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

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

[Out]

-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

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Rubi [A]
time = 0.20, antiderivative size = 171, normalized size of antiderivative = 1.00, 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 = {2442, 52, 65, 214, 2495} \begin {gather*} \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 (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 \sqrt {g+h x} (f g-e h)^2}{5 f^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} \end {gather*}

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 52

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 65

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 214

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 2442

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 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 (g+h x)^{3/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right ) \, dx &=\text {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}-\text {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}-\text {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}-\text {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}-\text {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}-\text {Subst}\left (\frac {\left (4 b (f g-e h)^3 p q\right ) \text {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*}

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Mathematica [A]
time = 0.39, size = 163, normalized size = 0.95 \begin {gather*} \frac {2 \left (30 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 )+\sqrt {f} \sqrt {g+h x} \left (15 a f^2 (g+h x)^2-2 b p q \left (15 e^2 h^2-5 e f h (7 g+h x)+f^2 \left (23 g^2+11 g h x+3 h^2 x^2\right )\right )+15 b f^2 (g+h x)^2 \log \left (c \left (d (e+f x)^p\right )^q\right )\right )\right )}{75 f^{5/2} h} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(30*b*(f*g - e*h)^(5/2)*p*q*ArcTanh[(Sqrt[f]*Sqrt[g + h*x])/Sqrt[f*g - e*h]] + Sqrt[f]*Sqrt[g + h*x]*(15*a*
f^2*(g + h*x)^2 - 2*b*p*q*(15*e^2*h^2 - 5*e*f*h*(7*g + h*x) + f^2*(23*g^2 + 11*g*h*x + 3*h^2*x^2)) + 15*b*f^2*
(g + h*x)^2*Log[c*(d*(e + f*x)^p)^q])))/(75*f^(5/2)*h)

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Maple [F]
time = 0.19, size = 0, normalized size = 0.00 \[\int \left (h x +g \right )^{\frac {3}{2}} \left (a +b \ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right )\right )\, dx\]

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)

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

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(%e*h-f*g>0)', see `assume?` fo
r more detai

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 318 vs. \(2 (148) = 296\).
time = 0.42, size = 647, normalized size = 3.78 \begin {gather*} \left [\frac {2 \, {\left (15 \, {\left (b f^{2} g^{2} p q - 2 \, b f g h p q e + b h^{2} p q e^{2}\right )} \sqrt {\frac {f g - h e}{f}} \log \left (\frac {f h x + 2 \, f g + 2 \, \sqrt {h x + g} f \sqrt {\frac {f g - h e}{f}} - h e}{f x + e}\right ) - {\left (46 \, b f^{2} g^{2} p q + 30 \, b h^{2} p q e^{2} - 15 \, a f^{2} g^{2} + 3 \, {\left (2 \, b f^{2} h^{2} p q - 5 \, a f^{2} h^{2}\right )} x^{2} + 2 \, {\left (11 \, b f^{2} g h p q - 15 \, a f^{2} g h\right )} x - 10 \, {\left (b f h^{2} p q x + 7 \, b f g h p q\right )} e - 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} p q - 2 \, b f g h p q e + b h^{2} p q e^{2}\right )} \sqrt {-\frac {f g - h e}{f}} \arctan \left (-\frac {\sqrt {h x + g} f \sqrt {-\frac {f g - h e}{f}}}{f g - h e}\right ) - {\left (46 \, b f^{2} g^{2} p q + 30 \, b h^{2} p q e^{2} - 15 \, a f^{2} g^{2} + 3 \, {\left (2 \, b f^{2} h^{2} p q - 5 \, a f^{2} h^{2}\right )} x^{2} + 2 \, {\left (11 \, b f^{2} g h p q - 15 \, a f^{2} g h\right )} x - 10 \, {\left (b f h^{2} p q x + 7 \, b f g h p q\right )} e - 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 {gather*}

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*p*q - 2*b*f*g*h*p*q*e + b*h^2*p*q*e^2)*sqrt((f*g - h*e)/f)*log((f*h*x + 2*f*g + 2*sqrt(h*
x + g)*f*sqrt((f*g - h*e)/f) - h*e)/(f*x + e)) - (46*b*f^2*g^2*p*q + 30*b*h^2*p*q*e^2 - 15*a*f^2*g^2 + 3*(2*b*
f^2*h^2*p*q - 5*a*f^2*h^2)*x^2 + 2*(11*b*f^2*g*h*p*q - 15*a*f^2*g*h)*x - 10*(b*f*h^2*p*q*x + 7*b*f*g*h*p*q)*e
- 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), 2/7
5*(30*(b*f^2*g^2*p*q - 2*b*f*g*h*p*q*e + b*h^2*p*q*e^2)*sqrt(-(f*g - h*e)/f)*arctan(-sqrt(h*x + g)*f*sqrt(-(f*
g - h*e)/f)/(f*g - h*e)) - (46*b*f^2*g^2*p*q + 30*b*h^2*p*q*e^2 - 15*a*f^2*g^2 + 3*(2*b*f^2*h^2*p*q - 5*a*f^2*
h^2)*x^2 + 2*(11*b*f^2*g*h*p*q - 15*a*f^2*g*h)*x - 10*(b*f*h^2*p*q*x + 7*b*f*g*h*p*q)*e - 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)]

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Sympy [A]
time = 58.76, size = 484, normalized size = 2.83 \begin {gather*} a g \left (\begin {cases} \sqrt {g} x & \text {for}\: h = 0 \\\frac {2 \left (g + h x\right )^{\frac {3}{2}}}{3 h} & \text {otherwise} \end {cases}\right ) + \frac {2 a \left (- \frac {g \left (g + h x\right )^{\frac {3}{2}}}{3} + \frac {\left (g + h x\right )^{\frac {5}{2}}}{5}\right )}{h} + \frac {2 b g \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 )}{h} + \frac {2 b \left (- \frac {2 f p q \left (\frac {h \left (g + h x\right )^{\frac {5}{2}}}{5 f} + \frac {\left (g + h x\right )^{\frac {3}{2}} \left (- e h^{2} + f g h\right )}{3 f^{2}} + \frac {\sqrt {g + h x} \left (e^{2} h^{3} - 2 e f g h^{2} + f^{2} g^{2} h\right )}{f^{3}} - \frac {h \left (e h - f g\right )^{3} \operatorname {atan}{\left (\frac {\sqrt {g + h x}}{\sqrt {\frac {e h - f g}{f}}} \right )}}{f^{4} \sqrt {\frac {e h - f g}{f}}}\right )}{5 h} - g \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 ) + \frac {\left (g + h x\right )^{\frac {5}{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 )}}{5}\right )}{h} \end {gather*}

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]

a*g*Piecewise((sqrt(g)*x, Eq(h, 0)), (2*(g + h*x)**(3/2)/(3*h), True)) + 2*a*(-g*(g + h*x)**(3/2)/3 + (g + h*x
)**(5/2)/5)/h + 2*b*g*(-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 + 2*b*(-2*f*p*q*(h*(g + h*x)**(5/2)/(5*f) + (g + h*x)**(3/2)*(-e*h**2 +
f*g*h)/(3*f**2) + sqrt(g + h*x)*(e**2*h**3 - 2*e*f*g*h**2 + f**2*g**2*h)/f**3 - h*(e*h - f*g)**3*atan(sqrt(g
+ h*x)/sqrt((e*h - f*g)/f))/(f**4*sqrt((e*h - f*g)/f)))/(5*h) - g*(-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) + (g + h*x)**(5/2)*log(c*(d*(
e - f*g/h + f*(g + h*x)/h)**p)**q)/5)/h

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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((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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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