3.586 \(\int e^{\log ^2((d+e x)^n)} (d+e x)^m \, dx\)

Optimal. Leaf size=76 \[ \frac {\sqrt {\pi } e^{-\frac {(m+1)^2}{4 n^2}} (d+e x)^{m+1} \left ((d+e x)^n\right )^{-\frac {m+1}{n}} \text {erfi}\left (\frac {2 n \log \left ((d+e x)^n\right )+m+1}{2 n}\right )}{2 e n} \]

[Out]

1/2*(e*x+d)^(1+m)*erfi(1/2*(1+m+2*n*ln((e*x+d)^n))/n)*Pi^(1/2)/e/exp(1/4*(1+m)^2/n^2)/n/(((e*x+d)^n)^((1+m)/n)
)

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Rubi [A]  time = 0.16, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2276, 2234, 2204} \[ \frac {\sqrt {\pi } e^{-\frac {(m+1)^2}{4 n^2}} (d+e x)^{m+1} \left ((d+e x)^n\right )^{-\frac {m+1}{n}} \text {Erfi}\left (\frac {2 n \log \left ((d+e x)^n\right )+m+1}{2 n}\right )}{2 e n} \]

Antiderivative was successfully verified.

[In]

Int[E^Log[(d + e*x)^n]^2*(d + e*x)^m,x]

[Out]

(Sqrt[Pi]*(d + e*x)^(1 + m)*Erfi[(1 + m + 2*n*Log[(d + e*x)^n])/(2*n)])/(2*e*E^((1 + m)^2/(4*n^2))*n*((d + e*x
)^n)^((1 + m)/n))

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2234

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[F^(a - b^2/(4*c)), Int[F^((b + 2*c*x)^2/(4*c))
, x], x] /; FreeQ[{F, a, b, c}, x]

Rule 2276

Int[(F_)^(((a_.) + Log[(c_.)*(x_)^(n_.)]^2*(b_.))*(d_.))*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[(e*x)^(m + 1)/(
e*n*(c*x^n)^((m + 1)/n)), Subst[Int[E^(a*d*Log[F] + ((m + 1)*x)/n + b*d*Log[F]*x^2), x], x, Log[c*x^n]], x] /;
FreeQ[{F, a, b, c, d, e, m, n}, x]

Rubi steps

\begin {align*} \int e^{\log ^2\left ((d+e x)^n\right )} (d+e x)^m \, dx &=\frac {\operatorname {Subst}\left (\int e^{\log ^2\left (x^n\right )} x^m \, dx,x,d+e x\right )}{e}\\ &=\frac {\left ((d+e x)^{1+m} \left ((d+e x)^n\right )^{-\frac {1+m}{n}}\right ) \operatorname {Subst}\left (\int e^{\frac {(1+m) x}{n}+x^2} \, dx,x,\log \left ((d+e x)^n\right )\right )}{e n}\\ &=\frac {\left (e^{-\frac {(1+m)^2}{4 n^2}} (d+e x)^{1+m} \left ((d+e x)^n\right )^{-\frac {1+m}{n}}\right ) \operatorname {Subst}\left (\int e^{\frac {1}{4} \left (\frac {1+m}{n}+2 x\right )^2} \, dx,x,\log \left ((d+e x)^n\right )\right )}{e n}\\ &=\frac {e^{-\frac {(1+m)^2}{4 n^2}} \sqrt {\pi } (d+e x)^{1+m} \left ((d+e x)^n\right )^{-\frac {1+m}{n}} \text {erfi}\left (\frac {1+m+2 n \log \left ((d+e x)^n\right )}{2 n}\right )}{2 e n}\\ \end {align*}

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Mathematica [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int e^{\log ^2\left ((d+e x)^n\right )} (d+e x)^m \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[E^Log[(d + e*x)^n]^2*(d + e*x)^m,x]

[Out]

Integrate[E^Log[(d + e*x)^n]^2*(d + e*x)^m, x]

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fricas [A]  time = 0.43, size = 55, normalized size = 0.72 \[ \frac {\sqrt {\pi } \sqrt {n^{2}} \operatorname {erfi}\left (\frac {{\left (2 \, n^{2} \log \left (e x + d\right ) + m + 1\right )} \sqrt {n^{2}}}{2 \, n^{2}}\right ) e^{\left (-\frac {m^{2} + 2 \, m + 1}{4 \, n^{2}}\right )}}{2 \, e n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(log((e*x+d)^n)^2)*(e*x+d)^m,x, algorithm="fricas")

[Out]

1/2*sqrt(pi)*sqrt(n^2)*erfi(1/2*(2*n^2*log(e*x + d) + m + 1)*sqrt(n^2)/n^2)*e^(-1/4*(m^2 + 2*m + 1)/n^2)/(e*n)

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giac [A]  time = 0.27, size = 56, normalized size = 0.74 \[ -\frac {\sqrt {\pi } i \operatorname {erf}\left (i n \log \left (x e + d\right ) + \frac {i m}{2 \, n} + \frac {i}{2 \, n}\right ) e^{\left (-\frac {m^{2}}{4 \, n^{2}} - \frac {m}{2 \, n^{2}} - \frac {1}{4 \, n^{2}} - 1\right )}}{2 \, n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(log((e*x+d)^n)^2)*(e*x+d)^m,x, algorithm="giac")

[Out]

-1/2*sqrt(pi)*i*erf(i*n*log(x*e + d) + 1/2*i*m/n + 1/2*i/n)*e^(-1/4*m^2/n^2 - 1/2*m/n^2 - 1/4/n^2 - 1)/n

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maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[ \int \left (e x +d \right )^{m} {\mathrm e}^{\ln \left (\left (e x +d \right )^{n}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(ln((e*x+d)^n)^2)*(e*x+d)^m,x)

[Out]

int(exp(ln((e*x+d)^n)^2)*(e*x+d)^m,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x + d\right )}^{m} e^{\left (\log \left ({\left (e x + d\right )}^{n}\right )^{2}\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(log((e*x+d)^n)^2)*(e*x+d)^m,x, algorithm="maxima")

[Out]

integrate((e*x + d)^m*e^(log((e*x + d)^n)^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(log((d + e*x)^n)^2)*(d + e*x)^m,x)

[Out]

int(exp(log((d + e*x)^n)^2)*(d + e*x)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(ln((e*x+d)**n)**2)*(e*x+d)**m,x)

[Out]

Integral((d + e*x)**m*exp(log((d + e*x)**n)**2), x)

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