3.5.62 \(\int \sqrt {a+b \log (c (d (e+f x)^p)^q)} \, dx\) [462]

Optimal. Leaf size=139 \[ -\frac {\sqrt {b} e^{-\frac {a}{b p q}} \sqrt {p} \sqrt {\pi } \sqrt {q} (e+f x) \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac {1}{p q}} \text {erfi}\left (\frac {\sqrt {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}}{\sqrt {b} \sqrt {p} \sqrt {q}}\right )}{2 f}+\frac {(e+f x) \sqrt {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}}{f} \]

[Out]

-1/2*(f*x+e)*erfi((a+b*ln(c*(d*(f*x+e)^p)^q))^(1/2)/b^(1/2)/p^(1/2)/q^(1/2))*b^(1/2)*p^(1/2)*Pi^(1/2)*q^(1/2)/
exp(a/b/p/q)/f/((c*(d*(f*x+e)^p)^q)^(1/p/q))+(f*x+e)*(a+b*ln(c*(d*(f*x+e)^p)^q))^(1/2)/f

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Rubi [A]
time = 0.16, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2436, 2333, 2337, 2211, 2235, 2495} \begin {gather*} \frac {(e+f x) \sqrt {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}}{f}-\frac {\sqrt {\pi } \sqrt {b} \sqrt {p} \sqrt {q} (e+f x) e^{-\frac {a}{b p q}} \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac {1}{p q}} \text {Erfi}\left (\frac {\sqrt {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}}{\sqrt {b} \sqrt {p} \sqrt {q}}\right )}{2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Log[c*(d*(e + f*x)^p)^q]],x]

[Out]

-1/2*(Sqrt[b]*Sqrt[p]*Sqrt[Pi]*Sqrt[q]*(e + f*x)*Erfi[Sqrt[a + b*Log[c*(d*(e + f*x)^p)^q]]/(Sqrt[b]*Sqrt[p]*Sq
rt[q])])/(E^(a/(b*p*q))*f*(c*(d*(e + f*x)^p)^q)^(1/(p*q))) + ((e + f*x)*Sqrt[a + b*Log[c*(d*(e + f*x)^p)^q]])/
f

Rule 2211

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[F^(g*(e - c*(
f/d)) + f*g*(x^2/d)), x], x, Sqrt[c + d*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2235

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 2333

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2337

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[E^(x/n)*(a +
b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2436

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

Rule 2495

Int[((a_.) + Log[(c_.)*((d_.)*((e_.) + (f_.)*(x_))^(m_.))^(n_)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Subst[Int[u*(
a + b*Log[c*d^n*(e + f*x)^(m*n)])^p, x], c*d^n*(e + f*x)^(m*n), c*(d*(e + f*x)^m)^n] /; FreeQ[{a, b, c, d, e,
f, m, n, p}, x] &&  !IntegerQ[n] &&  !(EqQ[d, 1] && EqQ[m, 1]) && IntegralFreeQ[IntHide[u*(a + b*Log[c*d^n*(e
+ f*x)^(m*n)])^p, x]]

Rubi steps

\begin {align*} \int \sqrt {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )} \, dx &=\text {Subst}\left (\int \sqrt {a+b \log \left (c d^q (e+f x)^{p q}\right )} \, dx,c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\text {Subst}\left (\frac {\text {Subst}\left (\int \sqrt {a+b \log \left (c d^q x^{p q}\right )} \, dx,x,e+f x\right )}{f},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac {(e+f x) \sqrt {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}}{f}-\text {Subst}\left (\frac {(b p q) \text {Subst}\left (\int \frac {1}{\sqrt {a+b \log \left (c d^q x^{p q}\right )}} \, dx,x,e+f x\right )}{2 f},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac {(e+f x) \sqrt {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}}{f}-\text {Subst}\left (\frac {\left (b (e+f x) \left (c d^q (e+f x)^{p q}\right )^{-\frac {1}{p q}}\right ) \text {Subst}\left (\int \frac {e^{\frac {x}{p q}}}{\sqrt {a+b x}} \, dx,x,\log \left (c d^q (e+f x)^{p q}\right )\right )}{2 f},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac {(e+f x) \sqrt {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}}{f}-\text {Subst}\left (\frac {\left ((e+f x) \left (c d^q (e+f x)^{p q}\right )^{-\frac {1}{p q}}\right ) \text {Subst}\left (\int e^{-\frac {a}{b p q}+\frac {x^2}{b p q}} \, dx,x,\sqrt {a+b \log \left (c d^q (e+f x)^{p q}\right )}\right )}{f},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=-\frac {\sqrt {b} e^{-\frac {a}{b p q}} \sqrt {p} \sqrt {\pi } \sqrt {q} (e+f x) \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac {1}{p q}} \text {erfi}\left (\frac {\sqrt {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}}{\sqrt {b} \sqrt {p} \sqrt {q}}\right )}{2 f}+\frac {(e+f x) \sqrt {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}}{f}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 134, normalized size = 0.96 \begin {gather*} \frac {(e+f x) \left (-\sqrt {b} e^{-\frac {a}{b p q}} \sqrt {p} \sqrt {\pi } \sqrt {q} \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac {1}{p q}} \text {erfi}\left (\frac {\sqrt {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}}{\sqrt {b} \sqrt {p} \sqrt {q}}\right )+2 \sqrt {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}\right )}{2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*Log[c*(d*(e + f*x)^p)^q]],x]

[Out]

((e + f*x)*(-((Sqrt[b]*Sqrt[p]*Sqrt[Pi]*Sqrt[q]*Erfi[Sqrt[a + b*Log[c*(d*(e + f*x)^p)^q]]/(Sqrt[b]*Sqrt[p]*Sqr
t[q])])/(E^(a/(b*p*q))*(c*(d*(e + f*x)^p)^q)^(1/(p*q)))) + 2*Sqrt[a + b*Log[c*(d*(e + f*x)^p)^q]]))/(2*f)

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \sqrt {a +b \ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*(d*(f*x+e)^p)^q))^(1/2),x)

[Out]

int((a+b*ln(c*(d*(f*x+e)^p)^q))^(1/2),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((a+b*log(c*(d*(f*x+e)^p)^q))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*log(((f*x + e)^p*d)^q*c) + a), x)

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d*(f*x+e)^p)^q))^(1/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a + b \log {\left (c \left (d \left (e + f x\right )^{p}\right )^{q} \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*(d*(f*x+e)**p)**q))**(1/2),x)

[Out]

Integral(sqrt(a + b*log(c*(d*(e + f*x)**p)**q)), x)

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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((a+b*log(c*(d*(f*x+e)^p)^q))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*log(((f*x + e)^p*d)^q*c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*log(c*(d*(e + f*x)^p)^q))^(1/2),x)

[Out]

int((a + b*log(c*(d*(e + f*x)^p)^q))^(1/2), x)

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