### 3.725 $$\int \frac{(d+e x)^m}{(a+c x^2)^2} \, dx$$

Optimal. Leaf size=304 $-\frac{(d+e x)^{m+1} \left (\sqrt{-a} \sqrt{c} d e m+a e^2 (1-m)+c d^2\right ) \, _2F_1\left (1,m+1;m+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )}{4 (-a)^{3/2} (m+1) \left (\sqrt{c} d-\sqrt{-a} e\right ) \left (a e^2+c d^2\right )}+\frac{(d+e x)^{m+1} \left (-\sqrt{-a} \sqrt{c} d e m+a e^2 (1-m)+c d^2\right ) \, _2F_1\left (1,m+1;m+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )}{4 (-a)^{3/2} (m+1) \left (\sqrt{-a} e+\sqrt{c} d\right ) \left (a e^2+c d^2\right )}+\frac{(d+e x)^{m+1} (a e+c d x)}{2 a \left (a+c x^2\right ) \left (a e^2+c d^2\right )}$

[Out]

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

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Rubi [A]  time = 0.380677, antiderivative size = 304, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.176, Rules used = {741, 831, 68} $-\frac{(d+e x)^{m+1} \left (\sqrt{-a} \sqrt{c} d e m+a e^2 (1-m)+c d^2\right ) \, _2F_1\left (1,m+1;m+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )}{4 (-a)^{3/2} (m+1) \left (\sqrt{c} d-\sqrt{-a} e\right ) \left (a e^2+c d^2\right )}+\frac{(d+e x)^{m+1} \left (-\sqrt{-a} \sqrt{c} d e m+a e^2 (1-m)+c d^2\right ) \, _2F_1\left (1,m+1;m+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )}{4 (-a)^{3/2} (m+1) \left (\sqrt{-a} e+\sqrt{c} d\right ) \left (a e^2+c d^2\right )}+\frac{(d+e x)^{m+1} (a e+c d x)}{2 a \left (a+c x^2\right ) \left (a e^2+c d^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^m/(a + c*x^2)^2,x]

[Out]

((a*e + c*d*x)*(d + e*x)^(1 + m))/(2*a*(c*d^2 + a*e^2)*(a + c*x^2)) - ((c*d^2 + a*e^2*(1 - m) + Sqrt[-a]*Sqrt[
c]*d*e*m)*(d + e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (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 + m)) + ((c*d^2 + a*e^2*(1 - m) - Sqrt[-a]*Sqrt[c]*d
*e*m)*(d + e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (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 + 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 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 )^2} \, dx &=\frac{(a e+c d x) (d+e x)^{1+m}}{2 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac{\int \frac{(d+e x)^m \left (-c d^2-a e^2 (1-m)+c d e m x\right )}{a+c x^2} \, dx}{2 a \left (c d^2+a e^2\right )}\\ &=\frac{(a e+c d x) (d+e x)^{1+m}}{2 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac{\int \left (\frac{\left (\sqrt{-a} \left (-c d^2-a e^2 (1-m)\right )-a \sqrt{c} d e m\right ) (d+e x)^m}{2 a \left (\sqrt{-a}-\sqrt{c} x\right )}+\frac{\left (\sqrt{-a} \left (-c d^2-a e^2 (1-m)\right )+a \sqrt{c} d e m\right ) (d+e x)^m}{2 a \left (\sqrt{-a}+\sqrt{c} x\right )}\right ) \, dx}{2 a \left (c d^2+a e^2\right )}\\ &=\frac{(a e+c d x) (d+e x)^{1+m}}{2 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )}+\frac{\left (c d^2+a e^2 (1-m)-\sqrt{-a} \sqrt{c} d e m\right ) \int \frac{(d+e x)^m}{\sqrt{-a}-\sqrt{c} x} \, dx}{4 (-a)^{3/2} \left (c d^2+a e^2\right )}+\frac{\left (c d^2+a e^2 (1-m)+\sqrt{-a} \sqrt{c} d e m\right ) \int \frac{(d+e x)^m}{\sqrt{-a}+\sqrt{c} x} \, dx}{4 (-a)^{3/2} \left (c d^2+a e^2\right )}\\ &=\frac{(a e+c d x) (d+e x)^{1+m}}{2 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac{\left (c d^2+a e^2 (1-m)+\sqrt{-a} \sqrt{c} d e m\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 )}{4 (-a)^{3/2} \left (\sqrt{c} d-\sqrt{-a} e\right ) \left (c d^2+a e^2\right ) (1+m)}+\frac{\left (c d^2+a e^2 (1-m)-\sqrt{-a} \sqrt{c} d e m\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 )}{4 (-a)^{3/2} \left (\sqrt{c} d+\sqrt{-a} e\right ) \left (c d^2+a e^2\right ) (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.361402, size = 253, normalized size = 0.83 $\frac{(d+e x)^{m+1} \left (\frac{\left (\sqrt{-a} \sqrt{c} d e m-a e^2 (m-1)+c d^2\right ) \, _2F_1\left (1,m+1;m+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )}{\sqrt{-a} (m+1) \left (\sqrt{c} d-\sqrt{-a} e\right )}+\frac{\left (\sqrt{-a} \sqrt{c} d e m+a e^2 (m-1)-c d^2\right ) \, _2F_1\left (1,m+1;m+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )}{\sqrt{-a} (m+1) \left (\sqrt{-a} e+\sqrt{c} d\right )}+\frac{2 (a e+c d x)}{a+c x^2}\right )}{4 a \left (a e^2+c d^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((d + e*x)^(1 + m)*((2*(a*e + c*d*x))/(a + c*x^2) + ((c*d^2 - a*e^2*(-1 + m) + Sqrt[-a]*Sqrt[c]*d*e*m)*Hyperge
ometric2F1[1, 1 + m, 2 + m, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(Sqrt[-a]*(Sqrt[c]*d - Sqrt[-a]*e)*
(1 + m)) + ((-(c*d^2) + a*e^2*(-1 + m) + Sqrt[-a]*Sqrt[c]*d*e*m)*Hypergeometric2F1[1, 1 + m, 2 + m, (Sqrt[c]*(
d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/(Sqrt[-a]*(Sqrt[c]*d + Sqrt[-a]*e)*(1 + m))))/(4*a*(c*d^2 + a*e^2))

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Maple [F]  time = 0.545, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex+d \right ) ^{m}}{ \left ( c{x}^{2}+a \right ) ^{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((e*x+d)^m/(c*x^2+a)^2,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 )}^{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x + d)^m/(c*x^2 + a)^2, 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^{2} x^{4} + 2 \, a c x^{2} + a^{2}}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e*x + d)^m/(c^2*x^4 + 2*a*c*x^2 + a^2), 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)**2,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 )}^{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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