3.373 \(\int \frac {x (d+e x)^n}{(a+c x^2)^2} \, dx\)

Optimal. Leaf size=279 \[ \frac {e n \left (\sqrt {-a} e+\sqrt {c} d\right ) (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{4 \sqrt {-a} \sqrt {c} (n+1) \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (a e^2+c d^2\right )}+\frac {e n \left (\sqrt {-a} \sqrt {c} d+a e\right ) (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{4 a \sqrt {c} (n+1) \left (\sqrt {-a} e+\sqrt {c} d\right ) \left (a e^2+c d^2\right )}-\frac {(d-e x) (d+e x)^{n+1}}{2 \left (a+c x^2\right ) \left (a e^2+c d^2\right )} \]

[Out]

-1/2*(-e*x+d)*(e*x+d)^(1+n)/(a*e^2+c*d^2)/(c*x^2+a)+1/4*e*n*(e*x+d)^(1+n)*hypergeom([1, 1+n],[2+n],(e*x+d)*c^(
1/2)/(-e*(-a)^(1/2)+d*c^(1/2)))*(e*(-a)^(1/2)+d*c^(1/2))/(a*e^2+c*d^2)/(1+n)/(-a)^(1/2)/c^(1/2)/(-e*(-a)^(1/2)
+d*c^(1/2))+1/4*e*n*(e*x+d)^(1+n)*hypergeom([1, 1+n],[2+n],(e*x+d)*c^(1/2)/(e*(-a)^(1/2)+d*c^(1/2)))*(a*e+d*(-
a)^(1/2)*c^(1/2))/a/(a*e^2+c*d^2)/(1+n)/c^(1/2)/(e*(-a)^(1/2)+d*c^(1/2))

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Rubi [A]  time = 0.30, antiderivative size = 279, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {823, 831, 68} \[ \frac {e n \left (\sqrt {-a} e+\sqrt {c} d\right ) (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{4 \sqrt {-a} \sqrt {c} (n+1) \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (a e^2+c d^2\right )}+\frac {e n \left (\sqrt {-a} \sqrt {c} d+a e\right ) (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{4 a \sqrt {c} (n+1) \left (\sqrt {-a} e+\sqrt {c} d\right ) \left (a e^2+c d^2\right )}-\frac {(d-e x) (d+e x)^{n+1}}{2 \left (a+c x^2\right ) \left (a e^2+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(x*(d + e*x)^n)/(a + c*x^2)^2,x]

[Out]

-((d - e*x)*(d + e*x)^(1 + n))/(2*(c*d^2 + a*e^2)*(a + c*x^2)) + (e*(Sqrt[c]*d + Sqrt[-a]*e)*n*(d + e*x)^(1 +
n)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(4*Sqrt[-a]*Sqrt[c]*(Sqrt
[c]*d - Sqrt[-a]*e)*(c*d^2 + a*e^2)*(1 + n)) + (e*(Sqrt[-a]*Sqrt[c]*d + a*e)*n*(d + e*x)^(1 + n)*Hypergeometri
c2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/(4*a*Sqrt[c]*(Sqrt[c]*d + Sqrt[-a]*e)*(c*
d^2 + a*e^2)*(1 + n))

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 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 831

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m, (f + g*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !Ration
alQ[m]

Rubi steps

\begin {align*} \int \frac {x (d+e x)^n}{\left (a+c x^2\right )^2} \, dx &=-\frac {(d-e x) (d+e x)^{1+n}}{2 \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac {\int \frac {(d+e x)^n \left (-a c d e n+a c e^2 n x\right )}{a+c x^2} \, dx}{2 a c \left (c d^2+a e^2\right )}\\ &=-\frac {(d-e x) (d+e x)^{1+n}}{2 \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac {\int \left (\frac {\left (-\sqrt {-a} a c d e n-a^2 \sqrt {c} e^2 n\right ) (d+e x)^n}{2 a \left (\sqrt {-a}-\sqrt {c} x\right )}+\frac {\left (-\sqrt {-a} a c d e n+a^2 \sqrt {c} e^2 n\right ) (d+e x)^n}{2 a \left (\sqrt {-a}+\sqrt {c} x\right )}\right ) \, dx}{2 a c \left (c d^2+a e^2\right )}\\ &=-\frac {(d-e x) (d+e x)^{1+n}}{2 \left (c d^2+a e^2\right ) \left (a+c x^2\right )}+\frac {\left (e \left (\sqrt {-a} \sqrt {c} d-a e\right ) n\right ) \int \frac {(d+e x)^n}{\sqrt {-a}+\sqrt {c} x} \, dx}{4 a \sqrt {c} \left (c d^2+a e^2\right )}+\frac {\left (e \left (\sqrt {-a} d+\frac {a e}{\sqrt {c}}\right ) n\right ) \int \frac {(d+e x)^n}{\sqrt {-a}-\sqrt {c} x} \, dx}{4 a \left (c d^2+a e^2\right )}\\ &=-\frac {(d-e x) (d+e x)^{1+n}}{2 \left (c d^2+a e^2\right ) \left (a+c x^2\right )}+\frac {e \left (\sqrt {c} d+\sqrt {-a} e\right ) n (d+e x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{4 \sqrt {-a} \sqrt {c} \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (c d^2+a e^2\right ) (1+n)}+\frac {e \left (\sqrt {-a} \sqrt {c} d+a e\right ) n (d+e x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{4 a \sqrt {c} \left (\sqrt {c} d+\sqrt {-a} e\right ) \left (c d^2+a e^2\right ) (1+n)}\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 230, normalized size = 0.82 \[ \frac {(d+e x)^{n+1} \left (-\frac {\left (\sqrt {-a} c d e n-a \sqrt {c} e^2 n\right ) \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{(n+1) \left (\sqrt {c} d-\sqrt {-a} e\right )}+\frac {\left (\sqrt {-a} c d e n+a \sqrt {c} e^2 n\right ) \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{(n+1) \left (\sqrt {-a} e+\sqrt {c} d\right )}-\frac {2 a c (d-e x)}{a+c x^2}\right )}{4 a c \left (a e^2+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*(d + e*x)^n)/(a + c*x^2)^2,x]

[Out]

((d + e*x)^(1 + n)*((-2*a*c*(d - e*x))/(a + c*x^2) - ((Sqrt[-a]*c*d*e*n - a*Sqrt[c]*e^2*n)*Hypergeometric2F1[1
, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/((Sqrt[c]*d - Sqrt[-a]*e)*(1 + n)) + ((Sqrt[-a]
*c*d*e*n + a*Sqrt[c]*e^2*n)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/
((Sqrt[c]*d + Sqrt[-a]*e)*(1 + n))))/(4*a*c*(c*d^2 + a*e^2))

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fricas [F]  time = 0.87, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e x + d\right )}^{n} x}{c^{2} x^{4} + 2 \, a c x^{2} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e*x + d)^n*x/(c^2*x^4 + 2*a*c*x^2 + a^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x + d)^n*x/(c*x^2 + a)^2, x)

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {x \left (e x +d \right )^{n}}{\left (c \,x^{2}+a \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(e*x+d)^n/(c*x^2+a)^2,x)

[Out]

int(x*(e*x+d)^n/(c*x^2+a)^2,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x + d)^n*x/(c*x^2 + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(d + e*x)^n)/(a + c*x^2)^2,x)

[Out]

int((x*(d + e*x)^n)/(a + c*x^2)^2, 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(x*(e*x+d)**n/(c*x**2+a)**2,x)

[Out]

Timed out

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