3.8.25 \(\int \frac {(d+e x)^m}{(a+c x^2)^2} \, dx\) [725]

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

[Out]

1/2*(c*d*x+a*e)*(e*x+d)^(1+m)/a/(a*e^2+c*d^2)/(c*x^2+a)+1/4*(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)))*(c*d^2+a*e^2*(1-m)-d*e*m*(-a)^(1/2)*c^(1/2))/(-a)^(3/2)/(a*e^2+c*d^2)/(1+m)/(e*
(-a)^(1/2)+d*c^(1/2))-1/4*(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)))*(c
*d^2+a*e^2*(1-m)+d*e*m*(-a)^(1/2)*c^(1/2))/(-a)^(3/2)/(a*e^2+c*d^2)/(1+m)/(-e*(-a)^(1/2)+d*c^(1/2))

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Rubi [A]
time = 0.25, antiderivative size = 304, normalized size of antiderivative = 1.00, 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 = {755, 845, 70} \begin {gather*} -\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 )} \end {gather*}

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 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 755

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 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 {(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*}

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Mathematica [A]
time = 0.34, size = 253, normalized size = 0.83 \begin {gather*} \frac {(d+e x)^{1+m} \left (\frac {2 (a e+c d x)}{a+c x^2}+\frac {\left (c d^2-a e^2 (-1+m)+\sqrt {-a} \sqrt {c} d e m\right ) \, _2F_1\left (1,1+m;2+m;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{\sqrt {-a} \left (\sqrt {c} d-\sqrt {-a} e\right ) (1+m)}+\frac {\left (-c d^2+a e^2 (-1+m)+\sqrt {-a} \sqrt {c} d e m\right ) \, _2F_1\left (1,1+m;2+m;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{\sqrt {-a} \left (\sqrt {c} d+\sqrt {-a} e\right ) (1+m)}\right )}{4 a \left (c d^2+a e^2\right )} \end {gather*}

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 = 0.13, size = 0, normalized size = 0.00 \[\int \frac {\left (e x +d \right )^{m}}{\left (c \,x^{2}+a \right )^{2}}\, dx\]

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

[Out]

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

[Out]

integral((x*e + d)^m/(c^2*x^4 + 2*a*c*x^2 + a^2), 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((e*x+d)**m/(c*x**2+a)**2,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((e*x+d)^m/(c*x^2+a)^2,x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^m/(a + c*x^2)^2,x)

[Out]

int((d + e*x)^m/(a + c*x^2)^2, x)

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