3.739 \(\int (d+e x)^{-2 p} (a+c x^2)^p \, dx\)

Optimal. Leaf size=160 \[ \frac {\left (a+c x^2\right )^p (d+e x)^{1-2 p} \left (1-\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}}\right )^{-p} \left (1-\frac {d+e x}{\frac {\sqrt {-a} e}{\sqrt {c}}+d}\right )^{-p} F_1\left (1-2 p;-p,-p;2-2 p;\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}},\frac {d+e x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}\right )}{e (1-2 p)} \]

[Out]

(e*x+d)^(1-2*p)*(c*x^2+a)^p*AppellF1(1-2*p,-p,-p,2-2*p,(e*x+d)/(d-e*(-a)^(1/2)/c^(1/2)),(e*x+d)/(d+e*(-a)^(1/2
)/c^(1/2)))/e/(1-2*p)/((1+(-e*x-d)/(d-e*(-a)^(1/2)/c^(1/2)))^p)/((1+(-e*x-d)/(d+e*(-a)^(1/2)/c^(1/2)))^p)

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Rubi [A]  time = 0.09, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {760, 133} \[ \frac {\left (a+c x^2\right )^p (d+e x)^{1-2 p} \left (1-\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}}\right )^{-p} \left (1-\frac {d+e x}{\frac {\sqrt {-a} e}{\sqrt {c}}+d}\right )^{-p} F_1\left (1-2 p;-p,-p;2-2 p;\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}},\frac {d+e x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}\right )}{e (1-2 p)} \]

Antiderivative was successfully verified.

[In]

Int[(a + c*x^2)^p/(d + e*x)^(2*p),x]

[Out]

((d + e*x)^(1 - 2*p)*(a + c*x^2)^p*AppellF1[1 - 2*p, -p, -p, 2 - 2*p, (d + e*x)/(d - (Sqrt[-a]*e)/Sqrt[c]), (d
+ e*x)/(d + (Sqrt[-a]*e)/Sqrt[c])])/(e*(1 - 2*p)*(1 - (d + e*x)/(d - (Sqrt[-a]*e)/Sqrt[c]))^p*(1 - (d + e*x)/
(d + (Sqrt[-a]*e)/Sqrt[c]))^p)

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 760

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[(a + c*x^
2)^p/(e*(1 - (d + e*x)/(d + (e*q)/c))^p*(1 - (d + e*x)/(d - (e*q)/c))^p), Subst[Int[x^m*Simp[1 - x/(d + (e*q)/
c), x]^p*Simp[1 - x/(d - (e*q)/c), x]^p, x], x, d + e*x], x]] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a
*e^2, 0] &&  !IntegerQ[p]

Rubi steps

\begin {align*} \int (d+e x)^{-2 p} \left (a+c x^2\right )^p \, dx &=\frac {\left (\left (a+c x^2\right )^p \left (1-\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}}\right )^{-p} \left (1-\frac {d+e x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}\right )^{-p}\right ) \operatorname {Subst}\left (\int x^{-2 p} \left (1-\frac {x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}}\right )^p \left (1-\frac {x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}\right )^p \, dx,x,d+e x\right )}{e}\\ &=\frac {(d+e x)^{1-2 p} \left (a+c x^2\right )^p \left (1-\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}}\right )^{-p} \left (1-\frac {d+e x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}\right )^{-p} F_1\left (1-2 p;-p,-p;2-2 p;\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}},\frac {d+e x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}\right )}{e (1-2 p)}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 166, normalized size = 1.04 \[ -\frac {\left (a+c x^2\right )^p (d+e x)^{1-2 p} \left (\frac {e \left (\sqrt {-\frac {a}{c}}-x\right )}{e \sqrt {-\frac {a}{c}}+d}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{c}}+x\right )}{e \sqrt {-\frac {a}{c}}-d}\right )^{-p} F_1\left (1-2 p;-p,-p;2-2 p;\frac {d+e x}{d-\sqrt {-\frac {a}{c}} e},\frac {d+e x}{d+\sqrt {-\frac {a}{c}} e}\right )}{e (2 p-1)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + c*x^2)^p/(d + e*x)^(2*p),x]

[Out]

-(((d + e*x)^(1 - 2*p)*(a + c*x^2)^p*AppellF1[1 - 2*p, -p, -p, 2 - 2*p, (d + e*x)/(d - Sqrt[-(a/c)]*e), (d + e
*x)/(d + Sqrt[-(a/c)]*e)])/(e*(-1 + 2*p)*((e*(Sqrt[-(a/c)] - x))/(d + Sqrt[-(a/c)]*e))^p*((e*(Sqrt[-(a/c)] + x
))/(-d + Sqrt[-(a/c)]*e))^p))

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fricas [F]  time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c x^{2} + a\right )}^{p}}{{\left (e x + d\right )}^{2 \, p}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*x^2 + a)^p/(e*x + d)^(2*p), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.72, size = 0, normalized size = 0.00 \[ \int \left (c \,x^{2}+a \right )^{p} \left (e x +d \right )^{-2 p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^2)^p/(d + e*x)^(2*p),x)

[Out]

int((a + c*x^2)^p/(d + e*x)^(2*p), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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