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

Optimal. Leaf size=213 \[ -\frac {\left (a+b x^2\right )^{p+1} \left (a e^2+b d^2 p\right ) \, _2F_1\left (1,p+1;p+2;\frac {b x^2}{a}+1\right )}{2 a^2 d^3 (p+1)}+\frac {e \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} F_1\left (-\frac {1}{2};-p,1;\frac {1}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^2 x}+\frac {e^4 \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 d^3 (p+1) \left (a e^2+b d^2\right )}-\frac {\left (a+b x^2\right )^{p+1}}{2 a d x^2} \]

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {959, 446, 103, 156, 65, 68, 511, 510} \[ -\frac {\left (a+b x^2\right )^{p+1} \left (a e^2+b d^2 p\right ) \, _2F_1\left (1,p+1;p+2;\frac {b x^2}{a}+1\right )}{2 a^2 d^3 (p+1)}+\frac {e \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} F_1\left (-\frac {1}{2};-p,1;\frac {1}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^2 x}+\frac {e^4 \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 d^3 (p+1) \left (a e^2+b d^2\right )}-\frac {\left (a+b x^2\right )^{p+1}}{2 a d x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^p/(x^3*(d + e*x)),x]

[Out]

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

Rule 65

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

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 446

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 510

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)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), 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 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[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 959

Int[(((g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Dist[(d*(g*x)^n)/x^n, In
t[(x^n*(a + c*x^2)^p)/(d^2 - e^2*x^2), x], x] - Dist[(e*(g*x)^n)/x^n, Int[(x^(n + 1)*(a + c*x^2)^p)/(d^2 - e^2
*x^2), x], x] /; FreeQ[{a, c, d, e, g, n, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] &&  !IntegersQ[n, 2
*p]

Rubi steps

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

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Mathematica [A]  time = 0.28, size = 256, normalized size = 1.20 \[ \frac {\left (a+b x^2\right )^p \left (-\frac {e^2 \left (\frac {e \left (x-\sqrt {-\frac {a}{b}}\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} F_1\left (-2 p;-p,-p;1-2 p;\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )}{p}+\left (\frac {a}{b x^2}+1\right )^{-p} \left (\frac {d^2 \, _2F_1\left (1-p,-p;2-p;-\frac {a}{b x^2}\right )}{(p-1) x^2}+\frac {e^2 \, _2F_1\left (-p,-p;1-p;-\frac {a}{b x^2}\right )}{p}\right )+\frac {2 d e \left (\frac {b x^2}{a}+1\right )^{-p} \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};-\frac {b x^2}{a}\right )}{x}\right )}{2 d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + b*x^2)^p*(-((e^2*AppellF1[-2*p, -p, -p, 1 - 2*p, (d - Sqrt[-(a/b)]*e)/(d + e*x), (d + Sqrt[-(a/b)]*e)/(d
+ e*x)])/(p*((e*(-Sqrt[-(a/b)] + x))/(d + e*x))^p*((e*(Sqrt[-(a/b)] + x))/(d + e*x))^p)) + (2*d*e*Hypergeomet
ric2F1[-1/2, -p, 1/2, -((b*x^2)/a)])/(x*(1 + (b*x^2)/a)^p) + ((d^2*Hypergeometric2F1[1 - p, -p, 2 - p, -(a/(b*
x^2))])/((-1 + p)*x^2) + (e^2*Hypergeometric2F1[-p, -p, 1 - p, -(a/(b*x^2))])/p)/(1 + a/(b*x^2))^p))/(2*d^3)

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fricas [F]  time = 0.85, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )}^{p}}{e x^{4} + d x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^2 + a)^p/(e*x^4 + d*x^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{2}+a \right )^{p}}{\left (e x +d \right ) x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^p/(x^3*(d + e*x)),x)

[Out]

int((a + b*x^2)^p/(x^3*(d + e*x)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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