3.370 \(\int \frac {x^4 (d+e x)^n}{(a+c x^2)^2} \, dx\)

Optimal. Leaf size=332 \[ \frac {(d+e x)^{n+1} \left (3 \sqrt {-a} c d^2+a \sqrt {c} d e n+\sqrt {-a} a e^2 (n+3)\right ) \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{4 c^2 (n+1) \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (a e^2+c d^2\right )}-\frac {(d+e x)^{n+1} \left (3 \sqrt {-a} c d^2-a \sqrt {c} d e n+\sqrt {-a} a e^2 (n+3)\right ) \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{4 c^2 (n+1) \left (\sqrt {-a} e+\sqrt {c} d\right ) \left (a e^2+c d^2\right )}+\frac {a (d+e x)^{n+1} (a e+c d x)}{2 c^2 \left (a+c x^2\right ) \left (a e^2+c d^2\right )}+\frac {(d+e x)^{n+1}}{c^2 e (n+1)} \]

[Out]

(e*x+d)^(1+n)/c^2/e/(1+n)+1/2*a*(c*d*x+a*e)*(e*x+d)^(1+n)/c^2/(a*e^2+c*d^2)/(c*x^2+a)-1/4*(e*x+d)^(1+n)*hyperg
eom([1, 1+n],[2+n],(e*x+d)*c^(1/2)/(e*(-a)^(1/2)+d*c^(1/2)))*(3*c*d^2*(-a)^(1/2)+a*e^2*(3+n)*(-a)^(1/2)-a*d*e*
n*c^(1/2))/c^2/(a*e^2+c*d^2)/(1+n)/(e*(-a)^(1/2)+d*c^(1/2))+1/4*(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)))*(3*c*d^2*(-a)^(1/2)+a*e^2*(3+n)*(-a)^(1/2)+a*d*e*n*c^(1/2))/c^2/(a*e^2+c*d
^2)/(1+n)/(-e*(-a)^(1/2)+d*c^(1/2))

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Rubi [A]  time = 0.45, antiderivative size = 332, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {1649, 1629, 68} \[ \frac {(d+e x)^{n+1} \left (3 \sqrt {-a} c d^2+a \sqrt {c} d e n+\sqrt {-a} a e^2 (n+3)\right ) \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{4 c^2 (n+1) \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (a e^2+c d^2\right )}-\frac {(d+e x)^{n+1} \left (3 \sqrt {-a} c d^2-a \sqrt {c} d e n+\sqrt {-a} a e^2 (n+3)\right ) \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{4 c^2 (n+1) \left (\sqrt {-a} e+\sqrt {c} d\right ) \left (a e^2+c d^2\right )}+\frac {a (d+e x)^{n+1} (a e+c d x)}{2 c^2 \left (a+c x^2\right ) \left (a e^2+c d^2\right )}+\frac {(d+e x)^{n+1}}{c^2 e (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d + e*x)^(1 + n)/(c^2*e*(1 + n)) + (a*(a*e + c*d*x)*(d + e*x)^(1 + n))/(2*c^2*(c*d^2 + a*e^2)*(a + c*x^2)) +
((3*Sqrt[-a]*c*d^2 + a*Sqrt[c]*d*e*n + Sqrt[-a]*a*e^2*(3 + n))*(d + e*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2
+ n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(4*c^2*(Sqrt[c]*d - Sqrt[-a]*e)*(c*d^2 + a*e^2)*(1 + n))
- ((3*Sqrt[-a]*c*d^2 - a*Sqrt[c]*d*e*n + Sqrt[-a]*a*e^2*(3 + n))*(d + e*x)^(1 + n)*Hypergeometric2F1[1, 1 + n,
2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/(4*c^2*(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 1629

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

Rule 1649

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a + c*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + c
*x^2, x], x, 1]}, -Simp[((d + e*x)^(m + 1)*(a + c*x^2)^(p + 1)*(a*(e*f - d*g) + (c*d*f + a*e*g)*x))/(2*a*(p +
1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSu
m[2*a*(p + 1)*(c*d^2 + a*e^2)*Q + c*d^2*f*(2*p + 3) - a*e*(d*g*m - e*f*(m + 2*p + 3)) + e*(c*d*f + a*e*g)*(m +
2*p + 4)*x, x], x], x]] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] &
&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

Rubi steps

\begin {align*} \int \frac {x^4 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx &=\frac {a (a e+c d x) (d+e x)^{1+n}}{2 c^2 \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac {\int \frac {(d+e x)^n \left (\frac {a^2 \left (c d^2+a e^2 (1+n)\right )}{c^2}+\frac {a^2 d e n x}{c}-2 a \left (d^2+\frac {a e^2}{c}\right ) x^2\right )}{a+c x^2} \, dx}{2 a \left (c d^2+a e^2\right )}\\ &=\frac {a (a e+c d x) (d+e x)^{1+n}}{2 c^2 \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac {\int \left (-\frac {2 a \left (c d^2+a e^2\right ) (d+e x)^n}{c^2}+\frac {\left (-\frac {a^3 d e n}{c^{3/2}}+\sqrt {-a} \left (\frac {3 a^2 d^2}{c}+\frac {3 a^3 e^2}{c^2}+\frac {a^3 e^2 n}{c^2}\right )\right ) (d+e x)^n}{2 a \left (\sqrt {-a}-\sqrt {c} x\right )}+\frac {\left (\frac {a^3 d e n}{c^{3/2}}+\sqrt {-a} \left (\frac {3 a^2 d^2}{c}+\frac {3 a^3 e^2}{c^2}+\frac {a^3 e^2 n}{c^2}\right )\right ) (d+e x)^n}{2 a \left (\sqrt {-a}+\sqrt {c} x\right )}\right ) \, dx}{2 a \left (c d^2+a e^2\right )}\\ &=\frac {(d+e x)^{1+n}}{c^2 e (1+n)}+\frac {a (a e+c d x) (d+e x)^{1+n}}{2 c^2 \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac {\left (3 \sqrt {-a} c d^2-a \sqrt {c} d e n+\sqrt {-a} a e^2 (3+n)\right ) \int \frac {(d+e x)^n}{\sqrt {-a}-\sqrt {c} x} \, dx}{4 c^2 \left (c d^2+a e^2\right )}-\frac {\left (3 \sqrt {-a} c d^2+a \sqrt {c} d e n+\sqrt {-a} a e^2 (3+n)\right ) \int \frac {(d+e x)^n}{\sqrt {-a}+\sqrt {c} x} \, dx}{4 c^2 \left (c d^2+a e^2\right )}\\ &=\frac {(d+e x)^{1+n}}{c^2 e (1+n)}+\frac {a (a e+c d x) (d+e x)^{1+n}}{2 c^2 \left (c d^2+a e^2\right ) \left (a+c x^2\right )}+\frac {\left (3 \sqrt {-a} c d^2+a \sqrt {c} d e n+\sqrt {-a} a e^2 (3+n)\right ) (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 c^2 \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (c d^2+a e^2\right ) (1+n)}-\frac {\left (3 \sqrt {-a} c d^2-a \sqrt {c} d e n+\sqrt {-a} a e^2 (3+n)\right ) (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 c^2 \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 = 0.83, size = 413, normalized size = 1.24 \[ \frac {(d+e x)^{n+1} \left (\frac {a \left (\frac {\left (\sqrt {-a} \sqrt {c} d e n-a e^2 (n-1)+c d^2\right ) \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{\sqrt {c} d-\sqrt {-a} e}-\frac {\left (-\sqrt {-a} \sqrt {c} d e n-a e^2 (n-1)+c d^2\right ) \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{\sqrt {-a} e+\sqrt {c} d}\right )}{\sqrt {-a} (n+1) \left (a e^2+c d^2\right )}+\frac {2 a (a e+c d x)}{\left (a+c x^2\right ) \left (a e^2+c d^2\right )}+\frac {4 \sqrt {-a} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{(n+1) \left (\sqrt {c} d-\sqrt {-a} e\right )}-\frac {4 \sqrt {-a} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{(n+1) \left (\sqrt {-a} e+\sqrt {c} d\right )}+\frac {4}{e n+e}\right )}{4 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)^(1 + n)*(4/(e + e*n) + (2*a*(a*e + c*d*x))/((c*d^2 + a*e^2)*(a + c*x^2)) + (4*Sqrt[-a]*Hypergeometr
ic2F1[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*
Sqrt[-a]*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)) + (a*(((c*d^2 - a*e^2*(-1 + n) + Sqrt[-a]*Sqrt[c]*d*e*n)*Hypergeometric2F1[1, 1 + n, 2 + n, (S
qrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(Sqrt[c]*d - Sqrt[-a]*e) - ((c*d^2 - a*e^2*(-1 + n) - Sqrt[-a]*Sq
rt[c]*d*e*n)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/(Sqrt[c]*d + Sq
rt[-a]*e)))/(Sqrt[-a]*(c*d^2 + a*e^2)*(1 + n))))/(4*c^2)

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fricas [F]  time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e x + d\right )}^{n} x^{4}}{c^{2} x^{4} + 2 \, a c x^{2} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)^n/(c*x^2+a)^2,x, algorithm="fricas")

[Out]

integral((e*x + d)^n*x^4/(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 \[ \int \frac {{\left (e x + d\right )}^{n} x^{4}}{{\left (c x^{2} + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)^n/(c*x^2+a)^2,x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \frac {x^{4} \left (e x +d \right )^{n}}{\left (c \,x^{2}+a \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^4*(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 \[ \int \frac {{\left (e x + d\right )}^{n} x^{4}}{{\left (c x^{2} + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)^n/(c*x^2+a)^2,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4\,{\left (d+e\,x\right )}^n}{{\left (c\,x^2+a\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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