3.5.82 \(\int (g+h x)^{3/2} (a+b \log (c (d (e+f x)^p)^q)) \, dx\) [482]

3.5.82.1 Optimal result

Integrand size = 28, antiderivative size = 171 \[ \int (g+h x)^{3/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right ) \, dx=-\frac {4 b (f g-e h)^2 p q \sqrt {g+h x}}{5 f^2 h}-\frac {4 b (f g-e h) p q (g+h x)^{3/2}}{15 f h}-\frac {4 b p q (g+h x)^{5/2}}{25 h}+\frac {4 b (f g-e h)^{5/2} p q \text {arctanh}\left (\frac {\sqrt {f} \sqrt {g+h x}}{\sqrt {f g-e h}}\right )}{5 f^{5/2} h}+\frac {2 (g+h x)^{5/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h} \]

output

-4/15*b*(-e*h+f*g)*p*q*(h*x+g)^(3/2)/f/h-4/25*b*p*q*(h*x+g)^(5/2)/h+4/5*b* 
(-e*h+f*g)^(5/2)*p*q*arctanh(f^(1/2)*(h*x+g)^(1/2)/(-e*h+f*g)^(1/2))/f^(5/ 
2)/h+2/5*(h*x+g)^(5/2)*(a+b*ln(c*(d*(f*x+e)^p)^q))/h-4/5*b*(-e*h+f*g)^2*p* 
q*(h*x+g)^(1/2)/f^2/h
 

3.5.82.2 Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.89 \[ \int (g+h x)^{3/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right ) \, dx=\frac {2 \left (\frac {1}{5} a (g+h x)^{5/2}-\frac {2}{75} b p q \left (3 (g+h x)^{5/2}+\frac {5 (f g-e h) \left (\sqrt {f} \sqrt {g+h x} (4 f g-3 e h+f h x)-3 (f g-e h)^{3/2} \text {arctanh}\left (\frac {\sqrt {f} \sqrt {g+h x}}{\sqrt {f g-e h}}\right )\right )}{f^{5/2}}\right )+\frac {1}{5} b (g+h x)^{5/2} \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h} \]

input

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

output

(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)*ArcTan 
h[(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
 

3.5.82.3 Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2895, 2842, 60, 60, 60, 73, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 2895

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

\(\Big \downarrow \) 2842

\(\displaystyle \frac {2 (g+h x)^{5/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h}-\frac {2 b f p q \int \frac {(g+h x)^{5/2}}{e+f x}dx}{5 h}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {2 (g+h x)^{5/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h}-\frac {2 b f p q \left (\frac {(f g-e h) \int \frac {(g+h x)^{3/2}}{e+f x}dx}{f}+\frac {2 (g+h x)^{5/2}}{5 f}\right )}{5 h}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {2 (g+h x)^{5/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h}-\frac {2 b f p q \left (\frac {(f g-e h) \left (\frac {(f g-e h) \int \frac {\sqrt {g+h x}}{e+f x}dx}{f}+\frac {2 (g+h x)^{3/2}}{3 f}\right )}{f}+\frac {2 (g+h x)^{5/2}}{5 f}\right )}{5 h}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {2 (g+h x)^{5/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h}-\frac {2 b f p q \left (\frac {(f g-e h) \left (\frac {(f g-e h) \left (\frac {(f g-e h) \int \frac {1}{(e+f x) \sqrt {g+h x}}dx}{f}+\frac {2 \sqrt {g+h x}}{f}\right )}{f}+\frac {2 (g+h x)^{3/2}}{3 f}\right )}{f}+\frac {2 (g+h x)^{5/2}}{5 f}\right )}{5 h}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 (g+h x)^{5/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h}-\frac {2 b f p q \left (\frac {(f g-e h) \left (\frac {(f g-e h) \left (\frac {2 (f g-e h) \int \frac {1}{e+\frac {f (g+h x)}{h}-\frac {f g}{h}}d\sqrt {g+h x}}{f h}+\frac {2 \sqrt {g+h x}}{f}\right )}{f}+\frac {2 (g+h x)^{3/2}}{3 f}\right )}{f}+\frac {2 (g+h x)^{5/2}}{5 f}\right )}{5 h}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 (g+h x)^{5/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h}-\frac {2 b f p q \left (\frac {(f g-e h) \left (\frac {(f g-e h) \left (\frac {2 \sqrt {g+h x}}{f}-\frac {2 \sqrt {f g-e h} \text {arctanh}\left (\frac {\sqrt {f} \sqrt {g+h x}}{\sqrt {f g-e h}}\right )}{f^{3/2}}\right )}{f}+\frac {2 (g+h x)^{3/2}}{3 f}\right )}{f}+\frac {2 (g+h x)^{5/2}}{5 f}\right )}{5 h}\)

input

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

output

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

3.5.82.3.1 Defintions of rubi rules used

rule 60

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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

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

rule 2842

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] - Simp[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 2895

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]]
 

3.5.82.4 Maple [F]

\[\int \left (h x +g \right )^{\frac {3}{2}} \left (a +b \ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right )\right )d x\]

input

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

output

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

3.5.82.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 307 vs. \(2 (143) = 286\).

Time = 0.34 (sec) , antiderivative size = 624, normalized size of antiderivative = 3.65 \[ \int (g+h x)^{3/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right ) \, dx=\left [\frac {2 \, {\left (15 \, {\left (b f^{2} g^{2} - 2 \, b e f g h + b e^{2} h^{2}\right )} p q \sqrt {\frac {f g - e h}{f}} \log \left (\frac {f h x + 2 \, f g - e h + 2 \, \sqrt {h x + g} f \sqrt {\frac {f g - e h}{f}}}{f x + e}\right ) + {\left (15 \, a f^{2} g^{2} - 2 \, {\left (23 \, b f^{2} g^{2} - 35 \, b e f g h + 15 \, b e^{2} h^{2}\right )} p q - 3 \, {\left (2 \, b f^{2} h^{2} p q - 5 \, a f^{2} h^{2}\right )} x^{2} + 2 \, {\left (15 \, a f^{2} g h - {\left (11 \, b f^{2} g h - 5 \, b e f h^{2}\right )} p q\right )} x + 15 \, {\left (b f^{2} h^{2} p q x^{2} + 2 \, b f^{2} g h p q x + b f^{2} g^{2} p q\right )} \log \left (f x + e\right ) + 15 \, {\left (b f^{2} h^{2} x^{2} + 2 \, b f^{2} g h x + b f^{2} g^{2}\right )} \log \left (c\right ) + 15 \, {\left (b f^{2} h^{2} q x^{2} + 2 \, b f^{2} g h q x + b f^{2} g^{2} q\right )} \log \left (d\right )\right )} \sqrt {h x + g}\right )}}{75 \, f^{2} h}, \frac {2 \, {\left (30 \, {\left (b f^{2} g^{2} - 2 \, b e f g h + b e^{2} h^{2}\right )} p q \sqrt {-\frac {f g - e h}{f}} \arctan \left (-\frac {\sqrt {h x + g} f \sqrt {-\frac {f g - e h}{f}}}{f g - e h}\right ) + {\left (15 \, a f^{2} g^{2} - 2 \, {\left (23 \, b f^{2} g^{2} - 35 \, b e f g h + 15 \, b e^{2} h^{2}\right )} p q - 3 \, {\left (2 \, b f^{2} h^{2} p q - 5 \, a f^{2} h^{2}\right )} x^{2} + 2 \, {\left (15 \, a f^{2} g h - {\left (11 \, b f^{2} g h - 5 \, b e f h^{2}\right )} p q\right )} x + 15 \, {\left (b f^{2} h^{2} p q x^{2} + 2 \, b f^{2} g h p q x + b f^{2} g^{2} p q\right )} \log \left (f x + e\right ) + 15 \, {\left (b f^{2} h^{2} x^{2} + 2 \, b f^{2} g h x + b f^{2} g^{2}\right )} \log \left (c\right ) + 15 \, {\left (b f^{2} h^{2} q x^{2} + 2 \, b f^{2} g h q x + b f^{2} g^{2} q\right )} \log \left (d\right )\right )} \sqrt {h x + g}\right )}}{75 \, f^{2} h}\right ] \]

input

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

output

[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)*lo 
g((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)*l 
og(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)]
 

3.5.82.6 Sympy [F]

\[ \int (g+h x)^{3/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right ) \, dx=\int \left (a + b \log {\left (c \left (d \left (e + f x\right )^{p}\right )^{q} \right )}\right ) \left (g + h x\right )^{\frac {3}{2}}\, dx \]

input

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

output

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

3.5.82.7 Maxima [F(-2)]

Exception generated. \[ \int (g+h x)^{3/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right ) \, dx=\text {Exception raised: ValueError} \]

input

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

output

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(e*h-f*g>0)', see `assume?` for m 
ore detail
 

3.5.82.8 Giac [F]

\[ \int (g+h x)^{3/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right ) \, dx=\int { {\left (h x + g\right )}^{\frac {3}{2}} {\left (b \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) + a\right )} \,d x } \]

input

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

output

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

3.5.82.9 Mupad [F(-1)]

Timed out. \[ \int (g+h x)^{3/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right ) \, dx=\int {\left (g+h\,x\right )}^{3/2}\,\left (a+b\,\ln \left (c\,{\left (d\,{\left (e+f\,x\right )}^p\right )}^q\right )\right ) \,d x \]

input

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

output

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