3.1492 \(\int \frac{(A+B x) (d+e x)^{1+m}}{a+c x^2} \, dx\)

Optimal. Leaf size=202 \[ -\frac{\left (\sqrt{-a} A \sqrt{c}+a B\right ) (d+e x)^{m+2} \, _2F_1\left (1,m+2;m+3;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )}{2 a \sqrt{c} (m+2) \left (\sqrt{c} d-\sqrt{-a} e\right )}-\frac{\left (\frac{\sqrt{-a} B}{\sqrt{c}}+A\right ) (d+e x)^{m+2} \, _2F_1\left (1,m+2;m+3;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )}{2 \sqrt{-a} (m+2) \left (\sqrt{-a} e+\sqrt{c} d\right )} \]

[Out]

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

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Rubi [A]  time = 0.18557, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {831, 68} \[ -\frac{\left (\sqrt{-a} A \sqrt{c}+a B\right ) (d+e x)^{m+2} \, _2F_1\left (1,m+2;m+3;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )}{2 a \sqrt{c} (m+2) \left (\sqrt{c} d-\sqrt{-a} e\right )}-\frac{\left (\frac{\sqrt{-a} B}{\sqrt{c}}+A\right ) (d+e x)^{m+2} \, _2F_1\left (1,m+2;m+3;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )}{2 \sqrt{-a} (m+2) \left (\sqrt{-a} e+\sqrt{c} d\right )} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(d + e*x)^(1 + m))/(a + c*x^2),x]

[Out]

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

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

Mathematica [A]  time = 0.172907, size = 182, normalized size = 0.9 \[ \frac{(d+e x)^{m+2} \left (\frac{\left (\sqrt{-a} A \sqrt{c}+a B\right ) \, _2F_1\left (1,m+2;m+3;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )}{\sqrt{-a} e-\sqrt{c} d}+\frac{\left (\sqrt{-a} A \sqrt{c}-a B\right ) \, _2F_1\left (1,m+2;m+3;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )}{\sqrt{-a} e+\sqrt{c} d}\right )}{2 a \sqrt{c} (m+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(d + e*x)^(1 + m))/(a + c*x^2),x]

[Out]

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

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Maple [F]  time = 0.044, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( Bx+A \right ) \left ( ex+d \right ) ^{1+m}}{c{x}^{2}+a}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^(1+m)/(c*x^2+a),x)

[Out]

int((B*x+A)*(e*x+d)^(1+m)/(c*x^2+a),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x + A\right )}{\left (e x + d\right )}^{m + 1}}{c x^{2} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(1+m)/(c*x^2+a),x, algorithm="maxima")

[Out]

integrate((B*x + A)*(e*x + d)^(m + 1)/(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 (B x + A\right )}{\left (e x + d\right )}^{m + 1}}{c x^{2} + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(1+m)/(c*x^2+a),x, algorithm="fricas")

[Out]

integral((B*x + A)*(e*x + d)^(m + 1)/(c*x^2 + a), 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((B*x+A)*(e*x+d)**(1+m)/(c*x**2+a),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x + A\right )}{\left (e x + d\right )}^{m + 1}}{c x^{2} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(1+m)/(c*x^2+a),x, algorithm="giac")

[Out]

integrate((B*x + A)*(e*x + d)^(m + 1)/(c*x^2 + a), x)