3.119 \(\int (a+b \log (c (d+e x)^n))^{5/2} \, dx\)

Optimal. Leaf size=179 \[ -\frac {15 \sqrt {\pi } b^{5/2} n^{5/2} e^{-\frac {a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text {erfi}\left (\frac {\sqrt {a+b \log \left (c (d+e x)^n\right )}}{\sqrt {b} \sqrt {n}}\right )}{8 e}+\frac {15 b^2 n^2 (d+e x) \sqrt {a+b \log \left (c (d+e x)^n\right )}}{4 e}+\frac {(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}}{e}-\frac {5 b n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}}{2 e} \]

[Out]

-5/2*b*n*(e*x+d)*(a+b*ln(c*(e*x+d)^n))^(3/2)/e+(e*x+d)*(a+b*ln(c*(e*x+d)^n))^(5/2)/e-15/8*b^(5/2)*n^(5/2)*(e*x
+d)*erfi((a+b*ln(c*(e*x+d)^n))^(1/2)/b^(1/2)/n^(1/2))*Pi^(1/2)/e/exp(a/b/n)/((c*(e*x+d)^n)^(1/n))+15/4*b^2*n^2
*(e*x+d)*(a+b*ln(c*(e*x+d)^n))^(1/2)/e

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Rubi [A]  time = 0.13, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {2389, 2296, 2300, 2180, 2204} \[ -\frac {15 \sqrt {\pi } b^{5/2} n^{5/2} e^{-\frac {a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text {Erfi}\left (\frac {\sqrt {a+b \log \left (c (d+e x)^n\right )}}{\sqrt {b} \sqrt {n}}\right )}{8 e}+\frac {15 b^2 n^2 (d+e x) \sqrt {a+b \log \left (c (d+e x)^n\right )}}{4 e}+\frac {(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}}{e}-\frac {5 b n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}}{2 e} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d + e*x)^n])^(5/2),x]

[Out]

(-15*b^(5/2)*n^(5/2)*Sqrt[Pi]*(d + e*x)*Erfi[Sqrt[a + b*Log[c*(d + e*x)^n]]/(Sqrt[b]*Sqrt[n])])/(8*e*E^(a/(b*n
))*(c*(d + e*x)^n)^n^(-1)) + (15*b^2*n^2*(d + e*x)*Sqrt[a + b*Log[c*(d + e*x)^n]])/(4*e) - (5*b*n*(d + e*x)*(a
+ b*Log[c*(d + e*x)^n])^(3/2))/(2*e) + ((d + e*x)*(a + b*Log[c*(d + e*x)^n])^(5/2))/e

Rule 2180

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[F^(g*(e - (c*
f)/d) + (f*g*x^2)/d), x], x, Sqrt[c + d*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2300

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[E^(x/n)*(a +
b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rubi steps

\begin {align*} \int \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2} \, dx &=\frac {\operatorname {Subst}\left (\int \left (a+b \log \left (c x^n\right )\right )^{5/2} \, dx,x,d+e x\right )}{e}\\ &=\frac {(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}}{e}-\frac {(5 b n) \operatorname {Subst}\left (\int \left (a+b \log \left (c x^n\right )\right )^{3/2} \, dx,x,d+e x\right )}{2 e}\\ &=-\frac {5 b n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}}{2 e}+\frac {(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}}{e}+\frac {\left (15 b^2 n^2\right ) \operatorname {Subst}\left (\int \sqrt {a+b \log \left (c x^n\right )} \, dx,x,d+e x\right )}{4 e}\\ &=\frac {15 b^2 n^2 (d+e x) \sqrt {a+b \log \left (c (d+e x)^n\right )}}{4 e}-\frac {5 b n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}}{2 e}+\frac {(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}}{e}-\frac {\left (15 b^3 n^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b \log \left (c x^n\right )}} \, dx,x,d+e x\right )}{8 e}\\ &=\frac {15 b^2 n^2 (d+e x) \sqrt {a+b \log \left (c (d+e x)^n\right )}}{4 e}-\frac {5 b n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}}{2 e}+\frac {(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}}{e}-\frac {\left (15 b^3 n^2 (d+e x) \left (c (d+e x)^n\right )^{-1/n}\right ) \operatorname {Subst}\left (\int \frac {e^{\frac {x}{n}}}{\sqrt {a+b x}} \, dx,x,\log \left (c (d+e x)^n\right )\right )}{8 e}\\ &=\frac {15 b^2 n^2 (d+e x) \sqrt {a+b \log \left (c (d+e x)^n\right )}}{4 e}-\frac {5 b n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}}{2 e}+\frac {(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}}{e}-\frac {\left (15 b^2 n^2 (d+e x) \left (c (d+e x)^n\right )^{-1/n}\right ) \operatorname {Subst}\left (\int e^{-\frac {a}{b n}+\frac {x^2}{b n}} \, dx,x,\sqrt {a+b \log \left (c (d+e x)^n\right )}\right )}{4 e}\\ &=-\frac {15 b^{5/2} e^{-\frac {a}{b n}} n^{5/2} \sqrt {\pi } (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text {erfi}\left (\frac {\sqrt {a+b \log \left (c (d+e x)^n\right )}}{\sqrt {b} \sqrt {n}}\right )}{8 e}+\frac {15 b^2 n^2 (d+e x) \sqrt {a+b \log \left (c (d+e x)^n\right )}}{4 e}-\frac {5 b n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}}{2 e}+\frac {(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}}{e}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 152, normalized size = 0.85 \[ \frac {(d+e x) \left (8 \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}-5 b n \left (3 \sqrt {\pi } b^{3/2} n^{3/2} e^{-\frac {a}{b n}} \left (c (d+e x)^n\right )^{-1/n} \text {erfi}\left (\frac {\sqrt {a+b \log \left (c (d+e x)^n\right )}}{\sqrt {b} \sqrt {n}}\right )+2 \sqrt {a+b \log \left (c (d+e x)^n\right )} \left (2 a+2 b \log \left (c (d+e x)^n\right )-3 b n\right )\right )\right )}{8 e} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(d + e*x)^n])^(5/2),x]

[Out]

((d + e*x)*(8*(a + b*Log[c*(d + e*x)^n])^(5/2) - 5*b*n*((3*b^(3/2)*n^(3/2)*Sqrt[Pi]*Erfi[Sqrt[a + b*Log[c*(d +
e*x)^n]]/(Sqrt[b]*Sqrt[n])])/(E^(a/(b*n))*(c*(d + e*x)^n)^n^(-1)) + 2*Sqrt[a + b*Log[c*(d + e*x)^n]]*(2*a - 3
*b*n + 2*b*Log[c*(d + e*x)^n]))))/(8*e)

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))^(5/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))^(5/2),x, algorithm="giac")

[Out]

integrate((b*log((e*x + d)^n*c) + a)^(5/2), x)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \left (b \ln \left (c \left (e x +d \right )^{n}\right )+a \right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*ln(c*(e*x+d)^n)+a)^(5/2),x)

[Out]

int((b*ln(c*(e*x+d)^n)+a)^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*log((e*x + d)^n*c) + a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*log(c*(d + e*x)^n))^(5/2),x)

[Out]

int((a + b*log(c*(d + e*x)^n))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \log {\left (c \left (d + e x\right )^{n} \right )}\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*(e*x+d)**n))**(5/2),x)

[Out]

Integral((a + b*log(c*(d + e*x)**n))**(5/2), x)

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