}{e}-\frac {b d (2 p+1) x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{e}}{a e^2+b d^2}+\frac {d \left (a+b x^2\right )^{p+1}}{(d+e x) \left (a e^2+b d^2\right )}\)

\(\Big \downarrow \) 78

\(\displaystyle \frac {\frac {\left (a e^2+b d^2 (2 p+1)\right ) \left (\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d}-\frac {e \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 (p+1) \left (a e^2+b d^2\right )}\right )}{e}-\frac {b d (2 p+1) x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{e}}{a e^2+b d^2}+\frac {d \left (a+b x^2\right )^{p+1}}{(d+e x) \left (a e^2+b d^2\right )}\)

input

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

output

(d*(a + b*x^2)^(1 + p))/((b*d^2 + a*e^2)*(d + e*x)) + (-((b*d*(1 + 2*p)*x* 
(a + b*x^2)^p*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(e*(1 + (b*x^ 
2)/a)^p)) + ((a*e^2 + b*d^2*(1 + 2*p))*((x*(a + b*x^2)^p*AppellF1[1/2, -p, 
 1, 3/2, -((b*x^2)/a), (e^2*x^2)/d^2])/(d*(1 + (b*x^2)/a)^p) - (e*(a + b*x 
^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, (e^2*(a + b*x^2))/(b*d^2 + 
a*e^2)])/(2*(b*d^2 + a*e^2)*(1 + p))))/e)/(b*d^2 + a*e^2)
 

3.5.19.3.1 Defintions of rubi rules used

rule 25

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

rule 78

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

rule 237

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[- 
p, 1/2, 1/2 + 1, (-b)*(x^2/a)], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p 
] && GtQ[a, 0]
 

rule 238

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2) 
^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(1 + b*(x^2/a))^p, x], x] / 
; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p] &&  !GtQ[a, 0]
 

rule 333

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F 
reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 
0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 334

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[ 
(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p, q}, x] && 
NeQ[b*c - a*d, 0] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 353

Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] 
 :> Simp[1/2   Subst[Int[(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]
 

rule 504

Int[((a_) + (b_.)*(x_)^2)^(p_)/((c_) + (d_.)*(x_)), x_Symbol] :> Simp[c   I 
nt[(a + b*x^2)^p/(c^2 - d^2*x^2), x], x] - Simp[d   Int[x*((a + b*x^2)^p/(c 
^2 - d^2*x^2)), x], x] /; FreeQ[{a, b, c, d, p}, x]
 

rule 594

Int[(x_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-c)*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/((n + 1)*(b*c^2 + a*d^2))) 
, x] + Simp[1/((n + 1)*(b*c^2 + a*d^2))   Int[(c + d*x)^(n + 1)*(a + b*x^2) 
^p*(a*d*(n + 1) + b*c*(n + 2*p + 3)*x), x], x] /; FreeQ[{a, b, c, d, p}, x] 
 && LtQ[n, -1] && NeQ[b*c^2 + a*d^2, 0]
 

rule 719

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] + 
Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, 
d, e, f, g, m, p}, x] &&  !IGtQ[m, 0]
 

3.5.19.4 Maple [F]

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

input

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

output

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

3.5.19.5 Fricas [F]

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

input

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

output

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

3.5.19.6 Sympy [F]

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

input

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

output

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

3.5.19.7 Maxima [F]

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

input

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

output

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

3.5.19.8 Giac [F]

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

input

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

output

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

3.5.19.9 Mupad [F(-1)]

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

input

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

output

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