3.15.92 \(\int \frac {(A+B x) (d+e x)^{1+m}}{a+c x^2} \, dx\) [1492]

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

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 202, normalized size of antiderivative = 1.00, 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 = {845, 70} \begin {gather*} -\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 )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]
time = 0.19, size = 182, normalized size = 0.90 \begin {gather*} \frac {(d+e x)^{2+m} \left (\frac {\left (a B+\sqrt {-a} A \sqrt {c}\right ) \, _2F_1\left (1,2+m;3+m;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{-\sqrt {c} d+\sqrt {-a} e}+\frac {\left (-a B+\sqrt {-a} A \sqrt {c}\right ) \, _2F_1\left (1,2+m;3+m;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 a \sqrt {c} (2+m)} \end {gather*}

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 = 0.23, size = 0, normalized size = 0.00 \[\int \frac {\left (B x +A \right ) \left (e x +d \right )^{1+m}}{c \,x^{2}+a}\, dx\]

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 = 0.00, 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((B*x+A)*(e*x+d)^(1+m)/(c*x^2+a),x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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