3.488 \(\int \frac{a+b \log (c (d (e+f x)^p)^q)}{(g+h x)^{9/2}} \, dx\)

Optimal. Leaf size=184 \[ -\frac{2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{7 h (g+h x)^{7/2}}+\frac{4 b f^3 p q}{7 h \sqrt{g+h x} (f g-e h)^3}+\frac{4 b f^2 p q}{21 h (g+h x)^{3/2} (f g-e h)^2}-\frac{4 b f^{7/2} p q \tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{g+h x}}{\sqrt{f g-e h}}\right )}{7 h (f g-e h)^{7/2}}+\frac{4 b f p q}{35 h (g+h x)^{5/2} (f g-e h)} \]

[Out]

(4*b*f*p*q)/(35*h*(f*g - e*h)*(g + h*x)^(5/2)) + (4*b*f^2*p*q)/(21*h*(f*g - e*h)^2*(g + h*x)^(3/2)) + (4*b*f^3
*p*q)/(7*h*(f*g - e*h)^3*Sqrt[g + h*x]) - (4*b*f^(7/2)*p*q*ArcTanh[(Sqrt[f]*Sqrt[g + h*x])/Sqrt[f*g - e*h]])/(
7*h*(f*g - e*h)^(7/2)) - (2*(a + b*Log[c*(d*(e + f*x)^p)^q]))/(7*h*(g + h*x)^(7/2))

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Rubi [A]  time = 0.277749, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ -\frac{2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{7 h (g+h x)^{7/2}}+\frac{4 b f^3 p q}{7 h \sqrt{g+h x} (f g-e h)^3}+\frac{4 b f^2 p q}{21 h (g+h x)^{3/2} (f g-e h)^2}-\frac{4 b f^{7/2} p q \tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{g+h x}}{\sqrt{f g-e h}}\right )}{7 h (f g-e h)^{7/2}}+\frac{4 b f p q}{35 h (g+h x)^{5/2} (f g-e h)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d*(e + f*x)^p)^q])/(g + h*x)^(9/2),x]

[Out]

(4*b*f*p*q)/(35*h*(f*g - e*h)*(g + h*x)^(5/2)) + (4*b*f^2*p*q)/(21*h*(f*g - e*h)^2*(g + h*x)^(3/2)) + (4*b*f^3
*p*q)/(7*h*(f*g - e*h)^3*Sqrt[g + h*x]) - (4*b*f^(7/2)*p*q*ArcTanh[(Sqrt[f]*Sqrt[g + h*x])/Sqrt[f*g - e*h]])/(
7*h*(f*g - e*h)^(7/2)) - (2*(a + b*Log[c*(d*(e + f*x)^p)^q]))/(7*h*(g + h*x)^(7/2))

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)^{9/2}} \, dx &=\operatorname{Subst}\left (\int \frac{a+b \log \left (c d^q (e+f x)^{p q}\right )}{(g+h x)^{9/2}} \, 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 )}{7 h (g+h x)^{7/2}}+\operatorname{Subst}\left (\frac{(2 b f p q) \int \frac{1}{(e+f x) (g+h x)^{7/2}} \, dx}{7 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}{35 h (f g-e h) (g+h x)^{5/2}}-\frac{2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{7 h (g+h x)^{7/2}}+\operatorname{Subst}\left (\frac{\left (2 b f^2 p q\right ) \int \frac{1}{(e+f x) (g+h x)^{5/2}} \, dx}{7 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}{35 h (f g-e h) (g+h x)^{5/2}}+\frac{4 b f^2 p q}{21 h (f g-e h)^2 (g+h x)^{3/2}}-\frac{2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{7 h (g+h x)^{7/2}}+\operatorname{Subst}\left (\frac{\left (2 b f^3 p q\right ) \int \frac{1}{(e+f x) (g+h x)^{3/2}} \, dx}{7 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}{35 h (f g-e h) (g+h x)^{5/2}}+\frac{4 b f^2 p q}{21 h (f g-e h)^2 (g+h x)^{3/2}}+\frac{4 b f^3 p q}{7 h (f g-e h)^3 \sqrt{g+h x}}-\frac{2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{7 h (g+h x)^{7/2}}+\operatorname{Subst}\left (\frac{\left (2 b f^4 p q\right ) \int \frac{1}{(e+f x) \sqrt{g+h x}} \, dx}{7 h (f g-e h)^3},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac{4 b f p q}{35 h (f g-e h) (g+h x)^{5/2}}+\frac{4 b f^2 p q}{21 h (f g-e h)^2 (g+h x)^{3/2}}+\frac{4 b f^3 p q}{7 h (f g-e h)^3 \sqrt{g+h x}}-\frac{2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{7 h (g+h x)^{7/2}}+\operatorname{Subst}\left (\frac{\left (4 b f^4 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 )}{7 h^2 (f g-e h)^3},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac{4 b f p q}{35 h (f g-e h) (g+h x)^{5/2}}+\frac{4 b f^2 p q}{21 h (f g-e h)^2 (g+h x)^{3/2}}+\frac{4 b f^3 p q}{7 h (f g-e h)^3 \sqrt{g+h x}}-\frac{4 b f^{7/2} p q \tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{g+h x}}{\sqrt{f g-e h}}\right )}{7 h (f g-e h)^{7/2}}-\frac{2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{7 h (g+h x)^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.102163, size = 91, normalized size = 0.49 \[ \frac{10 (f g-e h) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )-4 b f p q (g+h x) \, _2F_1\left (-\frac{5}{2},1;-\frac{3}{2};\frac{f (g+h x)}{f g-e h}\right )}{35 h (g+h x)^{7/2} (e h-f g)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(d*(e + f*x)^p)^q])/(g + h*x)^(9/2),x]

[Out]

(-4*b*f*p*q*(g + h*x)*Hypergeometric2F1[-5/2, 1, -3/2, (f*(g + h*x))/(f*g - e*h)] + 10*(f*g - e*h)*(a + b*Log[
c*(d*(e + f*x)^p)^q]))/(35*h*(-(f*g) + e*h)*(g + h*x)^(7/2))

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Maple [F]  time = 0.713, 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{9}{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)^(9/2),x)

[Out]

int((a+b*ln(c*(d*(f*x+e)^p)^q))/(h*x+g)^(9/2),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)^(9/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.17595, size = 2865, normalized size = 15.57 \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)^(9/2),x, algorithm="fricas")

[Out]

[-2/105*(15*(b*f^3*h^4*p*q*x^4 + 4*b*f^3*g*h^3*p*q*x^3 + 6*b*f^3*g^2*h^2*p*q*x^2 + 4*b*f^3*g^3*h*p*q*x + b*f^3
*g^4*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)) - (30*b*f^3*h^3*p*q*x^3 - 15*a*f^3*g^3 + 45*a*e*f^2*g^2*h - 45*a*e^2*f*g*h^2 + 15*a*e^3*h^3 + 10*(10*b*
f^3*g*h^2 - b*e*f^2*h^3)*p*q*x^2 + 2*(58*b*f^3*g^2*h - 16*b*e*f^2*g*h^2 + 3*b*e^2*f*h^3)*p*q*x - 15*(b*f^3*g^3
- 3*b*e*f^2*g^2*h + 3*b*e^2*f*g*h^2 - b*e^3*h^3)*p*q*log(f*x + e) + 2*(23*b*f^3*g^3 - 11*b*e*f^2*g^2*h + 3*b*
e^2*f*g*h^2)*p*q - 15*(b*f^3*g^3 - 3*b*e*f^2*g^2*h + 3*b*e^2*f*g*h^2 - b*e^3*h^3)*q*log(d) - 15*(b*f^3*g^3 - 3
*b*e*f^2*g^2*h + 3*b*e^2*f*g*h^2 - b*e^3*h^3)*log(c))*sqrt(h*x + g))/(f^3*g^7*h - 3*e*f^2*g^6*h^2 + 3*e^2*f*g^
5*h^3 - e^3*g^4*h^4 + (f^3*g^3*h^5 - 3*e*f^2*g^2*h^6 + 3*e^2*f*g*h^7 - e^3*h^8)*x^4 + 4*(f^3*g^4*h^4 - 3*e*f^2
*g^3*h^5 + 3*e^2*f*g^2*h^6 - e^3*g*h^7)*x^3 + 6*(f^3*g^5*h^3 - 3*e*f^2*g^4*h^4 + 3*e^2*f*g^3*h^5 - e^3*g^2*h^6
)*x^2 + 4*(f^3*g^6*h^2 - 3*e*f^2*g^5*h^3 + 3*e^2*f*g^4*h^4 - e^3*g^3*h^5)*x), -2/105*(30*(b*f^3*h^4*p*q*x^4 +
4*b*f^3*g*h^3*p*q*x^3 + 6*b*f^3*g^2*h^2*p*q*x^2 + 4*b*f^3*g^3*h*p*q*x + b*f^3*g^4*p*q)*sqrt(-f/(f*g - e*h))*ar
ctan(-(f*g - e*h)*sqrt(h*x + g)*sqrt(-f/(f*g - e*h))/(f*h*x + f*g)) - (30*b*f^3*h^3*p*q*x^3 - 15*a*f^3*g^3 + 4
5*a*e*f^2*g^2*h - 45*a*e^2*f*g*h^2 + 15*a*e^3*h^3 + 10*(10*b*f^3*g*h^2 - b*e*f^2*h^3)*p*q*x^2 + 2*(58*b*f^3*g^
2*h - 16*b*e*f^2*g*h^2 + 3*b*e^2*f*h^3)*p*q*x - 15*(b*f^3*g^3 - 3*b*e*f^2*g^2*h + 3*b*e^2*f*g*h^2 - b*e^3*h^3)
*p*q*log(f*x + e) + 2*(23*b*f^3*g^3 - 11*b*e*f^2*g^2*h + 3*b*e^2*f*g*h^2)*p*q - 15*(b*f^3*g^3 - 3*b*e*f^2*g^2*
h + 3*b*e^2*f*g*h^2 - b*e^3*h^3)*q*log(d) - 15*(b*f^3*g^3 - 3*b*e*f^2*g^2*h + 3*b*e^2*f*g*h^2 - b*e^3*h^3)*log
(c))*sqrt(h*x + g))/(f^3*g^7*h - 3*e*f^2*g^6*h^2 + 3*e^2*f*g^5*h^3 - e^3*g^4*h^4 + (f^3*g^3*h^5 - 3*e*f^2*g^2*
h^6 + 3*e^2*f*g*h^7 - e^3*h^8)*x^4 + 4*(f^3*g^4*h^4 - 3*e*f^2*g^3*h^5 + 3*e^2*f*g^2*h^6 - e^3*g*h^7)*x^3 + 6*(
f^3*g^5*h^3 - 3*e*f^2*g^4*h^4 + 3*e^2*f*g^3*h^5 - e^3*g^2*h^6)*x^2 + 4*(f^3*g^6*h^2 - 3*e*f^2*g^5*h^3 + 3*e^2*
f*g^4*h^4 - e^3*g^3*h^5)*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)**(9/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) + a}{{\left (h x + g\right )}^{\frac{9}{2}}}\,{d x} \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)^(9/2),x, algorithm="giac")

[Out]

integrate((b*log(((f*x + e)^p*d)^q*c) + a)/(h*x + g)^(9/2), x)