3.8.31 \(\int (d+e x)^m (a+c x^2)^p \, dx\) [731]

Optimal. Leaf size=152 \[ \frac {(d+e x)^{1+m} \left (a+c x^2\right )^p \left (1-\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}}\right )^{-p} \left (1-\frac {d+e x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}\right )^{-p} F_1\left (1+m;-p,-p;2+m;\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}},\frac {d+e x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}\right )}{e (1+m)} \]

[Out]

(e*x+d)^(1+m)*(c*x^2+a)^p*AppellF1(1+m,-p,-p,2+m,(e*x+d)/(d-e*(-a)^(1/2)/c^(1/2)),(e*x+d)/(d+e*(-a)^(1/2)/c^(1
/2)))/e/(1+m)/((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 = 0.04, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {774, 138} \begin {gather*} \frac {\left (a+c x^2\right )^p (d+e x)^{m+1} \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 (m+1;-p,-p;m+2;\frac {d+e x}{d-\frac {\sqrt {-a} e}{\sqrt {c}}},\frac {d+e x}{d+\frac {\sqrt {-a} e}{\sqrt {c}}}\right )}{e (m+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^m*(a + c*x^2)^p,x]

[Out]

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

Rule 138

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

Rule 774

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)^m \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 ) \text {Subst}\left (\int x^m \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+m} \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+m;-p,-p;2+m;\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+m)}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 157, normalized size = 1.03 \begin {gather*} \frac {\left (\frac {e \left (\sqrt {-\frac {a}{c}}-x\right )}{d+\sqrt {-\frac {a}{c}} e}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{c}}+x\right )}{-d+\sqrt {-\frac {a}{c}} e}\right )^{-p} (d+e x)^{1+m} \left (a+c x^2\right )^p F_1\left (1+m;-p,-p;2+m;\frac {d+e x}{d-\sqrt {-\frac {a}{c}} e},\frac {d+e x}{d+\sqrt {-\frac {a}{c}} e}\right )}{e (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.15, size = 0, normalized size = 0.00 \[\int \left (e x +d \right )^{m} \left (c \,x^{2}+a \right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (c\,x^2+a\right )}^p\,{\left (d+e\,x\right )}^m \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^2)^p*(d + e*x)^m,x)

[Out]

int((a + c*x^2)^p*(d + e*x)^m, x)

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