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

Optimal. Leaf size=207 \[ \frac {\sqrt {c} (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 a (n+1) \left (\sqrt {c} d-\sqrt {-a} e\right )}+\frac {\sqrt {c} (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 a (n+1) \left (\sqrt {-a} e+\sqrt {c} d\right )}-\frac {(d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {e x}{d}+1\right )}{a d (n+1)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {961, 65, 831, 68} \[ \frac {\sqrt {c} (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 a (n+1) \left (\sqrt {c} d-\sqrt {-a} e\right )}+\frac {\sqrt {c} (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 a (n+1) \left (\sqrt {-a} e+\sqrt {c} d\right )}-\frac {(d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {e x}{d}+1\right )}{a d (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^n/(x*(a + c*x^2)),x]

[Out]

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

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 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]

Rule 961

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (ILtQ[m, 0] && ILtQ[n, 0])) &&  !(IGtQ[m, 0] || IGtQ[n, 0])

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.12, size = 189, normalized size = 0.91 \[ \frac {(d+e x)^{n+1} \left (-2 \left (a e^2+c d^2\right ) \, _2F_1\left (1,n+1;n+2;\frac {e x}{d}+1\right )+\left (\sqrt {-a} \sqrt {c} d e+c d^2\right ) \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )+\left (c d^2-\sqrt {-a} \sqrt {c} d e\right ) \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )\right )}{2 a d (n+1) \left (a e^2+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.86, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e x + d\right )}^{n}}{c x^{3} + a x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x +d \right )^{n}}{\left (c \,x^{2}+a \right ) x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^n/(x*(a + c*x^2)),x)

[Out]

int((d + e*x)^n/(x*(a + c*x^2)), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________