### 3.724 $$\int \frac{(d+e x)^m}{a+c x^2} \, dx$$

Optimal. Leaf size=167 $\frac{(d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )}{2 \sqrt{-a} (m+1) \left (\sqrt{c} d-\sqrt{-a} e\right )}-\frac{(d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )}{2 \sqrt{-a} (m+1) \left (\sqrt{-a} e+\sqrt{c} d\right )}$

[Out]

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

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Rubi [A]  time = 0.208284, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.118, Rules used = {712, 68} $\frac{(d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )}{2 \sqrt{-a} (m+1) \left (\sqrt{c} d-\sqrt{-a} e\right )}-\frac{(d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )}{2 \sqrt{-a} (m+1) \left (\sqrt{-a} e+\sqrt{c} d\right )}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^m/(a + c*x^2),x]

[Out]

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

Rule 712

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m, 1/(a + c*x^2
), x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[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}{a+c x^2} \, dx &=\int \left (\frac{\sqrt{-a} (d+e x)^m}{2 a \left (\sqrt{-a}-\sqrt{c} x\right )}+\frac{\sqrt{-a} (d+e x)^m}{2 a \left (\sqrt{-a}+\sqrt{c} x\right )}\right ) \, dx\\ &=-\frac{\int \frac{(d+e x)^m}{\sqrt{-a}-\sqrt{c} x} \, dx}{2 \sqrt{-a}}-\frac{\int \frac{(d+e x)^m}{\sqrt{-a}+\sqrt{c} x} \, dx}{2 \sqrt{-a}}\\ &=\frac{(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 )}{2 \sqrt{-a} \left (\sqrt{c} d-\sqrt{-a} e\right ) (1+m)}-\frac{(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 )}{2 \sqrt{-a} \left (\sqrt{c} d+\sqrt{-a} e\right ) (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.157862, size = 145, normalized size = 0.87 $\frac{(d+e x)^{m+1} \left (\frac{\, _2F_1\left (1,m+1;m+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )}{\sqrt{c} d-\sqrt{-a} e}-\frac{\, _2F_1\left (1,m+1;m+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )}{\sqrt{-a} e+\sqrt{c} d}\right )}{2 \sqrt{-a} (m+1)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^m/(a + c*x^2),x]

[Out]

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

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Maple [F]  time = 0.587, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex+d \right ) ^{m}}{c{x}^{2}+a}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e*x + d)^m/(c*x^2 + a), x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x + d)^m/(c*x^2 + a), x)