3.487 \(\int \frac{a+b \log (c (d (e+f x)^p)^q)}{(g+h x)^{7/2}} \, dx\)

Optimal. Leaf size=152 \[ -\frac{2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h (g+h x)^{5/2}}+\frac{4 b f^2 p q}{5 h \sqrt{g+h x} (f g-e h)^2}-\frac{4 b f^{5/2} p q \tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{g+h x}}{\sqrt{f g-e h}}\right )}{5 h (f g-e h)^{5/2}}+\frac{4 b f p q}{15 h (g+h x)^{3/2} (f g-e h)} \]

[Out]

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

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Rubi [A]  time = 0.199417, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h (g+h x)^{5/2}}+\frac{4 b f^2 p q}{5 h \sqrt{g+h x} (f g-e h)^2}-\frac{4 b f^{5/2} p q \tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{g+h x}}{\sqrt{f g-e h}}\right )}{5 h (f g-e h)^{5/2}}+\frac{4 b f p q}{15 h (g+h x)^{3/2} (f g-e h)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d*(e + f*x)^p)^q])/(g + h*x)^(7/2),x]

[Out]

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

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)^{7/2}} \, dx &=\operatorname{Subst}\left (\int \frac{a+b \log \left (c d^q (e+f x)^{p q}\right )}{(g+h x)^{7/2}} \, 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 )}{5 h (g+h x)^{5/2}}+\operatorname{Subst}\left (\frac{(2 b f p q) \int \frac{1}{(e+f x) (g+h x)^{5/2}} \, 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 p q}{15 h (f g-e h) (g+h x)^{3/2}}-\frac{2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h (g+h x)^{5/2}}+\operatorname{Subst}\left (\frac{\left (2 b f^2 p q\right ) \int \frac{1}{(e+f x) (g+h x)^{3/2}} \, dx}{5 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}{15 h (f g-e h) (g+h x)^{3/2}}+\frac{4 b f^2 p q}{5 h (f g-e h)^2 \sqrt{g+h x}}-\frac{2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h (g+h x)^{5/2}}+\operatorname{Subst}\left (\frac{\left (2 b f^3 p q\right ) \int \frac{1}{(e+f x) \sqrt{g+h x}} \, dx}{5 h (f g-e h)^2},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac{4 b f p q}{15 h (f g-e h) (g+h x)^{3/2}}+\frac{4 b f^2 p q}{5 h (f g-e h)^2 \sqrt{g+h x}}-\frac{2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h (g+h x)^{5/2}}+\operatorname{Subst}\left (\frac{\left (4 b f^3 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 )}{5 h^2 (f g-e h)^2},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac{4 b f p q}{15 h (f g-e h) (g+h x)^{3/2}}+\frac{4 b f^2 p q}{5 h (f g-e h)^2 \sqrt{g+h x}}-\frac{4 b f^{5/2} p q \tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{g+h x}}{\sqrt{f g-e h}}\right )}{5 h (f g-e h)^{5/2}}-\frac{2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h (g+h x)^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0870892, size = 91, normalized size = 0.6 \[ \frac{6 (f g-e h) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )-4 b f p q (g+h x) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{f (g+h x)}{f g-e h}\right )}{15 h (g+h x)^{5/2} (e h-f g)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(d*(e + f*x)^p)^q])/(g + h*x)^(7/2),x]

[Out]

(-4*b*f*p*q*(g + h*x)*Hypergeometric2F1[-3/2, 1, -1/2, (f*(g + h*x))/(f*g - e*h)] + 6*(f*g - e*h)*(a + b*Log[c
*(d*(e + f*x)^p)^q]))/(15*h*(-(f*g) + e*h)*(g + h*x)^(5/2))

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Maple [F]  time = 0.711, 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{7}{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)^(7/2),x)

[Out]

int((a+b*ln(c*(d*(f*x+e)^p)^q))/(h*x+g)^(7/2),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)^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.85264, size = 1859, normalized size = 12.23 \begin{align*} \text{result too large to display} \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)^(7/2),x, algorithm="fricas")

[Out]

[2/15*(3*(b*f^2*h^3*p*q*x^3 + 3*b*f^2*g*h^2*p*q*x^2 + 3*b*f^2*g^2*h*p*q*x + b*f^2*g^3*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)) + (6*b*f^2*h^2*p*q*x^2
- 3*a*f^2*g^2 + 6*a*e*f*g*h - 3*a*e^2*h^2 + 2*(7*b*f^2*g*h - b*e*f*h^2)*p*q*x - 3*(b*f^2*g^2 - 2*b*e*f*g*h +
b*e^2*h^2)*p*q*log(f*x + e) + 2*(4*b*f^2*g^2 - b*e*f*g*h)*p*q - 3*(b*f^2*g^2 - 2*b*e*f*g*h + b*e^2*h^2)*q*log(
d) - 3*(b*f^2*g^2 - 2*b*e*f*g*h + b*e^2*h^2)*log(c))*sqrt(h*x + g))/(f^2*g^5*h - 2*e*f*g^4*h^2 + e^2*g^3*h^3 +
(f^2*g^2*h^4 - 2*e*f*g*h^5 + e^2*h^6)*x^3 + 3*(f^2*g^3*h^3 - 2*e*f*g^2*h^4 + e^2*g*h^5)*x^2 + 3*(f^2*g^4*h^2
- 2*e*f*g^3*h^3 + e^2*g^2*h^4)*x), -2/15*(6*(b*f^2*h^3*p*q*x^3 + 3*b*f^2*g*h^2*p*q*x^2 + 3*b*f^2*g^2*h*p*q*x +
b*f^2*g^3*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)) - (
6*b*f^2*h^2*p*q*x^2 - 3*a*f^2*g^2 + 6*a*e*f*g*h - 3*a*e^2*h^2 + 2*(7*b*f^2*g*h - b*e*f*h^2)*p*q*x - 3*(b*f^2*g
^2 - 2*b*e*f*g*h + b*e^2*h^2)*p*q*log(f*x + e) + 2*(4*b*f^2*g^2 - b*e*f*g*h)*p*q - 3*(b*f^2*g^2 - 2*b*e*f*g*h
+ b*e^2*h^2)*q*log(d) - 3*(b*f^2*g^2 - 2*b*e*f*g*h + b*e^2*h^2)*log(c))*sqrt(h*x + g))/(f^2*g^5*h - 2*e*f*g^4*
h^2 + e^2*g^3*h^3 + (f^2*g^2*h^4 - 2*e*f*g*h^5 + e^2*h^6)*x^3 + 3*(f^2*g^3*h^3 - 2*e*f*g^2*h^4 + e^2*g*h^5)*x^
2 + 3*(f^2*g^4*h^2 - 2*e*f*g^3*h^3 + e^2*g^2*h^4)*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)**(7/2),x)

[Out]

Timed out

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Giac [B]  time = 1.42114, size = 510, normalized size = 3.36 \begin{align*} \frac{4 \, b f^{3} h p q \arctan \left (\frac{\sqrt{h x + g} f}{\sqrt{-f^{2} g + f h e}}\right )}{5 \,{\left (f^{2} g^{2} h^{2} - 2 \, f g h^{3} e + h^{4} e^{2}\right )} \sqrt{-f^{2} g + f h e}} - \frac{2 \,{\left (3 \, b f^{2} g^{2} p q \log \left ({\left (h x + g\right )} f - f g + h e\right ) - 6 \, b f g h p q e \log \left ({\left (h x + g\right )} f - f g + h e\right ) - 3 \, b f^{2} g^{2} p q \log \left (h\right ) + 6 \, b f g h p q e \log \left (h\right ) - 6 \,{\left (h x + g\right )}^{2} b f^{2} p q - 2 \,{\left (h x + g\right )} b f^{2} g p q + 2 \,{\left (h x + g\right )} b f h p q e + 3 \, b h^{2} p q e^{2} \log \left ({\left (h x + g\right )} f - f g + h e\right ) + 3 \, b f^{2} g^{2} q \log \left (d\right ) - 6 \, b f g h q e \log \left (d\right ) - 3 \, b h^{2} p q e^{2} \log \left (h\right ) + 3 \, b f^{2} g^{2} \log \left (c\right ) - 6 \, b f g h e \log \left (c\right ) + 3 \, b h^{2} q e^{2} \log \left (d\right ) + 3 \, a f^{2} g^{2} - 6 \, a f g h e + 3 \, b h^{2} e^{2} \log \left (c\right ) + 3 \, a h^{2} e^{2}\right )}}{15 \,{\left ({\left (h x + g\right )}^{\frac{5}{2}} f^{2} g^{2} h - 2 \,{\left (h x + g\right )}^{\frac{5}{2}} f g h^{2} e +{\left (h x + g\right )}^{\frac{5}{2}} h^{3} e^{2}\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)^(7/2),x, algorithm="giac")

[Out]

4/5*b*f^3*h*p*q*arctan(sqrt(h*x + g)*f/sqrt(-f^2*g + f*h*e))/((f^2*g^2*h^2 - 2*f*g*h^3*e + h^4*e^2)*sqrt(-f^2*
g + f*h*e)) - 2/15*(3*b*f^2*g^2*p*q*log((h*x + g)*f - f*g + h*e) - 6*b*f*g*h*p*q*e*log((h*x + g)*f - f*g + h*e
) - 3*b*f^2*g^2*p*q*log(h) + 6*b*f*g*h*p*q*e*log(h) - 6*(h*x + g)^2*b*f^2*p*q - 2*(h*x + g)*b*f^2*g*p*q + 2*(h
*x + g)*b*f*h*p*q*e + 3*b*h^2*p*q*e^2*log((h*x + g)*f - f*g + h*e) + 3*b*f^2*g^2*q*log(d) - 6*b*f*g*h*q*e*log(
d) - 3*b*h^2*p*q*e^2*log(h) + 3*b*f^2*g^2*log(c) - 6*b*f*g*h*e*log(c) + 3*b*h^2*q*e^2*log(d) + 3*a*f^2*g^2 - 6
*a*f*g*h*e + 3*b*h^2*e^2*log(c) + 3*a*h^2*e^2)/((h*x + g)^(5/2)*f^2*g^2*h - 2*(h*x + g)^(5/2)*f*g*h^2*e + (h*x
+ g)^(5/2)*h^3*e^2)