3.143 \(\int \frac{a+b \log (c (d+e x)^n)}{(f+g x)^{7/2}} \, dx\)

Optimal. Leaf size=145 \[ -\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g (f+g x)^{5/2}}+\frac{4 b e^2 n}{5 g \sqrt{f+g x} (e f-d g)^2}-\frac{4 b e^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )}{5 g (e f-d g)^{5/2}}+\frac{4 b e n}{15 g (f+g x)^{3/2} (e f-d g)} \]

[Out]

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

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Rubi [A]  time = 0.110727, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2395, 51, 63, 208} \[ -\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g (f+g x)^{5/2}}+\frac{4 b e^2 n}{5 g \sqrt{f+g x} (e f-d g)^2}-\frac{4 b e^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )}{5 g (e f-d g)^{5/2}}+\frac{4 b e n}{15 g (f+g x)^{3/2} (e f-d g)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [C]  time = 0.0460885, size = 78, normalized size = 0.54 \[ \frac{2 \left (\frac{2 b e n (f+g x) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{e (f+g x)}{e f-d g}\right )}{e f-d g}-3 \left (a+b \log \left (c (d+e x)^n\right )\right )\right )}{15 g (f+g x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.921, size = 0, normalized size = 0. \begin{align*} \int{(a+b\ln \left ( c \left ( ex+d \right ) ^{n} \right ) ) \left ( gx+f \right ) ^{-{\frac{7}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*(e*x+d)^n))/(g*x+f)^(7/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.10924, size = 1678, normalized size = 11.57 \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*(e*x+d)^n))/(g*x+f)^(7/2),x, algorithm="fricas")

[Out]

[2/15*(3*(b*e^2*g^3*n*x^3 + 3*b*e^2*f*g^2*n*x^2 + 3*b*e^2*f^2*g*n*x + b*e^2*f^3*n)*sqrt(e/(e*f - d*g))*log((e*
g*x + 2*e*f - d*g - 2*(e*f - d*g)*sqrt(g*x + f)*sqrt(e/(e*f - d*g)))/(e*x + d)) + (6*b*e^2*g^2*n*x^2 - 3*a*e^2
*f^2 + 6*a*d*e*f*g - 3*a*d^2*g^2 + 2*(7*b*e^2*f*g - b*d*e*g^2)*n*x - 3*(b*e^2*f^2 - 2*b*d*e*f*g + b*d^2*g^2)*n
*log(e*x + d) + 2*(4*b*e^2*f^2 - b*d*e*f*g)*n - 3*(b*e^2*f^2 - 2*b*d*e*f*g + b*d^2*g^2)*log(c))*sqrt(g*x + f))
/(e^2*f^5*g - 2*d*e*f^4*g^2 + d^2*f^3*g^3 + (e^2*f^2*g^4 - 2*d*e*f*g^5 + d^2*g^6)*x^3 + 3*(e^2*f^3*g^3 - 2*d*e
*f^2*g^4 + d^2*f*g^5)*x^2 + 3*(e^2*f^4*g^2 - 2*d*e*f^3*g^3 + d^2*f^2*g^4)*x), -2/15*(6*(b*e^2*g^3*n*x^3 + 3*b*
e^2*f*g^2*n*x^2 + 3*b*e^2*f^2*g*n*x + b*e^2*f^3*n)*sqrt(-e/(e*f - d*g))*arctan(-(e*f - d*g)*sqrt(g*x + f)*sqrt
(-e/(e*f - d*g))/(e*g*x + e*f)) - (6*b*e^2*g^2*n*x^2 - 3*a*e^2*f^2 + 6*a*d*e*f*g - 3*a*d^2*g^2 + 2*(7*b*e^2*f*
g - b*d*e*g^2)*n*x - 3*(b*e^2*f^2 - 2*b*d*e*f*g + b*d^2*g^2)*n*log(e*x + d) + 2*(4*b*e^2*f^2 - b*d*e*f*g)*n -
3*(b*e^2*f^2 - 2*b*d*e*f*g + b*d^2*g^2)*log(c))*sqrt(g*x + f))/(e^2*f^5*g - 2*d*e*f^4*g^2 + d^2*f^3*g^3 + (e^2
*f^2*g^4 - 2*d*e*f*g^5 + d^2*g^6)*x^3 + 3*(e^2*f^3*g^3 - 2*d*e*f^2*g^4 + d^2*f*g^5)*x^2 + 3*(e^2*f^4*g^2 - 2*d
*e*f^3*g^3 + d^2*f^2*g^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*(e*x+d)**n))/(g*x+f)**(7/2),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 (e x + d\right )}^{n} c\right ) + a}{{\left (g x + f\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))/(g*x+f)^(7/2),x, algorithm="giac")

[Out]

integrate((b*log((e*x + d)^n*c) + a)/(g*x + f)^(7/2), x)