3.726 \(\int \frac{(d+e x)^m}{(a+c x^2)^3} \, dx\)

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

[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))

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Rubi [A]  time = 0.81205, antiderivative size = 472, normalized size of antiderivative = 1., 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 = {741, 823, 831, 68} \[ \frac{(d+e x)^{m+1} \left (a \sqrt{c} d e m \left (a e^2 (5-2 m)+3 c d^2\right )-\sqrt{-a} \left (a^2 e^4 \left (m^2-4 m+3\right )+a c d^2 e^2 \left (-m^2-2 m+6\right )+3 c^2 d^4\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )}{16 a^3 (m+1) \left (\sqrt{c} d-\sqrt{-a} e\right ) \left (a e^2+c d^2\right )^2}+\frac{(d+e x)^{m+1} \left (\sqrt{-a} \left (a^2 e^4 \left (m^2-4 m+3\right )+a c d^2 e^2 \left (-m^2-2 m+6\right )+3 c^2 d^4\right )+a \sqrt{c} d e m \left (a e^2 (5-2 m)+3 c d^2\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )}{16 a^3 (m+1) \left (\sqrt{-a} e+\sqrt{c} d\right ) \left (a e^2+c d^2\right )^2}+\frac{(d+e x)^{m+1} \left (c d x \left (a e^2 (5-2 m)+3 c d^2\right )+a e \left (a e^2 (3-m)+c d^2 (m+1)\right )\right )}{8 a^2 \left (a+c x^2\right ) \left (a e^2+c d^2\right )^2}+\frac{(d+e x)^{m+1} (a e+c d x)}{4 a \left (a+c x^2\right )^2 \left (a e^2+c d^2\right )} \]

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 741

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 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 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]

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*}

Mathematica [A]  time = 0.923583, size = 396, normalized size = 0.84 \[ \frac{(d+e x)^{m+1} \left (\frac{\frac{\left (\sqrt{-a} \left (-a^2 e^4 \left (m^2-4 m+3\right )+a c d^2 e^2 \left (m^2+2 m-6\right )-3 c^2 d^4\right )+a \sqrt{c} d e m \left (a e^2 (5-2 m)+3 c d^2\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )}{\sqrt{c} d-\sqrt{-a} e}+\frac{\left (\sqrt{-a} \left (a^2 e^4 \left (m^2-4 m+3\right )-a c d^2 e^2 \left (m^2+2 m-6\right )+3 c^2 d^4\right )+a \sqrt{c} d e m \left (a e^2 (5-2 m)+3 c d^2\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )}{\sqrt{-a} e+\sqrt{c} d}}{a (m+1)}+\frac{2 \left (-a^2 e^3 (m-3)+a c d e (d (m+1)+e (5-2 m) x)+3 c^2 d^3 x\right )}{a+c x^2}+\frac{4 a \left (a e^2+c d^2\right ) (a e+c d x)}{\left (a+c x^2\right )^2}\right )}{16 a^2 \left (a e^2+c d^2\right )^2} \]

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 = 0.604, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex+d \right ) ^{m}}{ \left ( c{x}^{2}+a \right ) ^{3}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{m}}{{\left (c x^{2} + a\right )}^{3}}\,{d x} \end{align*}

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((e*x + d)^m/(c*x^2 + a)^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (e x + d\right )}^{m}}{c^{3} x^{6} + 3 \, a c^{2} x^{4} + 3 \, a^{2} c x^{2} + a^{3}}, x\right ) \end{align*}

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((e*x + 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)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{m}}{{\left (c x^{2} + a\right )}^{3}}\,{d x} \end{align*}

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((e*x + d)^m/(c*x^2 + a)^3, x)