3.5.86 \(\int \frac {a+b \log (c (d (e+f x)^p)^q)}{(g+h x)^{5/2}} \, dx\) [486]

Optimal. Leaf size=120 \[ \frac {4 b f p q}{3 h (f g-e h) \sqrt {g+h x}}-\frac {4 b f^{3/2} p q \tanh ^{-1}\left (\frac {\sqrt {f} \sqrt {g+h x}}{\sqrt {f g-e h}}\right )}{3 h (f g-e h)^{3/2}}-\frac {2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{3 h (g+h x)^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {2442, 53, 65, 214, 2495} \begin {gather*} -\frac {2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{3 h (g+h x)^{3/2}}-\frac {4 b f^{3/2} p q \tanh ^{-1}\left (\frac {\sqrt {f} \sqrt {g+h x}}{\sqrt {f g-e h}}\right )}{3 h (f g-e h)^{3/2}}+\frac {4 b f p q}{3 h \sqrt {g+h x} (f g-e h)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d*(e + f*x)^p)^q])/(g + h*x)^(5/2),x]

[Out]

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

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 \frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{(g+h x)^{5/2}} \, dx &=\text {Subst}\left (\int \frac {a+b \log \left (c d^q (e+f x)^{p q}\right )}{(g+h x)^{5/2}} \, 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 )}{3 h (g+h x)^{3/2}}+\text {Subst}\left (\frac {(2 b f p q) \int \frac {1}{(e+f x) (g+h x)^{3/2}} \, dx}{3 h},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac {4 b f p q}{3 h (f g-e h) \sqrt {g+h x}}-\frac {2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{3 h (g+h x)^{3/2}}+\text {Subst}\left (\frac {\left (2 b f^2 p q\right ) \int \frac {1}{(e+f x) \sqrt {g+h x}} \, dx}{3 h (f g-e h)},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac {4 b f p q}{3 h (f g-e h) \sqrt {g+h x}}-\frac {2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{3 h (g+h x)^{3/2}}+\text {Subst}\left (\frac {\left (4 b f^2 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 )}{3 h^2 (f g-e h)},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac {4 b f p q}{3 h (f g-e h) \sqrt {g+h x}}-\frac {4 b f^{3/2} p q \tanh ^{-1}\left (\frac {\sqrt {f} \sqrt {g+h x}}{\sqrt {f g-e h}}\right )}{3 h (f g-e h)^{3/2}}-\frac {2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{3 h (g+h x)^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.32, size = 124, normalized size = 1.03 \begin {gather*} \frac {2 \left (-\frac {2 b f^{3/2} p q \tanh ^{-1}\left (\frac {\sqrt {f} \sqrt {g+h x}}{\sqrt {f g-e h}}\right )}{(f g-e h)^{3/2}}+\frac {a (-f g+e h)+2 b f p q (g+h x)+b (-f g+e h) \log \left (c \left (d (e+f x)^p\right )^q\right )}{(f g-e h) (g+h x)^{3/2}}\right )}{3 h} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(d*(e + f*x)^p)^q])/(g + h*x)^(5/2),x]

[Out]

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

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Maple [F]
time = 0.20, size = 0, normalized size = 0.00 \[\int \frac {a +b \ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right )}{\left (h x +g \right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*(d*(f*x+e)^p)^q))/(h*x+g)^(5/2),x)

[Out]

int((a+b*ln(c*(d*(f*x+e)^p)^q))/(h*x+g)^(5/2),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((a+b*log(c*(d*(f*x+e)^p)^q))/(h*x+g)^(5/2),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?` fo
r more detai

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 240 vs. \(2 (104) = 208\).
time = 0.43, size = 491, normalized size = 4.09 \begin {gather*} \left [-\frac {2 \, {\left ({\left (b f h^{2} p q x^{2} + 2 \, b f g h p q x + b f g^{2} p q\right )} \sqrt {\frac {f}{f g - h e}} \log \left (\frac {f h x + 2 \, f g + 2 \, {\left (f g - h e\right )} \sqrt {h x + g} \sqrt {\frac {f}{f g - h e}} - h e}{f x + e}\right ) - {\left (2 \, b f h p q x + 2 \, b f g p q - a f g + a h e - {\left (b f g p q - b h p q e\right )} \log \left (f x + e\right ) - {\left (b f g - b h e\right )} \log \left (c\right ) - {\left (b f g q - b h q e\right )} \log \left (d\right )\right )} \sqrt {h x + g}\right )}}{3 \, {\left (f g h^{3} x^{2} + 2 \, f g^{2} h^{2} x + f g^{3} h - {\left (h^{4} x^{2} + 2 \, g h^{3} x + g^{2} h^{2}\right )} e\right )}}, -\frac {2 \, {\left (2 \, {\left (b f h^{2} p q x^{2} + 2 \, b f g h p q x + b f g^{2} p q\right )} \sqrt {-\frac {f}{f g - h e}} \arctan \left (-\frac {{\left (f g - h e\right )} \sqrt {h x + g} \sqrt {-\frac {f}{f g - h e}}}{f h x + f g}\right ) - {\left (2 \, b f h p q x + 2 \, b f g p q - a f g + a h e - {\left (b f g p q - b h p q e\right )} \log \left (f x + e\right ) - {\left (b f g - b h e\right )} \log \left (c\right ) - {\left (b f g q - b h q e\right )} \log \left (d\right )\right )} \sqrt {h x + g}\right )}}{3 \, {\left (f g h^{3} x^{2} + 2 \, f g^{2} h^{2} x + f g^{3} h - {\left (h^{4} x^{2} + 2 \, g h^{3} x + g^{2} h^{2}\right )} e\right )}}\right ] \end {gather*}

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)^(5/2),x, algorithm="fricas")

[Out]

[-2/3*((b*f*h^2*p*q*x^2 + 2*b*f*g*h*p*q*x + b*f*g^2*p*q)*sqrt(f/(f*g - h*e))*log((f*h*x + 2*f*g + 2*(f*g - h*e
)*sqrt(h*x + g)*sqrt(f/(f*g - h*e)) - h*e)/(f*x + e)) - (2*b*f*h*p*q*x + 2*b*f*g*p*q - a*f*g + a*h*e - (b*f*g*
p*q - b*h*p*q*e)*log(f*x + e) - (b*f*g - b*h*e)*log(c) - (b*f*g*q - b*h*q*e)*log(d))*sqrt(h*x + g))/(f*g*h^3*x
^2 + 2*f*g^2*h^2*x + f*g^3*h - (h^4*x^2 + 2*g*h^3*x + g^2*h^2)*e), -2/3*(2*(b*f*h^2*p*q*x^2 + 2*b*f*g*h*p*q*x
+ b*f*g^2*p*q)*sqrt(-f/(f*g - h*e))*arctan(-(f*g - h*e)*sqrt(h*x + g)*sqrt(-f/(f*g - h*e))/(f*h*x + f*g)) - (2
*b*f*h*p*q*x + 2*b*f*g*p*q - a*f*g + a*h*e - (b*f*g*p*q - b*h*p*q*e)*log(f*x + e) - (b*f*g - b*h*e)*log(c) - (
b*f*g*q - b*h*q*e)*log(d))*sqrt(h*x + g))/(f*g*h^3*x^2 + 2*f*g^2*h^2*x + f*g^3*h - (h^4*x^2 + 2*g*h^3*x + g^2*
h^2)*e)]

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Sympy [A]
time = 95.07, size = 122, normalized size = 1.02 \begin {gather*} \frac {- \frac {2 a}{3 \left (g + h x\right )^{\frac {3}{2}}} + 2 b \left (\frac {2 f p q \left (- \frac {h}{\sqrt {g + h x} \left (e h - f g\right )} - \frac {h \operatorname {atan}{\left (\frac {\sqrt {g + h x}}{\sqrt {\frac {e h - f g}{f}}} \right )}}{\sqrt {\frac {e h - f g}{f}} \left (e h - f g\right )}\right )}{3 h} - \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 )}}{3 \left (g + h x\right )^{\frac {3}{2}}}\right )}{h} \end {gather*}

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)**(5/2),x)

[Out]

(-2*a/(3*(g + h*x)**(3/2)) + 2*b*(2*f*p*q*(-h/(sqrt(g + h*x)*(e*h - f*g)) - h*atan(sqrt(g + h*x)/sqrt((e*h - f
*g)/f))/(sqrt((e*h - f*g)/f)*(e*h - f*g)))/(3*h) - log(c*(d*(e - f*g/h + f*(g + h*x)/h)**p)**q)/(3*(g + h*x)**
(3/2))))/h

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (104) = 208\).
time = 2.73, size = 209, normalized size = 1.74 \begin {gather*} \frac {4 \, b f^{2} p q \arctan \left (\frac {\sqrt {h x + g} f}{\sqrt {-f^{2} g + f h e}}\right )}{3 \, \sqrt {-f^{2} g + f h e} {\left (f g h - h^{2} e\right )}} - \frac {2 \, {\left (b f g p q \log \left ({\left (h x + g\right )} f - f g + h e\right ) - b h p q e \log \left ({\left (h x + g\right )} f - f g + h e\right ) - b f g p q \log \left (h\right ) + b h p q e \log \left (h\right ) - 2 \, {\left (h x + g\right )} b f p q + b f g q \log \left (d\right ) - b h q e \log \left (d\right ) + b f g \log \left (c\right ) - b h e \log \left (c\right ) + a f g - a h e\right )}}{3 \, {\left ({\left (h x + g\right )}^{\frac {3}{2}} f g h - {\left (h x + g\right )}^{\frac {3}{2}} h^{2} e\right )}} \end {gather*}

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)^(5/2),x, algorithm="giac")

[Out]

4/3*b*f^2*p*q*arctan(sqrt(h*x + g)*f/sqrt(-f^2*g + f*h*e))/(sqrt(-f^2*g + f*h*e)*(f*g*h - h^2*e)) - 2/3*(b*f*g
*p*q*log((h*x + g)*f - f*g + h*e) - b*h*p*q*e*log((h*x + g)*f - f*g + h*e) - b*f*g*p*q*log(h) + b*h*p*q*e*log(
h) - 2*(h*x + g)*b*f*p*q + b*f*g*q*log(d) - b*h*q*e*log(d) + b*f*g*log(c) - b*h*e*log(c) + a*f*g - a*h*e)/((h*
x + g)^(3/2)*f*g*h - (h*x + g)^(3/2)*h^2*e)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+b\,\ln \left (c\,{\left (d\,{\left (e+f\,x\right )}^p\right )}^q\right )}{{\left (g+h\,x\right )}^{5/2}} \,d x \end {gather*}

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)^(5/2),x)

[Out]

int((a + b*log(c*(d*(e + f*x)^p)^q))/(g + h*x)^(5/2), x)

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