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]

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

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Rubi [A]  time = 0.21, antiderivative size = 167, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.15, 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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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e x + d\right )}^{m}}{c x^{2} + a}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x + d\right )}^{m}}{c x^{2} + a}\,{d x} \]

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)

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maple [F]  time = 0.78, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x +d \right )^{m}}{c \,x^{2}+a}\, dx \]

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

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d+e\,x\right )}^m}{c\,x^2+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^m/(a + c*x^2),x)

[Out]

int((d + e*x)^m/(a + c*x^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d + e x\right )^{m}}{a + c x^{2}}\, dx \]

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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