3.482 \(\int (g+h x)^{3/2} (a+b \log (c (d (e+f x)^p)^q)) \, dx\)

Optimal. Leaf size=171 \[ \frac{2 (g+h x)^{5/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h}-\frac{4 b p q \sqrt{g+h x} (f g-e h)^2}{5 f^2 h}+\frac{4 b p q (f g-e h)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{g+h x}}{\sqrt{f g-e h}}\right )}{5 f^{5/2} h}-\frac{4 b p q (g+h x)^{3/2} (f g-e h)}{15 f h}-\frac{4 b p q (g+h x)^{5/2}}{25 h} \]

[Out]

(-4*b*(f*g - e*h)^2*p*q*Sqrt[g + h*x])/(5*f^2*h) - (4*b*(f*g - e*h)*p*q*(g + h*x)^(3/2))/(15*f*h) - (4*b*p*q*(
g + h*x)^(5/2))/(25*h) + (4*b*(f*g - e*h)^(5/2)*p*q*ArcTanh[(Sqrt[f]*Sqrt[g + h*x])/Sqrt[f*g - e*h]])/(5*f^(5/
2)*h) + (2*(g + h*x)^(5/2)*(a + b*Log[c*(d*(e + f*x)^p)^q]))/(5*h)

________________________________________________________________________________________

Rubi [A]  time = 0.33455, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{2 (g+h x)^{5/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h}-\frac{4 b p q \sqrt{g+h x} (f g-e h)^2}{5 f^2 h}+\frac{4 b p q (f g-e h)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{g+h x}}{\sqrt{f g-e h}}\right )}{5 f^{5/2} h}-\frac{4 b p q (g+h x)^{3/2} (f g-e h)}{15 f h}-\frac{4 b p q (g+h x)^{5/2}}{25 h} \]

Antiderivative was successfully verified.

[In]

Int[(g + h*x)^(3/2)*(a + b*Log[c*(d*(e + f*x)^p)^q]),x]

[Out]

(-4*b*(f*g - e*h)^2*p*q*Sqrt[g + h*x])/(5*f^2*h) - (4*b*(f*g - e*h)*p*q*(g + h*x)^(3/2))/(15*f*h) - (4*b*p*q*(
g + h*x)^(5/2))/(25*h) + (4*b*(f*g - e*h)^(5/2)*p*q*ArcTanh[(Sqrt[f]*Sqrt[g + h*x])/Sqrt[f*g - e*h]])/(5*f^(5/
2)*h) + (2*(g + h*x)^(5/2)*(a + b*Log[c*(d*(e + f*x)^p)^q]))/(5*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 (g+h x)^{3/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right ) \, dx &=\operatorname{Subst}\left (\int (g+h x)^{3/2} \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)^{5/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h}-\operatorname{Subst}\left (\frac{(2 b f p q) \int \frac{(g+h x)^{5/2}}{e+f x} \, dx}{5 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)^{5/2}}{25 h}+\frac{2 (g+h x)^{5/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h}-\operatorname{Subst}\left (\frac{(2 b (f g-e h) p q) \int \frac{(g+h x)^{3/2}}{e+f x} \, dx}{5 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 (g+h x)^{3/2}}{15 f h}-\frac{4 b p q (g+h x)^{5/2}}{25 h}+\frac{2 (g+h x)^{5/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h}-\operatorname{Subst}\left (\frac{\left (2 b (f g-e h)^2 p q\right ) \int \frac{\sqrt{g+h x}}{e+f x} \, dx}{5 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)^2 p q \sqrt{g+h x}}{5 f^2 h}-\frac{4 b (f g-e h) p q (g+h x)^{3/2}}{15 f h}-\frac{4 b p q (g+h x)^{5/2}}{25 h}+\frac{2 (g+h x)^{5/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h}-\operatorname{Subst}\left (\frac{\left (2 b (f g-e h)^3 p q\right ) \int \frac{1}{(e+f x) \sqrt{g+h x}} \, dx}{5 f^2 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)^2 p q \sqrt{g+h x}}{5 f^2 h}-\frac{4 b (f g-e h) p q (g+h x)^{3/2}}{15 f h}-\frac{4 b p q (g+h x)^{5/2}}{25 h}+\frac{2 (g+h x)^{5/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h}-\operatorname{Subst}\left (\frac{\left (4 b (f g-e h)^3 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 )}{5 f^2 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)^2 p q \sqrt{g+h x}}{5 f^2 h}-\frac{4 b (f g-e h) p q (g+h x)^{3/2}}{15 f h}-\frac{4 b p q (g+h x)^{5/2}}{25 h}+\frac{4 b (f g-e h)^{5/2} p q \tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{g+h x}}{\sqrt{f g-e h}}\right )}{5 f^{5/2} h}+\frac{2 (g+h x)^{5/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h}\\ \end{align*}

Mathematica [A]  time = 0.37766, size = 153, normalized size = 0.89 \[ \frac{2 \left (\frac{1}{5} a (g+h x)^{5/2}+\frac{1}{5} b (g+h x)^{5/2} \log \left (c \left (d (e+f x)^p\right )^q\right )-\frac{2}{75} b p q \left (\frac{5 (f g-e h) \left (\sqrt{f} \sqrt{g+h x} (-3 e h+4 f g+f h x)-3 (f g-e h)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{g+h x}}{\sqrt{f g-e h}}\right )\right )}{f^{5/2}}+3 (g+h x)^{5/2}\right )\right )}{h} \]

Antiderivative was successfully verified.

[In]

Integrate[(g + h*x)^(3/2)*(a + b*Log[c*(d*(e + f*x)^p)^q]),x]

[Out]

(2*((a*(g + h*x)^(5/2))/5 - (2*b*p*q*(3*(g + h*x)^(5/2) + (5*(f*g - e*h)*(Sqrt[f]*Sqrt[g + h*x]*(4*f*g - 3*e*h
+ f*h*x) - 3*(f*g - e*h)^(3/2)*ArcTanh[(Sqrt[f]*Sqrt[g + h*x])/Sqrt[f*g - e*h]]))/f^(5/2)))/75 + (b*(g + h*x)
^(5/2)*Log[c*(d*(e + f*x)^p)^q])/5))/h

________________________________________________________________________________________

Maple [F]  time = 0.956, size = 0, normalized size = 0. \begin{align*} \int \left ( hx+g \right ) ^{{\frac{3}{2}}} \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)^(3/2)*(a+b*ln(c*(d*(f*x+e)^p)^q)),x)

[Out]

int((h*x+g)^(3/2)*(a+b*ln(c*(d*(f*x+e)^p)^q)),x)

________________________________________________________________________________________

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)^(3/2)*(a+b*log(c*(d*(f*x+e)^p)^q)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.76127, size = 1412, normalized size = 8.26 \begin{align*} \left [\frac{2 \,{\left (15 \,{\left (b f^{2} g^{2} - 2 \, b e f g h + b e^{2} h^{2}\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 (15 \, a f^{2} g^{2} - 2 \,{\left (23 \, b f^{2} g^{2} - 35 \, b e f g h + 15 \, b e^{2} h^{2}\right )} p q - 3 \,{\left (2 \, b f^{2} h^{2} p q - 5 \, a f^{2} h^{2}\right )} x^{2} + 2 \,{\left (15 \, a f^{2} g h -{\left (11 \, b f^{2} g h - 5 \, b e f h^{2}\right )} p q\right )} x + 15 \,{\left (b f^{2} h^{2} p q x^{2} + 2 \, b f^{2} g h p q x + b f^{2} g^{2} p q\right )} \log \left (f x + e\right ) + 15 \,{\left (b f^{2} h^{2} x^{2} + 2 \, b f^{2} g h x + b f^{2} g^{2}\right )} \log \left (c\right ) + 15 \,{\left (b f^{2} h^{2} q x^{2} + 2 \, b f^{2} g h q x + b f^{2} g^{2} q\right )} \log \left (d\right )\right )} \sqrt{h x + g}\right )}}{75 \, f^{2} h}, \frac{2 \,{\left (30 \,{\left (b f^{2} g^{2} - 2 \, b e f g h + b e^{2} h^{2}\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 (15 \, a f^{2} g^{2} - 2 \,{\left (23 \, b f^{2} g^{2} - 35 \, b e f g h + 15 \, b e^{2} h^{2}\right )} p q - 3 \,{\left (2 \, b f^{2} h^{2} p q - 5 \, a f^{2} h^{2}\right )} x^{2} + 2 \,{\left (15 \, a f^{2} g h -{\left (11 \, b f^{2} g h - 5 \, b e f h^{2}\right )} p q\right )} x + 15 \,{\left (b f^{2} h^{2} p q x^{2} + 2 \, b f^{2} g h p q x + b f^{2} g^{2} p q\right )} \log \left (f x + e\right ) + 15 \,{\left (b f^{2} h^{2} x^{2} + 2 \, b f^{2} g h x + b f^{2} g^{2}\right )} \log \left (c\right ) + 15 \,{\left (b f^{2} h^{2} q x^{2} + 2 \, b f^{2} g h q x + b f^{2} g^{2} q\right )} \log \left (d\right )\right )} \sqrt{h x + g}\right )}}{75 \, f^{2} h}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x+g)^(3/2)*(a+b*log(c*(d*(f*x+e)^p)^q)),x, algorithm="fricas")

[Out]

[2/75*(15*(b*f^2*g^2 - 2*b*e*f*g*h + b*e^2*h^2)*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)) + (15*a*f^2*g^2 - 2*(23*b*f^2*g^2 - 35*b*e*f*g*h + 15*b*e^2*h^2)*p*q -
3*(2*b*f^2*h^2*p*q - 5*a*f^2*h^2)*x^2 + 2*(15*a*f^2*g*h - (11*b*f^2*g*h - 5*b*e*f*h^2)*p*q)*x + 15*(b*f^2*h^2*
p*q*x^2 + 2*b*f^2*g*h*p*q*x + b*f^2*g^2*p*q)*log(f*x + e) + 15*(b*f^2*h^2*x^2 + 2*b*f^2*g*h*x + b*f^2*g^2)*log
(c) + 15*(b*f^2*h^2*q*x^2 + 2*b*f^2*g*h*q*x + b*f^2*g^2*q)*log(d))*sqrt(h*x + g))/(f^2*h), 2/75*(30*(b*f^2*g^2
- 2*b*e*f*g*h + b*e^2*h^2)*p*q*sqrt(-(f*g - e*h)/f)*arctan(-sqrt(h*x + g)*f*sqrt(-(f*g - e*h)/f)/(f*g - e*h))
+ (15*a*f^2*g^2 - 2*(23*b*f^2*g^2 - 35*b*e*f*g*h + 15*b*e^2*h^2)*p*q - 3*(2*b*f^2*h^2*p*q - 5*a*f^2*h^2)*x^2
+ 2*(15*a*f^2*g*h - (11*b*f^2*g*h - 5*b*e*f*h^2)*p*q)*x + 15*(b*f^2*h^2*p*q*x^2 + 2*b*f^2*g*h*p*q*x + b*f^2*g^
2*p*q)*log(f*x + e) + 15*(b*f^2*h^2*x^2 + 2*b*f^2*g*h*x + b*f^2*g^2)*log(c) + 15*(b*f^2*h^2*q*x^2 + 2*b*f^2*g*
h*q*x + b*f^2*g^2*q)*log(d))*sqrt(h*x + g))/(f^2*h)]

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x+g)**(3/2)*(a+b*ln(c*(d*(f*x+e)**p)**q)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (h x + g\right )}^{\frac{3}{2}}{\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)^(3/2)*(a+b*log(c*(d*(f*x+e)^p)^q)),x, algorithm="giac")

[Out]

integrate((h*x + g)^(3/2)*(b*log(((f*x + e)^p*d)^q*c) + a), x)