3.22.6 \(\int (d+e x)^{-p} (a d e+(c d^2+a e^2) x+c d e x^2)^p \, dx\) [2106]

Optimal. Leaf size=52 \[ \frac {(d+e x)^{-1-p} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{1+p}}{c d (1+p)} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {662} \begin {gather*} \frac {(d+e x)^{-p-1} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{p+1}}{c d (p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 662

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*
((a + b*x + c*x^2)^(p + 1)/(c*(p + 1))), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c
*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 41, normalized size = 0.79 \begin {gather*} \frac {(d+e x)^{-1-p} ((a e+c d x) (d+e x))^{1+p}}{c d (1+p)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.75, size = 56, normalized size = 1.08

method	result	size
gosper	\(\frac {\left (c d x +a e \right ) \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{p} \left (e x +d \right )^{-p}}{c d \left (1+p \right )}\)	\(56\)
norman	\(\left (\frac {x \,{\mathrm e}^{p \ln \left (a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}\right )}}{1+p}+\frac {a e \,{\mathrm e}^{p \ln \left (a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}\right )}}{c d \left (1+p \right )}\right ) {\mathrm e}^{-p \ln \left (e x +d \right )}\)	\(93\)
risch	\(\frac {\left (c d x +a e \right ) \left (e x +d \right )^{-p} {\mathrm e}^{\frac {p \left (-i \mathrm {csgn}\left (i \left (e x +d \right ) \left (c d x +a e \right )\right )^{3} \pi +i \mathrm {csgn}\left (i \left (e x +d \right ) \left (c d x +a e \right )\right )^{2} \mathrm {csgn}\left (i \left (e x +d \right )\right ) \pi +i \mathrm {csgn}\left (i \left (e x +d \right ) \left (c d x +a e \right )\right )^{2} \mathrm {csgn}\left (i \left (c d x +a e \right )\right ) \pi -i \mathrm {csgn}\left (i \left (e x +d \right ) \left (c d x +a e \right )\right ) \mathrm {csgn}\left (i \left (e x +d \right )\right ) \mathrm {csgn}\left (i \left (c d x +a e \right )\right ) \pi +2 \ln \left (e x +d \right )+2 \ln \left (c d x +a e \right )\right )}{2}}}{c d \left (1+p \right )}\)	\(186\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 32, normalized size = 0.62 \begin {gather*} \frac {{\left (c d x + a e\right )} {\left (c d x + a e\right )}^{p}}{c d {\left (p + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*d*x + a*e)*(c*d*x + a*e)^p/(c*d*(p + 1))

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Fricas [A]
time = 2.74, size = 57, normalized size = 1.10 \begin {gather*} \frac {{\left (c d x + a e\right )} {\left (c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e\right )}^{p}}{{\left (c d p + c d\right )} {\left (x e + d\right )}^{p}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [A]
time = 1.66, size = 85, normalized size = 1.63 \begin {gather*} \frac {\frac {c d x e^{\left (p \log \left (c d x + a e\right ) + p \log \left (x e + d\right )\right )}}{{\left (x e + d\right )}^{p}} + \frac {a e^{\left (p \log \left (c d x + a e\right ) + p \log \left (x e + d\right ) + 1\right )}}{{\left (x e + d\right )}^{p}}}{c d p + c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*d*x*e^(p*log(c*d*x + a*e) + p*log(x*e + d))/(x*e + d)^p + a*e^(p*log(c*d*x + a*e) + p*log(x*e + d) + 1)/(x*
e + d)^p)/(c*d*p + c*d)

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Mupad [B]
time = 0.75, size = 56, normalized size = 1.08 \begin {gather*} \frac {\left (a\,e+c\,d\,x\right )\,{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^p}{c\,d\,\left (p+1\right )\,{\left (d+e\,x\right )}^p} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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