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

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

[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))

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Rubi [A]  time = 0.159532, antiderivative size = 120, normalized size of antiderivative = 1., 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 = {2395, 51, 63, 208, 2445} \[ -\frac{2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{3 h (g+h x)^{3/2}}-\frac{4 b f^{3/2} p q \tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{g+h x}}{\sqrt{f g-e h}}\right )}{3 h (f g-e h)^{3/2}}+\frac{4 b f p q}{3 h \sqrt{g+h x} (f g-e h)} \]

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 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 51

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 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 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)^{5/2}} \, dx &=\operatorname{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}}+\operatorname{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}}+\operatorname{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}}+\operatorname{Subst}\left (\frac{\left (4 b f^2 p q\right ) \operatorname{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*}

Mathematica [C]  time = 0.0988886, size = 91, normalized size = 0.76 \[ \frac{2 (f g-e h) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )-4 b f p q (g+h x) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{f (g+h x)}{f g-e h}\right )}{3 h (g+h x)^{3/2} (e h-f g)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*b*f*p*q*(g + h*x)*Hypergeometric2F1[-1/2, 1, 1/2, (f*(g + h*x))/(f*g - e*h)] + 2*(f*g - e*h)*(a + b*Log[c*
(d*(e + f*x)^p)^q]))/(3*h*(-(f*g) + e*h)*(g + h*x)^(3/2))

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Maple [F]  time = 0.677, size = 0, normalized size = 0. \begin{align*} \int{(a+b\ln \left ( c \left ( d \left ( fx+e \right ) ^{p} \right ) ^{q} \right ) ) \left ( hx+g \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [B]  time = 2.68841, size = 1034, normalized size = 8.62 \begin{align*} \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 - 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 (2 \, b f h p q x + 2 \, b f g p q -{\left (b f g - b e h\right )} p q \log \left (f x + e\right ) - a f g + a e h -{\left (b f g - b e h\right )} q \log \left (d\right ) -{\left (b f g - b e h\right )} \log \left (c\right )\right )} \sqrt{h x + g}\right )}}{3 \,{\left (f g^{3} h - e g^{2} h^{2} +{\left (f g h^{3} - e h^{4}\right )} x^{2} + 2 \,{\left (f g^{2} h^{2} - e g h^{3}\right )} x\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 - 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 (2 \, b f h p q x + 2 \, b f g p q -{\left (b f g - b e h\right )} p q \log \left (f x + e\right ) - a f g + a e h -{\left (b f g - b e h\right )} q \log \left (d\right ) -{\left (b f g - b e h\right )} \log \left (c\right )\right )} \sqrt{h x + g}\right )}}{3 \,{\left (f g^{3} h - e g^{2} h^{2} +{\left (f g h^{3} - e h^{4}\right )} x^{2} + 2 \,{\left (f g^{2} h^{2} - e g h^{3}\right )} x\right )}}\right ] \end{align*}

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [B]  time = 1.34433, size = 286, normalized size = 2.38 \begin{align*} \frac{4 \, b f^{2} h p q \arctan \left (\frac{\sqrt{h x + g} f}{\sqrt{-f^{2} g + f h e}}\right )}{3 \,{\left (f g h^{2} - h^{3} e\right )} \sqrt{-f^{2} g + f h e}} - \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{align*}

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*h*p*q*arctan(sqrt(h*x + g)*f/sqrt(-f^2*g + f*h*e))/((f*g*h^2 - h^3*e)*sqrt(-f^2*g + f*h*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)