3.5.27 \(\int \frac {(a+b x^2)^p}{(d+e x)^3} \, dx\) [427]

Optimal. Leaf size=322 \[ -\frac {d^2 e \left (a+b x^2\right )^{1+p}}{4 \left (b d^2+a e^2\right ) \left (d^2-e^2 x^2\right )^2}+\frac {x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} F_1\left (\frac {1}{2};-p,3;\frac {3}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^3}+\frac {e^2 x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} F_1\left (\frac {3}{2};-p,3;\frac {5}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^5}+\frac {b e \left (2 a e^2+b d^2 (1+p)\right ) \left (a+b x^2\right )^{1+p} \, _2F_1\left (2,1+p;2+p;\frac {e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{4 \left (b d^2+a e^2\right )^3 (1+p)}-\frac {3 b^2 d^2 e \left (a+b x^2\right )^{1+p} \, _2F_1\left (3,1+p;2+p;\frac {e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{2 \left (b d^2+a e^2\right )^3 (1+p)} \]

[Out]

-1/4*d^2*e*(b*x^2+a)^(1+p)/(a*e^2+b*d^2)/(-e^2*x^2+d^2)^2+x*(b*x^2+a)^p*AppellF1(1/2,3,-p,3/2,e^2*x^2/d^2,-b*x
^2/a)/d^3/((1+b*x^2/a)^p)+e^2*x^3*(b*x^2+a)^p*AppellF1(3/2,3,-p,5/2,e^2*x^2/d^2,-b*x^2/a)/d^5/((1+b*x^2/a)^p)+
1/4*b*e*(2*a*e^2+b*d^2*(1+p))*(b*x^2+a)^(1+p)*hypergeom([2, 1+p],[2+p],e^2*(b*x^2+a)/(a*e^2+b*d^2))/(a*e^2+b*d
^2)^3/(1+p)-3/2*b^2*d^2*e*(b*x^2+a)^(1+p)*hypergeom([3, 1+p],[2+p],e^2*(b*x^2+a)/(a*e^2+b*d^2))/(a*e^2+b*d^2)^
3/(1+p)

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Rubi [A]
time = 0.23, antiderivative size = 322, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {771, 441, 440, 455, 70, 525, 524, 457, 79} \begin {gather*} \frac {e^2 x^3 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} F_1\left (\frac {3}{2};-p,3;\frac {5}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^5}+\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} F_1\left (\frac {1}{2};-p,3;\frac {3}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^3}-\frac {3 b^2 d^2 e \left (a+b x^2\right )^{p+1} \, _2F_1\left (3,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 )^3}+\frac {b e \left (a+b x^2\right )^{p+1} \left (2 a e^2+b d^2 (p+1)\right ) \, _2F_1\left (2,p+1;p+2;\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{4 (p+1) \left (a e^2+b d^2\right )^3}-\frac {d^2 e \left (a+b x^2\right )^{p+1}}{4 \left (d^2-e^2 x^2\right )^2 \left (a e^2+b d^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^p/(d + e*x)^3,x]

[Out]

-1/4*(d^2*e*(a + b*x^2)^(1 + p))/((b*d^2 + a*e^2)*(d^2 - e^2*x^2)^2) + (x*(a + b*x^2)^p*AppellF1[1/2, -p, 3, 3
/2, -((b*x^2)/a), (e^2*x^2)/d^2])/(d^3*(1 + (b*x^2)/a)^p) + (e^2*x^3*(a + b*x^2)^p*AppellF1[3/2, -p, 3, 5/2, -
((b*x^2)/a), (e^2*x^2)/d^2])/(d^5*(1 + (b*x^2)/a)^p) + (b*e*(2*a*e^2 + b*d^2*(1 + p))*(a + b*x^2)^(1 + p)*Hype
rgeometric2F1[2, 1 + p, 2 + p, (e^2*(a + b*x^2))/(b*d^2 + a*e^2)])/(4*(b*d^2 + a*e^2)^3*(1 + p)) - (3*b^2*d^2*
e*(a + b*x^2)^(1 + p)*Hypergeometric2F1[3, 1 + p, 2 + p, (e^2*(a + b*x^2))/(b*d^2 + a*e^2)])/(2*(b*d^2 + a*e^2
)^3*(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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(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] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

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 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]

Rule 524

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

Rule 525

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

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^p}{(d+e x)^3} \, dx &=\int \left (\frac {d^3 \left (a+b x^2\right )^p}{\left (d^2-e^2 x^2\right )^3}-\frac {3 d^2 e x \left (a+b x^2\right )^p}{\left (d^2-e^2 x^2\right )^3}+\frac {3 d e^2 x^2 \left (a+b x^2\right )^p}{\left (d^2-e^2 x^2\right )^3}+\frac {e^3 x^3 \left (a+b x^2\right )^p}{\left (-d^2+e^2 x^2\right )^3}\right ) \, dx\\ &=d^3 \int \frac {\left (a+b x^2\right )^p}{\left (d^2-e^2 x^2\right )^3} \, dx-\left (3 d^2 e\right ) \int \frac {x \left (a+b x^2\right )^p}{\left (d^2-e^2 x^2\right )^3} \, dx+\left (3 d e^2\right ) \int \frac {x^2 \left (a+b x^2\right )^p}{\left (d^2-e^2 x^2\right )^3} \, dx+e^3 \int \frac {x^3 \left (a+b x^2\right )^p}{\left (-d^2+e^2 x^2\right )^3} \, dx\\ &=-\left (\frac {1}{2} \left (3 d^2 e\right ) \text {Subst}\left (\int \frac {(a+b x)^p}{\left (d^2-e^2 x\right )^3} \, dx,x,x^2\right )\right )+\frac {1}{2} e^3 \text {Subst}\left (\int \frac {x (a+b x)^p}{\left (-d^2+e^2 x\right )^3} \, dx,x,x^2\right )+\left (d^3 \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}{\left (d^2-e^2 x^2\right )^3} \, dx+\left (3 d e^2 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int \frac {x^2 \left (1+\frac {b x^2}{a}\right )^p}{\left (d^2-e^2 x^2\right )^3} \, dx\\ &=-\frac {d^2 e \left (a+b x^2\right )^{1+p}}{4 \left (b d^2+a e^2\right ) \left (d^2-e^2 x^2\right )^2}+\frac {x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} F_1\left (\frac {1}{2};-p,3;\frac {3}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^3}+\frac {e^2 x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} F_1\left (\frac {3}{2};-p,3;\frac {5}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^5}-\frac {3 b^2 d^2 e \left (a+b x^2\right )^{1+p} \, _2F_1\left (3,1+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 )^3 (1+p)}+\frac {\left (e \left (2 a e^2+b d^2 (1+p)\right )\right ) \text {Subst}\left (\int \frac {(a+b x)^p}{\left (-d^2+e^2 x\right )^2} \, dx,x,x^2\right )}{4 \left (b d^2+a e^2\right )}\\ &=-\frac {d^2 e \left (a+b x^2\right )^{1+p}}{4 \left (b d^2+a e^2\right ) \left (d^2-e^2 x^2\right )^2}+\frac {x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} F_1\left (\frac {1}{2};-p,3;\frac {3}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^3}+\frac {e^2 x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} F_1\left (\frac {3}{2};-p,3;\frac {5}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^5}+\frac {b e \left (2 a e^2+b d^2 (1+p)\right ) \left (a+b x^2\right )^{1+p} \, _2F_1\left (2,1+p;2+p;\frac {e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{4 \left (b d^2+a e^2\right )^3 (1+p)}-\frac {3 b^2 d^2 e \left (a+b x^2\right )^{1+p} \, _2F_1\left (3,1+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 )^3 (1+p)}\\ \end {align*}

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Mathematica [A]
time = 0.28, size = 142, normalized size = 0.44 \begin {gather*} \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 F_1\left (2-2 p;-p,-p;3-2 p;\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )}{2 e (-1+p) (d+e x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (b \,x^{2}+a \right )^{p}}{\left (e x +d \right )^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x^{2}\right )^{p}}{\left (d + e x\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (b\,x^2+a\right )}^p}{{\left (d+e\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^p/(d + e*x)^3,x)

[Out]

int((a + b*x^2)^p/(d + e*x)^3, x)

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