3.8.26 \(\int \frac {(d+e x)^m}{(a+c x^2)^3} \, dx\) [726]

Optimal. Leaf size=472 \[ \frac {(a e+c d x) (d+e x)^{1+m}}{4 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )^2}+\frac {(d+e x)^{1+m} \left (a e \left (a e^2 (3-m)+c d^2 (1+m)\right )+c d \left (3 c d^2+a e^2 (5-2 m)\right ) x\right )}{8 a^2 \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}+\frac {\left (a \sqrt {c} d e \left (3 c d^2+a e^2 (5-2 m)\right ) m-\sqrt {-a} \left (3 c^2 d^4+a c d^2 e^2 \left (6-2 m-m^2\right )+a^2 e^4 \left (3-4 m+m^2\right )\right )\right ) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{16 a^3 \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (c d^2+a e^2\right )^2 (1+m)}+\frac {\left (a \sqrt {c} d e \left (3 c d^2+a e^2 (5-2 m)\right ) m+\sqrt {-a} \left (3 c^2 d^4+a c d^2 e^2 \left (6-2 m-m^2\right )+a^2 e^4 \left (3-4 m+m^2\right )\right )\right ) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{16 a^3 \left (\sqrt {c} d+\sqrt {-a} e\right ) \left (c d^2+a e^2\right )^2 (1+m)} \]

[Out]

1/4*(c*d*x+a*e)*(e*x+d)^(1+m)/a/(a*e^2+c*d^2)/(c*x^2+a)^2+1/8*(e*x+d)^(1+m)*(a*e*(a*e^2*(3-m)+c*d^2*(1+m))+c*d
*(3*c*d^2+a*e^2*(5-2*m))*x)/a^2/(a*e^2+c*d^2)^2/(c*x^2+a)+1/16*(e*x+d)^(1+m)*hypergeom([1, 1+m],[2+m],(e*x+d)*
c^(1/2)/(-e*(-a)^(1/2)+d*c^(1/2)))*(-(3*c^2*d^4+a*c*d^2*e^2*(-m^2-2*m+6)+a^2*e^4*(m^2-4*m+3))*(-a)^(1/2)+a*d*e
*(3*c*d^2+a*e^2*(5-2*m))*m*c^(1/2))/a^3/(a*e^2+c*d^2)^2/(1+m)/(-e*(-a)^(1/2)+d*c^(1/2))+1/16*(e*x+d)^(1+m)*hyp
ergeom([1, 1+m],[2+m],(e*x+d)*c^(1/2)/(e*(-a)^(1/2)+d*c^(1/2)))*((3*c^2*d^4+a*c*d^2*e^2*(-m^2-2*m+6)+a^2*e^4*(
m^2-4*m+3))*(-a)^(1/2)+a*d*e*(3*c*d^2+a*e^2*(5-2*m))*m*c^(1/2))/a^3/(a*e^2+c*d^2)^2/(1+m)/(e*(-a)^(1/2)+d*c^(1
/2))

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Rubi [A]
time = 0.56, antiderivative size = 472, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {755, 837, 845, 70} \begin {gather*} \frac {(d+e x)^{m+1} \left (c d x \left (a e^2 (5-2 m)+3 c d^2\right )+a e \left (a e^2 (3-m)+c d^2 (m+1)\right )\right )}{8 a^2 \left (a+c x^2\right ) \left (a e^2+c d^2\right )^2}+\frac {(d+e x)^{m+1} \left (a \sqrt {c} d e m \left (a e^2 (5-2 m)+3 c d^2\right )-\sqrt {-a} \left (a^2 e^4 \left (m^2-4 m+3\right )+a c d^2 e^2 \left (-m^2-2 m+6\right )+3 c^2 d^4\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{16 a^3 (m+1) \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (a e^2+c d^2\right )^2}+\frac {(d+e x)^{m+1} \left (\sqrt {-a} \left (a^2 e^4 \left (m^2-4 m+3\right )+a c d^2 e^2 \left (-m^2-2 m+6\right )+3 c^2 d^4\right )+a \sqrt {c} d e m \left (a e^2 (5-2 m)+3 c d^2\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{16 a^3 (m+1) \left (\sqrt {-a} e+\sqrt {c} d\right ) \left (a e^2+c d^2\right )^2}+\frac {(d+e x)^{m+1} (a e+c d x)}{4 a \left (a+c x^2\right )^2 \left (a e^2+c d^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^m/(a + c*x^2)^3,x]

[Out]

((a*e + c*d*x)*(d + e*x)^(1 + m))/(4*a*(c*d^2 + a*e^2)*(a + c*x^2)^2) + ((d + e*x)^(1 + m)*(a*e*(a*e^2*(3 - m)
+ c*d^2*(1 + m)) + c*d*(3*c*d^2 + a*e^2*(5 - 2*m))*x))/(8*a^2*(c*d^2 + a*e^2)^2*(a + c*x^2)) + ((a*Sqrt[c]*d*
e*(3*c*d^2 + a*e^2*(5 - 2*m))*m - Sqrt[-a]*(3*c^2*d^4 + a*c*d^2*e^2*(6 - 2*m - m^2) + a^2*e^4*(3 - 4*m + m^2))
)*(d + e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(16*a^3*
(Sqrt[c]*d - Sqrt[-a]*e)*(c*d^2 + a*e^2)^2*(1 + m)) + ((a*Sqrt[c]*d*e*(3*c*d^2 + a*e^2*(5 - 2*m))*m + Sqrt[-a]
*(3*c^2*d^4 + a*c*d^2*e^2*(6 - 2*m - m^2) + a^2*e^4*(3 - 4*m + m^2)))*(d + e*x)^(1 + m)*Hypergeometric2F1[1, 1
+ m, 2 + m, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/(16*a^3*(Sqrt[c]*d + Sqrt[-a]*e)*(c*d^2 + a*e^2)^2
*(1 + m))

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 755

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(m + 1))*(a*e + c*d*x)*
((a + c*x^2)^(p + 1)/(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*Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[
{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 837

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] +
Dist[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 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]

Rubi steps

\begin {align*} \int \frac {(d+e x)^m}{\left (a+c x^2\right )^3} \, dx &=\frac {(a e+c d x) (d+e x)^{1+m}}{4 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )^2}-\frac {\int \frac {(d+e x)^m \left (-3 c d^2-a e^2 (3-m)-c d e (2-m) x\right )}{\left (a+c x^2\right )^2} \, dx}{4 a \left (c d^2+a e^2\right )}\\ &=\frac {(a e+c d x) (d+e x)^{1+m}}{4 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )^2}+\frac {(d+e x)^{1+m} \left (a e \left (a e^2 (3-m)+c d^2 (1+m)\right )+c d \left (3 c d^2+a e^2 (5-2 m)\right ) x\right )}{8 a^2 \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}+\frac {\int \frac {(d+e x)^m \left (c \left (3 c^2 d^4+a c d^2 e^2 \left (6-2 m-m^2\right )+a^2 e^4 \left (3-4 m+m^2\right )\right )-c^2 d e \left (3 c d^2+a e^2 (5-2 m)\right ) m x\right )}{a+c x^2} \, dx}{8 a^2 c \left (c d^2+a e^2\right )^2}\\ &=\frac {(a e+c d x) (d+e x)^{1+m}}{4 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )^2}+\frac {(d+e x)^{1+m} \left (a e \left (a e^2 (3-m)+c d^2 (1+m)\right )+c d \left (3 c d^2+a e^2 (5-2 m)\right ) x\right )}{8 a^2 \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}+\frac {\int \left (\frac {\left (a c^{3/2} d e \left (3 c d^2+a e^2 (5-2 m)\right ) m+\sqrt {-a} c \left (3 c^2 d^4+a c d^2 e^2 \left (6-2 m-m^2\right )+a^2 e^4 \left (3-4 m+m^2\right )\right )\right ) (d+e x)^m}{2 a \left (\sqrt {-a}-\sqrt {c} x\right )}+\frac {\left (-a c^{3/2} d e \left (3 c d^2+a e^2 (5-2 m)\right ) m+\sqrt {-a} c \left (3 c^2 d^4+a c d^2 e^2 \left (6-2 m-m^2\right )+a^2 e^4 \left (3-4 m+m^2\right )\right )\right ) (d+e x)^m}{2 a \left (\sqrt {-a}+\sqrt {c} x\right )}\right ) \, dx}{8 a^2 c \left (c d^2+a e^2\right )^2}\\ &=\frac {(a e+c d x) (d+e x)^{1+m}}{4 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )^2}+\frac {(d+e x)^{1+m} \left (a e \left (a e^2 (3-m)+c d^2 (1+m)\right )+c d \left (3 c d^2+a e^2 (5-2 m)\right ) x\right )}{8 a^2 \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}-\frac {\left (a \sqrt {c} d e \left (3 c d^2+a e^2 (5-2 m)\right ) m-\sqrt {-a} \left (3 c^2 d^4+a c d^2 e^2 \left (6-2 m-m^2\right )+a^2 e^4 \left (3-4 m+m^2\right )\right )\right ) \int \frac {(d+e x)^m}{\sqrt {-a}+\sqrt {c} x} \, dx}{16 a^3 \left (c d^2+a e^2\right )^2}+\frac {\left (a \sqrt {c} d e \left (3 c d^2+a e^2 (5-2 m)\right ) m+\sqrt {-a} \left (3 c^2 d^4+a c d^2 e^2 \left (6-2 m-m^2\right )+a^2 e^4 \left (3-4 m+m^2\right )\right )\right ) \int \frac {(d+e x)^m}{\sqrt {-a}-\sqrt {c} x} \, dx}{16 a^3 \left (c d^2+a e^2\right )^2}\\ &=\frac {(a e+c d x) (d+e x)^{1+m}}{4 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )^2}+\frac {(d+e x)^{1+m} \left (a e \left (a e^2 (3-m)+c d^2 (1+m)\right )+c d \left (3 c d^2+a e^2 (5-2 m)\right ) x\right )}{8 a^2 \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}+\frac {\left (a \sqrt {c} d e \left (3 c d^2+a e^2 (5-2 m)\right ) m-\sqrt {-a} \left (3 c^2 d^4+a c d^2 e^2 \left (6-2 m-m^2\right )+a^2 e^4 \left (3-4 m+m^2\right )\right )\right ) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{16 a^3 \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (c d^2+a e^2\right )^2 (1+m)}+\frac {\left (a \sqrt {c} d e \left (3 c d^2+a e^2 (5-2 m)\right ) m+\sqrt {-a} \left (3 c^2 d^4+a c d^2 e^2 \left (6-2 m-m^2\right )+a^2 e^4 \left (3-4 m+m^2\right )\right )\right ) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{16 a^3 \left (\sqrt {c} d+\sqrt {-a} e\right ) \left (c d^2+a e^2\right )^2 (1+m)}\\ \end {align*}

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Mathematica [A]
time = 0.77, size = 396, normalized size = 0.84 \begin {gather*} \frac {(d+e x)^{1+m} \left (\frac {4 a \left (c d^2+a e^2\right ) (a e+c d x)}{\left (a+c x^2\right )^2}+\frac {2 \left (-a^2 e^3 (-3+m)+3 c^2 d^3 x+a c d e (d (1+m)+e (5-2 m) x)\right )}{a+c x^2}+\frac {\frac {\left (a \sqrt {c} d e \left (3 c d^2+a e^2 (5-2 m)\right ) m+\sqrt {-a} \left (-3 c^2 d^4-a^2 e^4 \left (3-4 m+m^2\right )+a c d^2 e^2 \left (-6+2 m+m^2\right )\right )\right ) \, _2F_1\left (1,1+m;2+m;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{\sqrt {c} d-\sqrt {-a} e}+\frac {\left (a \sqrt {c} d e \left (3 c d^2+a e^2 (5-2 m)\right ) m+\sqrt {-a} \left (3 c^2 d^4+a^2 e^4 \left (3-4 m+m^2\right )-a c d^2 e^2 \left (-6+2 m+m^2\right )\right )\right ) \, _2F_1\left (1,1+m;2+m;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{\sqrt {c} d+\sqrt {-a} e}}{a (1+m)}\right )}{16 a^2 \left (c d^2+a e^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)^(1 + m)*((4*a*(c*d^2 + a*e^2)*(a*e + c*d*x))/(a + c*x^2)^2 + (2*(-(a^2*e^3*(-3 + m)) + 3*c^2*d^3*x
+ a*c*d*e*(d*(1 + m) + e*(5 - 2*m)*x)))/(a + c*x^2) + (((a*Sqrt[c]*d*e*(3*c*d^2 + a*e^2*(5 - 2*m))*m + Sqrt[-a
]*(-3*c^2*d^4 - a^2*e^4*(3 - 4*m + m^2) + a*c*d^2*e^2*(-6 + 2*m + m^2)))*Hypergeometric2F1[1, 1 + m, 2 + m, (S
qrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(Sqrt[c]*d - Sqrt[-a]*e) + ((a*Sqrt[c]*d*e*(3*c*d^2 + a*e^2*(5 -
2*m))*m + Sqrt[-a]*(3*c^2*d^4 + a^2*e^4*(3 - 4*m + m^2) - a*c*d^2*e^2*(-6 + 2*m + m^2)))*Hypergeometric2F1[1,
1 + m, 2 + m, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/(Sqrt[c]*d + Sqrt[-a]*e))/(a*(1 + m))))/(16*a^2*(
c*d^2 + a*e^2)^2)

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Maple [F]
time = 0.16, size = 0, normalized size = 0.00 \[\int \frac {\left (e x +d \right )^{m}}{\left (c \,x^{2}+a \right )^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((x*e + d)^m/(c^3*x^6 + 3*a*c^2*x^4 + 3*a^2*c*x^2 + a^3), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^m}{{\left (c\,x^2+a\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^m/(a + c*x^2)^3,x)

[Out]

int((d + e*x)^m/(a + c*x^2)^3, x)

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