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

Optimal. Leaf size=250 \[ \frac {\left (c d^2-a e^2\right ) (d+e x)^{n+1}}{c^2 e^3 (n+1)}+\frac {(-a)^{3/2} (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 c^2 (n+1) \left (\sqrt {c} d-\sqrt {-a} e\right )}-\frac {(-a)^{3/2} (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 c^2 (n+1) \left (\sqrt {-a} e+\sqrt {c} d\right )}-\frac {2 d (d+e x)^{n+2}}{c e^3 (n+2)}+\frac {(d+e x)^{n+3}}{c e^3 (n+3)} \]

[Out]

(-a*e^2+c*d^2)*(e*x+d)^(1+n)/c^2/e^3/(1+n)-2*d*(e*x+d)^(2+n)/c/e^3/(2+n)+(e*x+d)^(3+n)/c/e^3/(3+n)+1/2*(-a)^(3
/2)*(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)))/c^2/(1+n)/(-e*(-a)^(1/2)
+d*c^(1/2))-1/2*(-a)^(3/2)*(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)))/c^
2/(1+n)/(e*(-a)^(1/2)+d*c^(1/2))

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Rubi [A]  time = 0.40, antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {1629, 712, 68} \[ \frac {\left (c d^2-a e^2\right ) (d+e x)^{n+1}}{c^2 e^3 (n+1)}+\frac {(-a)^{3/2} (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 c^2 (n+1) \left (\sqrt {c} d-\sqrt {-a} e\right )}-\frac {(-a)^{3/2} (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 c^2 (n+1) \left (\sqrt {-a} e+\sqrt {c} d\right )}-\frac {2 d (d+e x)^{n+2}}{c e^3 (n+2)}+\frac {(d+e x)^{n+3}}{c e^3 (n+3)} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(d + e*x)^n)/(a + c*x^2),x]

[Out]

((c*d^2 - a*e^2)*(d + e*x)^(1 + n))/(c^2*e^3*(1 + n)) - (2*d*(d + e*x)^(2 + n))/(c*e^3*(2 + n)) + (d + e*x)^(3
+ n)/(c*e^3*(3 + n)) + ((-a)^(3/2)*(d + e*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(
Sqrt[c]*d - Sqrt[-a]*e)])/(2*c^2*(Sqrt[c]*d - Sqrt[-a]*e)*(1 + n)) - ((-a)^(3/2)*(d + e*x)^(1 + n)*Hypergeomet
ric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/(2*c^2*(Sqrt[c]*d + Sqrt[-a]*e)*(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 712

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m, 1/(a + c*x^2
), x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[m]

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]

Rubi steps

\begin {align*} \int \frac {x^4 (d+e x)^n}{a+c x^2} \, dx &=\int \left (\frac {\left (c d^2-a e^2\right ) (d+e x)^n}{c^2 e^2}-\frac {2 d (d+e x)^{1+n}}{c e^2}+\frac {(d+e x)^{2+n}}{c e^2}+\frac {a^2 (d+e x)^n}{c^2 \left (a+c x^2\right )}\right ) \, dx\\ &=\frac {\left (c d^2-a e^2\right ) (d+e x)^{1+n}}{c^2 e^3 (1+n)}-\frac {2 d (d+e x)^{2+n}}{c e^3 (2+n)}+\frac {(d+e x)^{3+n}}{c e^3 (3+n)}+\frac {a^2 \int \frac {(d+e x)^n}{a+c x^2} \, dx}{c^2}\\ &=\frac {\left (c d^2-a e^2\right ) (d+e x)^{1+n}}{c^2 e^3 (1+n)}-\frac {2 d (d+e x)^{2+n}}{c e^3 (2+n)}+\frac {(d+e x)^{3+n}}{c e^3 (3+n)}+\frac {a^2 \int \left (\frac {\sqrt {-a} (d+e x)^n}{2 a \left (\sqrt {-a}-\sqrt {c} x\right )}+\frac {\sqrt {-a} (d+e x)^n}{2 a \left (\sqrt {-a}+\sqrt {c} x\right )}\right ) \, dx}{c^2}\\ &=\frac {\left (c d^2-a e^2\right ) (d+e x)^{1+n}}{c^2 e^3 (1+n)}-\frac {2 d (d+e x)^{2+n}}{c e^3 (2+n)}+\frac {(d+e x)^{3+n}}{c e^3 (3+n)}-\frac {(-a)^{3/2} \int \frac {(d+e x)^n}{\sqrt {-a}-\sqrt {c} x} \, dx}{2 c^2}-\frac {(-a)^{3/2} \int \frac {(d+e x)^n}{\sqrt {-a}+\sqrt {c} x} \, dx}{2 c^2}\\ &=\frac {\left (c d^2-a e^2\right ) (d+e x)^{1+n}}{c^2 e^3 (1+n)}-\frac {2 d (d+e x)^{2+n}}{c e^3 (2+n)}+\frac {(d+e x)^{3+n}}{c e^3 (3+n)}+\frac {(-a)^{3/2} (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 )}{2 c^2 \left (\sqrt {c} d-\sqrt {-a} e\right ) (1+n)}-\frac {(-a)^{3/2} (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 )}{2 c^2 \left (\sqrt {c} d+\sqrt {-a} e\right ) (1+n)}\\ \end {align*}

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Mathematica [A]  time = 0.50, size = 217, normalized size = 0.87 \[ \frac {(d+e x)^{n+1} \left (\frac {2 \left (c d^2-a e^2\right )}{e^3 (n+1)}+\frac {(-a)^{3/2} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{(n+1) \left (\sqrt {c} d-\sqrt {-a} e\right )}+\frac {\sqrt {-a} a \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{(n+1) \left (\sqrt {-a} e+\sqrt {c} d\right )}+\frac {2 c (d+e x)^2}{e^3 (n+3)}-\frac {4 c d (d+e x)}{e^3 (n+2)}\right )}{2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.97, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e x + d\right )}^{n} x^{4}}{c x^{2} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {x^{4} \left (e x +d \right )^{n}}{c \,x^{2}+a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(d + e*x)^n)/(a + c*x^2),x)

[Out]

int((x^4*(d + e*x)^n)/(a + c*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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