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

Optimal. Leaf size=86 \[ -\frac {2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h \sqrt {g+h x}}-\frac {4 b \sqrt {f} p q \tanh ^{-1}\left (\frac {\sqrt {f} \sqrt {g+h x}}{\sqrt {f g-e h}}\right )}{h \sqrt {f g-e h}} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2395, 63, 208, 2445} \[ -\frac {2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h \sqrt {g+h x}}-\frac {4 b \sqrt {f} p q \tanh ^{-1}\left (\frac {\sqrt {f} \sqrt {g+h x}}{\sqrt {f g-e h}}\right )}{h \sqrt {f g-e h}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.17, size = 84, normalized size = 0.98 \[ \frac {-\frac {2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{\sqrt {g+h x}}-\frac {4 b \sqrt {f} p q \tanh ^{-1}\left (\frac {\sqrt {f} \sqrt {g+h x}}{\sqrt {f g-e h}}\right )}{\sqrt {f g-e h}}}{h} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.55, size = 240, normalized size = 2.79 \[ \left [\frac {2 \, {\left ({\left (b h p q x + b g p q\right )} \sqrt {\frac {f}{f g - e h}} \log \left (\frac {f h x + 2 \, f g - e h - 2 \, {\left (f g - e h\right )} \sqrt {h x + g} \sqrt {\frac {f}{f g - e h}}}{f x + e}\right ) - {\left (b p q \log \left (f x + e\right ) + b q \log \relax (d) + b \log \relax (c) + a\right )} \sqrt {h x + g}\right )}}{h^{2} x + g h}, -\frac {2 \, {\left (2 \, {\left (b h p q x + b g p q\right )} \sqrt {-\frac {f}{f g - e h}} \arctan \left (-\frac {{\left (f g - e h\right )} \sqrt {h x + g} \sqrt {-\frac {f}{f g - e h}}}{f h x + f g}\right ) + {\left (b p q \log \left (f x + e\right ) + b q \log \relax (d) + b \log \relax (c) + a\right )} \sqrt {h x + g}\right )}}{h^{2} x + g h}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.22, size = 99, normalized size = 1.15 \[ \frac {4 \, b f p q \arctan \left (\frac {\sqrt {h x + g} f}{\sqrt {-f^{2} g + f h e}}\right )}{\sqrt {-f^{2} g + f h e} h} - \frac {2 \, {\left (b p q \log \left ({\left (h x + g\right )} f - f g + h e\right ) - b p q \log \relax (h) + b q \log \relax (d) + b \log \relax (c) + a\right )}}{\sqrt {h x + g} h} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.35, size = 0, normalized size = 0.00 \[ \int \frac {b \ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right )+a}{\left (h x +g \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d*(f*x+e)^p)^q))/(h*x+g)^(3/2),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?` for
more details)Is e*h-f*g positive or negative?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\ln \left (c\,{\left (d\,{\left (e+f\,x\right )}^p\right )}^q\right )}{{\left (g+h\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 21.45, size = 90, normalized size = 1.05 \[ \frac {- \frac {2 a}{\sqrt {g + h x}} + 2 b \left (\frac {2 p q \operatorname {atan}{\left (\frac {\sqrt {g + h x}}{\sqrt {\frac {h \left (e - \frac {f g}{h}\right )}{f}}} \right )}}{\sqrt {\frac {h \left (e - \frac {f g}{h}\right )}{f}}} - \frac {\log {\left (c \left (d \left (e - \frac {f g}{h} + \frac {f \left (g + h x\right )}{h}\right )^{p}\right )^{q} \right )}}{\sqrt {g + h x}}\right )}{h} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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