3.483 \(\int \sqrt{g+h x} (a+b \log (c (d (e+f x)^p)^q)) \, dx\)

Optimal. Leaf size=139 \[ \frac{2 (g+h x)^{3/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{3 h}+\frac{4 b p q (f g-e h)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{g+h x}}{\sqrt{f g-e h}}\right )}{3 f^{3/2} h}-\frac{4 b p q \sqrt{g+h x} (f g-e h)}{3 f h}-\frac{4 b p q (g+h x)^{3/2}}{9 h} \]

[Out]

(-4*b*(f*g - e*h)*p*q*Sqrt[g + h*x])/(3*f*h) - (4*b*p*q*(g + h*x)^(3/2))/(9*h) + (4*b*(f*g - e*h)^(3/2)*p*q*Ar
cTanh[(Sqrt[f]*Sqrt[g + h*x])/Sqrt[f*g - e*h]])/(3*f^(3/2)*h) + (2*(g + h*x)^(3/2)*(a + b*Log[c*(d*(e + f*x)^p
)^q]))/(3*h)

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Rubi [A]  time = 0.182752, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {2395, 50, 63, 208, 2445} \[ \frac{2 (g+h x)^{3/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{3 h}+\frac{4 b p q (f g-e h)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{g+h x}}{\sqrt{f g-e h}}\right )}{3 f^{3/2} h}-\frac{4 b p q \sqrt{g+h x} (f g-e h)}{3 f h}-\frac{4 b p q (g+h x)^{3/2}}{9 h} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*b*(f*g - e*h)*p*q*Sqrt[g + h*x])/(3*f*h) - (4*b*p*q*(g + h*x)^(3/2))/(9*h) + (4*b*(f*g - e*h)^(3/2)*p*q*Ar
cTanh[(Sqrt[f]*Sqrt[g + h*x])/Sqrt[f*g - e*h]])/(3*f^(3/2)*h) + (2*(g + h*x)^(3/2)*(a + b*Log[c*(d*(e + f*x)^p
)^q]))/(3*h)

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 2445

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{g+h x} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right ) \, dx &=\operatorname{Subst}\left (\int \sqrt{g+h x} \left (a+b \log \left (c d^q (e+f x)^{p q}\right )\right ) \, dx,c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac{2 (g+h x)^{3/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{3 h}-\operatorname{Subst}\left (\frac{(2 b f p q) \int \frac{(g+h x)^{3/2}}{e+f x} \, dx}{3 h},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=-\frac{4 b p q (g+h x)^{3/2}}{9 h}+\frac{2 (g+h x)^{3/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{3 h}-\operatorname{Subst}\left (\frac{(2 b (f g-e h) p q) \int \frac{\sqrt{g+h x}}{e+f x} \, dx}{3 h},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=-\frac{4 b (f g-e h) p q \sqrt{g+h x}}{3 f h}-\frac{4 b p q (g+h x)^{3/2}}{9 h}+\frac{2 (g+h x)^{3/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{3 h}-\operatorname{Subst}\left (\frac{\left (2 b (f g-e h)^2 p q\right ) \int \frac{1}{(e+f x) \sqrt{g+h x}} \, dx}{3 f h},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=-\frac{4 b (f g-e h) p q \sqrt{g+h x}}{3 f h}-\frac{4 b p q (g+h x)^{3/2}}{9 h}+\frac{2 (g+h x)^{3/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{3 h}-\operatorname{Subst}\left (\frac{\left (4 b (f g-e h)^2 p q\right ) \operatorname{Subst}\left (\int \frac{1}{e-\frac{f g}{h}+\frac{f x^2}{h}} \, dx,x,\sqrt{g+h x}\right )}{3 f h^2},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=-\frac{4 b (f g-e h) p q \sqrt{g+h x}}{3 f h}-\frac{4 b p q (g+h x)^{3/2}}{9 h}+\frac{4 b (f g-e h)^{3/2} p q \tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{g+h x}}{\sqrt{f g-e h}}\right )}{3 f^{3/2} h}+\frac{2 (g+h x)^{3/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{3 h}\\ \end{align*}

Mathematica [A]  time = 0.183829, size = 124, normalized size = 0.89 \[ \frac{2 \left (\sqrt{f} \sqrt{g+h x} \left (3 a f (g+h x)+3 b f (g+h x) \log \left (c \left (d (e+f x)^p\right )^q\right )-2 b p q (-3 e h+4 f g+f h x)\right )+6 b p q (f g-e h)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{g+h x}}{\sqrt{f g-e h}}\right )\right )}{9 f^{3/2} h} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(6*b*(f*g - e*h)^(3/2)*p*q*ArcTanh[(Sqrt[f]*Sqrt[g + h*x])/Sqrt[f*g - e*h]] + Sqrt[f]*Sqrt[g + h*x]*(3*a*f*
(g + h*x) - 2*b*p*q*(4*f*g - 3*e*h + f*h*x) + 3*b*f*(g + h*x)*Log[c*(d*(e + f*x)^p)^q])))/(9*f^(3/2)*h)

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Maple [F]  time = 0.688, size = 0, normalized size = 0. \begin{align*} \int \sqrt{hx+g} \left ( a+b\ln \left ( c \left ( d \left ( fx+e \right ) ^{p} \right ) ^{q} \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.64266, size = 844, normalized size = 6.07 \begin{align*} \left [-\frac{2 \,{\left (3 \,{\left (b f g - b e h\right )} p q \sqrt{\frac{f g - e h}{f}} \log \left (\frac{f h x + 2 \, f g - e h - 2 \, \sqrt{h x + g} f \sqrt{\frac{f g - e h}{f}}}{f x + e}\right ) -{\left (3 \, a f g - 2 \,{\left (4 \, b f g - 3 \, b e h\right )} p q -{\left (2 \, b f h p q - 3 \, a f h\right )} x + 3 \,{\left (b f h p q x + b f g p q\right )} \log \left (f x + e\right ) + 3 \,{\left (b f h x + b f g\right )} \log \left (c\right ) + 3 \,{\left (b f h q x + b f g q\right )} \log \left (d\right )\right )} \sqrt{h x + g}\right )}}{9 \, f h}, \frac{2 \,{\left (6 \,{\left (b f g - b e h\right )} p q \sqrt{-\frac{f g - e h}{f}} \arctan \left (-\frac{\sqrt{h x + g} f \sqrt{-\frac{f g - e h}{f}}}{f g - e h}\right ) +{\left (3 \, a f g - 2 \,{\left (4 \, b f g - 3 \, b e h\right )} p q -{\left (2 \, b f h p q - 3 \, a f h\right )} x + 3 \,{\left (b f h p q x + b f g p q\right )} \log \left (f x + e\right ) + 3 \,{\left (b f h x + b f g\right )} \log \left (c\right ) + 3 \,{\left (b f h q x + b f g q\right )} \log \left (d\right )\right )} \sqrt{h x + g}\right )}}{9 \, f h}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-2/9*(3*(b*f*g - b*e*h)*p*q*sqrt((f*g - e*h)/f)*log((f*h*x + 2*f*g - e*h - 2*sqrt(h*x + g)*f*sqrt((f*g - e*h)
/f))/(f*x + e)) - (3*a*f*g - 2*(4*b*f*g - 3*b*e*h)*p*q - (2*b*f*h*p*q - 3*a*f*h)*x + 3*(b*f*h*p*q*x + b*f*g*p*
q)*log(f*x + e) + 3*(b*f*h*x + b*f*g)*log(c) + 3*(b*f*h*q*x + b*f*g*q)*log(d))*sqrt(h*x + g))/(f*h), 2/9*(6*(b
*f*g - b*e*h)*p*q*sqrt(-(f*g - e*h)/f)*arctan(-sqrt(h*x + g)*f*sqrt(-(f*g - e*h)/f)/(f*g - e*h)) + (3*a*f*g -
2*(4*b*f*g - 3*b*e*h)*p*q - (2*b*f*h*p*q - 3*a*f*h)*x + 3*(b*f*h*p*q*x + b*f*g*p*q)*log(f*x + e) + 3*(b*f*h*x
+ b*f*g)*log(c) + 3*(b*f*h*q*x + b*f*g*q)*log(d))*sqrt(h*x + g))/(f*h)]

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Sympy [A]  time = 7.49032, size = 144, normalized size = 1.04 \begin{align*} \frac{2 \left (\frac{a \left (g + h x\right )^{\frac{3}{2}}}{3} + b \left (- \frac{2 f p q \left (\frac{h \left (g + h x\right )^{\frac{3}{2}}}{3 f} + \frac{\sqrt{g + h x} \left (- e h^{2} + f g h\right )}{f^{2}} + \frac{h \left (e h - f g\right )^{2} \operatorname{atan}{\left (\frac{\sqrt{g + h x}}{\sqrt{\frac{e h - f g}{f}}} \right )}}{f^{3} \sqrt{\frac{e h - f g}{f}}}\right )}{3 h} + \frac{\left (g + h x\right )^{\frac{3}{2}} \log{\left (c \left (d \left (e - \frac{f g}{h} + \frac{f \left (g + h x\right )}{h}\right )^{p}\right )^{q} \right )}}{3}\right )\right )}{h} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(a*(g + h*x)**(3/2)/3 + b*(-2*f*p*q*(h*(g + h*x)**(3/2)/(3*f) + sqrt(g + h*x)*(-e*h**2 + f*g*h)/f**2 + h*(e*
h - f*g)**2*atan(sqrt(g + h*x)/sqrt((e*h - f*g)/f))/(f**3*sqrt((e*h - f*g)/f)))/(3*h) + (g + h*x)**(3/2)*log(c
*(d*(e - f*g/h + f*(g + h*x)/h)**p)**q)/3))/h

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{h x + g}{\left (b \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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