\(\int \frac {x^3 (d+e x)^n}{a+c x^2} \, dx\) [364]

Optimal result

Integrand size = 20, antiderivative size = 209 \[ \int \frac {x^3 (d+e x)^n}{a+c x^2} \, dx=-\frac {d (d+e x)^{1+n}}{c e^2 (1+n)}+\frac {(d+e x)^{2+n}}{c e^2 (2+n)}+\frac {a (d+e x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 c^{3/2} \left (\sqrt {c} d-\sqrt {-a} e\right ) (1+n)}+\frac {a (d+e x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 c^{3/2} \left (\sqrt {c} d+\sqrt {-a} e\right ) (1+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {1643, 845, 70} \[ \int \frac {x^3 (d+e x)^n}{a+c x^2} \, dx=\frac {a (d+e x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 c^{3/2} (n+1) \left (\sqrt {c} d-\sqrt {-a} e\right )}+\frac {a (d+e x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 c^{3/2} (n+1) \left (\sqrt {-a} e+\sqrt {c} d\right )}-\frac {d (d+e x)^{n+1}}{c e^2 (n+1)}+\frac {(d+e x)^{n+2}}{c e^2 (n+2)} \]

[In]

Int[(x^3*(d + e*x)^n)/(a + c*x^2),x]

[Out]

-((d*(d + e*x)^(1 + n))/(c*e^2*(1 + n))) + (d + e*x)^(2 + n)/(c*e^2*(2 + n)) + (a*(d + e*x)^(1 + n)*Hypergeome
tric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(2*c^(3/2)*(Sqrt[c]*d - Sqrt[-a]*e)*(1
+ n)) + (a*(d + e*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)]
)/(2*c^(3/2)*(Sqrt[c]*d + Sqrt[-a]*e)*(1 + n))

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 845

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]

Rule 1643

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.80 \[ \int \frac {x^3 (d+e x)^n}{a+c x^2} \, dx=\frac {(d+e x)^{1+n} \left (-\frac {2 \sqrt {c} (d-e (1+n) x)}{e^2 (2+n)}+\frac {a \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{\sqrt {c} d-\sqrt {-a} e}+\frac {a \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 c^{3/2} (1+n)} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {x^{3} \left (e x +d \right )^{n}}{c \,x^{2}+a}d x\]

[In]

int(x^3*(e*x+d)^n/(c*x^2+a),x)

[Out]

int(x^3*(e*x+d)^n/(c*x^2+a),x)

Fricas [F]

\[ \int \frac {x^3 (d+e x)^n}{a+c x^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} x^{3}}{c x^{2} + a} \,d x } \]

[In]

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

[Out]

integral((e*x + d)^n*x^3/(c*x^2 + a), x)

Sympy [F]

\[ \int \frac {x^3 (d+e x)^n}{a+c x^2} \, dx=\int \frac {x^{3} \left (d + e x\right )^{n}}{a + c x^{2}}\, dx \]

[In]

integrate(x**3*(e*x+d)**n/(c*x**2+a),x)

[Out]

Integral(x**3*(d + e*x)**n/(a + c*x**2), x)

Maxima [F]

\[ \int \frac {x^3 (d+e x)^n}{a+c x^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} x^{3}}{c x^{2} + a} \,d x } \]

[In]

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

[Out]

integrate((e*x + d)^n*x^3/(c*x^2 + a), x)

Giac [F]

\[ \int \frac {x^3 (d+e x)^n}{a+c x^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} x^{3}}{c x^{2} + a} \,d x } \]

[In]

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

[Out]

integrate((e*x + d)^n*x^3/(c*x^2 + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 (d+e x)^n}{a+c x^2} \, dx=\int \frac {x^3\,{\left (d+e\,x\right )}^n}{c\,x^2+a} \,d x \]

[In]

int((x^3*(d + e*x)^n)/(a + c*x^2),x)

[Out]

int((x^3*(d + e*x)^n)/(a + c*x^2), x)