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

Optimal. Leaf size=489 \[ -\frac {\sqrt {c} e n \left (\sqrt {-a} \sqrt {c} d+a e\right ) (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{4 a^2 (n+1) \left (\sqrt {-a} e+\sqrt {c} d\right ) \left (a e^2+c d^2\right )}+\frac {\sqrt {c} (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 a^2 (n+1) \left (\sqrt {c} d-\sqrt {-a} e\right )}+\frac {\sqrt {c} (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 a^2 (n+1) \left (\sqrt {-a} e+\sqrt {c} d\right )}-\frac {(d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {e x}{d}+1\right )}{a^2 d (n+1)}+\frac {\sqrt {c} e n \left (\sqrt {-a} e+\sqrt {c} d\right ) (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{4 (-a)^{3/2} (n+1) \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (a e^2+c d^2\right )}+\frac {c (d-e x) (d+e x)^{n+1}}{2 a \left (a+c x^2\right ) \left (a e^2+c d^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.60, antiderivative size = 489, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {961, 65, 823, 831, 68} \[ -\frac {\sqrt {c} e n \left (\sqrt {-a} \sqrt {c} d+a e\right ) (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{4 a^2 (n+1) \left (\sqrt {-a} e+\sqrt {c} d\right ) \left (a e^2+c d^2\right )}+\frac {\sqrt {c} (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 a^2 (n+1) \left (\sqrt {c} d-\sqrt {-a} e\right )}+\frac {\sqrt {c} (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 a^2 (n+1) \left (\sqrt {-a} e+\sqrt {c} d\right )}-\frac {(d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {e x}{d}+1\right )}{a^2 d (n+1)}+\frac {\sqrt {c} e n \left (\sqrt {-a} e+\sqrt {c} d\right ) (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{4 (-a)^{3/2} (n+1) \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (a e^2+c d^2\right )}+\frac {c (d-e x) (d+e x)^{n+1}}{2 a \left (a+c x^2\right ) \left (a e^2+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^n/(x*(a + c*x^2)^2),x]

[Out]

(c*(d - e*x)*(d + e*x)^(1 + n))/(2*a*(c*d^2 + a*e^2)*(a + c*x^2)) + (Sqrt[c]*(d + e*x)^(1 + n)*Hypergeometric2
F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(2*a^2*(Sqrt[c]*d - Sqrt[-a]*e)*(1 + n)) +
(Sqrt[c]*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*(-a)^(3/2)*(Sqrt[c]*d - Sqrt[-a]*e)*(c*d^2 + a*e^2)*(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^2*(Sqrt[c]*d
+ Sqrt[-a]*e)*(1 + n)) - (Sqrt[c]*e*(Sqrt[-a]*Sqrt[c]*d + 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*a^2*(Sqrt[c]*d + Sqrt[-a]*e)*(c*d^2 + a*e^2)*(1 + n
)) - ((d + e*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, 1 + (e*x)/d])/(a^2*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 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]

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

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Mathematica [A]  time = 0.89, size = 391, normalized size = 0.80 \[ \frac {(d+e x)^{n+1} \left (\frac {\sqrt {c} e n \left (\left (\sqrt {-a} c d^2-2 a \sqrt {c} d e+(-a)^{3/2} e^2\right ) \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )+\left (-\sqrt {-a} c d^2-2 a \sqrt {c} d e+\sqrt {-a} a e^2\right ) \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )\right )}{(n+1) \left (a e^2+c d^2\right )^2}+\frac {2 a c (d-e x)}{\left (a+c x^2\right ) \left (a e^2+c d^2\right )}+\frac {2 \sqrt {c} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{(n+1) \left (\sqrt {c} d-\sqrt {-a} e\right )}+\frac {2 \sqrt {c} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{(n+1) \left (\sqrt {-a} e+\sqrt {c} d\right )}-\frac {4 \, _2F_1\left (1,n+1;n+2;\frac {d+e x}{d}\right )}{d n+d}\right )}{4 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.98, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e x + d\right )}^{n}}{c^{2} x^{5} + 2 \, a c x^{3} + a^{2} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e*x + d)^n/(c^2*x^5 + 2*a*c*x^3 + a^2*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^n/(x*(a + c*x^2)^2),x)

[Out]

int((d + e*x)^n/(x*(a + c*x^2)^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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