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

Optimal result

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

[Out]

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)*hypergeom([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)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {849, 858, 252, 251, 771, 441, 440, 455, 70} \[ \int \frac {x \left (a+b x^2\right )^p}{(d+e x)^2} \, dx=\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (a e^2+b d^2 (2 p+1)\right ) \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d e \left (a e^2+b d^2\right )}-\frac {b d (2 p+1) x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{e \left (a e^2+b d^2\right )}-\frac {\left (a+b x^2\right )^{p+1} \left (a e^2+b d^2 (2 p+1)\right ) \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 (p+1) \left (a e^2+b d^2\right )^2}+\frac {d \left (a+b x^2\right )^{p+1}}{(d+e x) \left (a e^2+b d^2\right )} \]

[In]

Int[(x*(a + b*x^2)^p)/(d + e*x)^2,x]

[Out]

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

Rule 70

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] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 251

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

Rule 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[p]), Int[(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 440

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

Rule 441

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

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 771

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^2)^p, (d/(d
^2 - e^2*x^2) - e*(x/(d^2 - e^2*x^2)))^(-m), x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&
!IntegerQ[p] && ILtQ[m, 0]

Rule 849

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*f - d*g)*
(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1)/((m + 1)*(c*d^2 + a*e^2))), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rule 858

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(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] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {d \left (a+b x^2\right )^{1+p}}{\left (b d^2+a e^2\right ) (d+e x)}-\frac {\int \frac {(-a e+b d (1+2 p) x) \left (a+b x^2\right )^p}{d+e x} \, dx}{b d^2+a e^2} \\ & = \frac {d \left (a+b x^2\right )^{1+p}}{\left (b d^2+a e^2\right ) (d+e x)}-\frac {(b d (1+2 p)) \int \left (a+b x^2\right )^p \, dx}{e \left (b d^2+a e^2\right )}+\frac {\left (a e^2+b d^2 (1+2 p)\right ) \int \frac {\left (a+b x^2\right )^p}{d+e x} \, dx}{e \left (b d^2+a e^2\right )} \\ & = \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 ) \int \left (\frac {d \left (a+b x^2\right )^p}{d^2-e^2 x^2}+\frac {e x \left (a+b x^2\right )^p}{-d^2+e^2 x^2}\right ) \, dx}{e \left (b d^2+a e^2\right )}-\frac {\left (b d (1+2 p) \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int \left (1+\frac {b x^2}{a}\right )^p \, dx}{e \left (b d^2+a e^2\right )} \\ & = \frac {d \left (a+b x^2\right )^{1+p}}{\left (b d^2+a e^2\right ) (d+e x)}-\frac {b d (1+2 p) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\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 ) \int \frac {x \left (a+b x^2\right )^p}{-d^2+e^2 x^2} \, dx}{b d^2+a e^2}+\frac {\left (d \left (a e^2+b d^2 (1+2 p)\right )\right ) \int \frac {\left (a+b x^2\right )^p}{d^2-e^2 x^2} \, dx}{e \left (b d^2+a e^2\right )} \\ & = \frac {d \left (a+b x^2\right )^{1+p}}{\left (b d^2+a e^2\right ) (d+e x)}-\frac {b d (1+2 p) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\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 ) \text {Subst}\left (\int \frac {(a+b x)^p}{-d^2+e^2 x} \, dx,x,x^2\right )}{2 \left (b d^2+a e^2\right )}+\frac {\left (d \left (a e^2+b d^2 (1+2 p)\right ) \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int \frac {\left (1+\frac {b x^2}{a}\right )^p}{d^2-e^2 x^2} \, dx}{e \left (b d^2+a e^2\right )} \\ & = \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} F_1\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} \, _2F_1\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} \, _2F_1\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)} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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

[Out]

((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)*AppellF1[-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))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((x*(a + b*x^2)^p)/(d + e*x)^2,x)

[Out]

int((x*(a + b*x^2)^p)/(d + e*x)^2, x)