3.2.38 \(\int (f+g x)^{3/2} (a+b \log (c (d+e x)^n)) \, dx\) [138]

Optimal. Leaf size=163 \[ -\frac {4 b (e f-d g)^2 n \sqrt {f+g x}}{5 e^2 g}-\frac {4 b (e f-d g) n (f+g x)^{3/2}}{15 e g}-\frac {4 b n (f+g x)^{5/2}}{25 g}+\frac {4 b (e f-d g)^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{5 e^{5/2} g}+\frac {2 (f+g x)^{5/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2442, 52, 65, 214} \begin {gather*} \frac {2 (f+g x)^{5/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g}+\frac {4 b n (e f-d g)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{5 e^{5/2} g}-\frac {4 b n \sqrt {f+g x} (e f-d g)^2}{5 e^2 g}-\frac {4 b n (f+g x)^{3/2} (e f-d g)}{15 e g}-\frac {4 b n (f+g x)^{5/2}}{25 g} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rubi steps

\begin {align*} \int (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx &=\frac {2 (f+g x)^{5/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g}-\frac {(2 b e n) \int \frac {(f+g x)^{5/2}}{d+e x} \, dx}{5 g}\\ &=-\frac {4 b n (f+g x)^{5/2}}{25 g}+\frac {2 (f+g x)^{5/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g}-\frac {(2 b (e f-d g) n) \int \frac {(f+g x)^{3/2}}{d+e x} \, dx}{5 g}\\ &=-\frac {4 b (e f-d g) n (f+g x)^{3/2}}{15 e g}-\frac {4 b n (f+g x)^{5/2}}{25 g}+\frac {2 (f+g x)^{5/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g}-\frac {\left (2 b (e f-d g)^2 n\right ) \int \frac {\sqrt {f+g x}}{d+e x} \, dx}{5 e g}\\ &=-\frac {4 b (e f-d g)^2 n \sqrt {f+g x}}{5 e^2 g}-\frac {4 b (e f-d g) n (f+g x)^{3/2}}{15 e g}-\frac {4 b n (f+g x)^{5/2}}{25 g}+\frac {2 (f+g x)^{5/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g}-\frac {\left (2 b (e f-d g)^3 n\right ) \int \frac {1}{(d+e x) \sqrt {f+g x}} \, dx}{5 e^2 g}\\ &=-\frac {4 b (e f-d g)^2 n \sqrt {f+g x}}{5 e^2 g}-\frac {4 b (e f-d g) n (f+g x)^{3/2}}{15 e g}-\frac {4 b n (f+g x)^{5/2}}{25 g}+\frac {2 (f+g x)^{5/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g}-\frac {\left (4 b (e f-d g)^3 n\right ) \text {Subst}\left (\int \frac {1}{d-\frac {e f}{g}+\frac {e x^2}{g}} \, dx,x,\sqrt {f+g x}\right )}{5 e^2 g^2}\\ &=-\frac {4 b (e f-d g)^2 n \sqrt {f+g x}}{5 e^2 g}-\frac {4 b (e f-d g) n (f+g x)^{3/2}}{15 e g}-\frac {4 b n (f+g x)^{5/2}}{25 g}+\frac {4 b (e f-d g)^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{5 e^{5/2} g}+\frac {2 (f+g x)^{5/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g}\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 137, normalized size = 0.84 \begin {gather*} \frac {2 \left (-\frac {2}{5} b n (f+g x)^{5/2}-\frac {2 b (e f-d g) n \left (\sqrt {e} \sqrt {f+g x} (4 e f-3 d g+e g x)-3 (e f-d g)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )\right )}{3 e^{5/2}}+(f+g x)^{5/2} \left (a+b \log \left (c (d+e x)^n\right )\right )\right )}{5 g} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.14, size = 0, normalized size = 0.00 \[\int \left (g x +f \right )^{\frac {3}{2}} \left (a +b \ln \left (c \left (e x +d \right )^{n}\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((g*x+f)^(3/2)*(a+b*ln(c*(e*x+d)^n)),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((g*x+f)^(3/2)*(a+b*log(c*(e*x+d)^n)),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(4*%e^2*f-4*%e*d*g>0)', see `as
sume?` for m

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Fricas [A]
time = 0.43, size = 479, normalized size = 2.94 \begin {gather*} \left [\frac {2 \, {\left (15 \, {\left (b d^{2} g^{2} n - 2 \, b d f g n e + b f^{2} n e^{2}\right )} \sqrt {-{\left (d g - f e\right )} e^{\left (-1\right )}} \log \left (-\frac {d g - {\left (g x + 2 \, f\right )} e - 2 \, \sqrt {g x + f} \sqrt {-{\left (d g - f e\right )} e^{\left (-1\right )}} e}{x e + d}\right ) - {\left (30 \, b d^{2} g^{2} n - 15 \, {\left (b g^{2} n x^{2} + 2 \, b f g n x + b f^{2} n\right )} e^{2} \log \left (x e + d\right ) - 15 \, {\left (b g^{2} x^{2} + 2 \, b f g x + b f^{2}\right )} e^{2} \log \left (c\right ) + {\left (46 \, b f^{2} n - 15 \, a f^{2} + 3 \, {\left (2 \, b g^{2} n - 5 \, a g^{2}\right )} x^{2} + 2 \, {\left (11 \, b f g n - 15 \, a f g\right )} x\right )} e^{2} - 10 \, {\left (b d g^{2} n x + 7 \, b d f g n\right )} e\right )} \sqrt {g x + f}\right )} e^{\left (-2\right )}}{75 \, g}, -\frac {2 \, {\left (30 \, {\left (b d^{2} g^{2} n - 2 \, b d f g n e + b f^{2} n e^{2}\right )} \sqrt {d g - f e} \arctan \left (-\frac {\sqrt {g x + f} e^{\frac {1}{2}}}{\sqrt {d g - f e}}\right ) e^{\left (-\frac {1}{2}\right )} + {\left (30 \, b d^{2} g^{2} n - 15 \, {\left (b g^{2} n x^{2} + 2 \, b f g n x + b f^{2} n\right )} e^{2} \log \left (x e + d\right ) - 15 \, {\left (b g^{2} x^{2} + 2 \, b f g x + b f^{2}\right )} e^{2} \log \left (c\right ) + {\left (46 \, b f^{2} n - 15 \, a f^{2} + 3 \, {\left (2 \, b g^{2} n - 5 \, a g^{2}\right )} x^{2} + 2 \, {\left (11 \, b f g n - 15 \, a f g\right )} x\right )} e^{2} - 10 \, {\left (b d g^{2} n x + 7 \, b d f g n\right )} e\right )} \sqrt {g x + f}\right )} e^{\left (-2\right )}}{75 \, g}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2/75*(15*(b*d^2*g^2*n - 2*b*d*f*g*n*e + b*f^2*n*e^2)*sqrt(-(d*g - f*e)*e^(-1))*log(-(d*g - (g*x + 2*f)*e - 2*
sqrt(g*x + f)*sqrt(-(d*g - f*e)*e^(-1))*e)/(x*e + d)) - (30*b*d^2*g^2*n - 15*(b*g^2*n*x^2 + 2*b*f*g*n*x + b*f^
2*n)*e^2*log(x*e + d) - 15*(b*g^2*x^2 + 2*b*f*g*x + b*f^2)*e^2*log(c) + (46*b*f^2*n - 15*a*f^2 + 3*(2*b*g^2*n
- 5*a*g^2)*x^2 + 2*(11*b*f*g*n - 15*a*f*g)*x)*e^2 - 10*(b*d*g^2*n*x + 7*b*d*f*g*n)*e)*sqrt(g*x + f))*e^(-2)/g,
-2/75*(30*(b*d^2*g^2*n - 2*b*d*f*g*n*e + b*f^2*n*e^2)*sqrt(d*g - f*e)*arctan(-sqrt(g*x + f)*e^(1/2)/sqrt(d*g
- f*e))*e^(-1/2) + (30*b*d^2*g^2*n - 15*(b*g^2*n*x^2 + 2*b*f*g*n*x + b*f^2*n)*e^2*log(x*e + d) - 15*(b*g^2*x^2
+ 2*b*f*g*x + b*f^2)*e^2*log(c) + (46*b*f^2*n - 15*a*f^2 + 3*(2*b*g^2*n - 5*a*g^2)*x^2 + 2*(11*b*f*g*n - 15*a
*f*g)*x)*e^2 - 10*(b*d*g^2*n*x + 7*b*d*f*g*n)*e)*sqrt(g*x + f))*e^(-2)/g]

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Sympy [A]
time = 30.76, size = 469, normalized size = 2.88 \begin {gather*} a f \left (\begin {cases} \sqrt {f} x & \text {for}\: g = 0 \\\frac {2 \left (f + g x\right )^{\frac {3}{2}}}{3 g} & \text {otherwise} \end {cases}\right ) + \frac {2 a \left (- \frac {f \left (f + g x\right )^{\frac {3}{2}}}{3} + \frac {\left (f + g x\right )^{\frac {5}{2}}}{5}\right )}{g} + \frac {2 b f \left (- \frac {2 e n \left (\frac {g \left (f + g x\right )^{\frac {3}{2}}}{3 e} + \frac {\sqrt {f + g x} \left (- d g^{2} + e f g\right )}{e^{2}} + \frac {g \left (d g - e f\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {f + g x}}{\sqrt {\frac {d g - e f}{e}}} \right )}}{e^{3} \sqrt {\frac {d g - e f}{e}}}\right )}{3 g} + \frac {\left (f + g x\right )^{\frac {3}{2}} \log {\left (c \left (d - \frac {e f}{g} + \frac {e \left (f + g x\right )}{g}\right )^{n} \right )}}{3}\right )}{g} + \frac {2 b \left (- \frac {2 e n \left (\frac {g \left (f + g x\right )^{\frac {5}{2}}}{5 e} + \frac {\left (f + g x\right )^{\frac {3}{2}} \left (- d g^{2} + e f g\right )}{3 e^{2}} + \frac {\sqrt {f + g x} \left (d^{2} g^{3} - 2 d e f g^{2} + e^{2} f^{2} g\right )}{e^{3}} - \frac {g \left (d g - e f\right )^{3} \operatorname {atan}{\left (\frac {\sqrt {f + g x}}{\sqrt {\frac {d g - e f}{e}}} \right )}}{e^{4} \sqrt {\frac {d g - e f}{e}}}\right )}{5 g} - f \left (- \frac {2 e n \left (\frac {g \left (f + g x\right )^{\frac {3}{2}}}{3 e} + \frac {\sqrt {f + g x} \left (- d g^{2} + e f g\right )}{e^{2}} + \frac {g \left (d g - e f\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {f + g x}}{\sqrt {\frac {d g - e f}{e}}} \right )}}{e^{3} \sqrt {\frac {d g - e f}{e}}}\right )}{3 g} + \frac {\left (f + g x\right )^{\frac {3}{2}} \log {\left (c \left (d - \frac {e f}{g} + \frac {e \left (f + g x\right )}{g}\right )^{n} \right )}}{3}\right ) + \frac {\left (f + g x\right )^{\frac {5}{2}} \log {\left (c \left (d - \frac {e f}{g} + \frac {e \left (f + g x\right )}{g}\right )^{n} \right )}}{5}\right )}{g} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (f+g\,x\right )}^{3/2}\,\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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