3.62 \(\int (f x)^m \log (c (d+\frac{e}{\sqrt{x}})^p) \, dx\)

Optimal. Leaf size=70 \[ \frac{(f x)^{m+1} \log \left (c \left (d+\frac{e}{\sqrt{x}}\right )^p\right )}{f (m+1)}+\frac{p x (f x)^m \, _2F_1\left (1,2 (m+1);2 m+3;-\frac{d \sqrt{x}}{e}\right )}{2 (m+1)^2} \]

[Out]

(p*x*(f*x)^m*Hypergeometric2F1[1, 2*(1 + m), 3 + 2*m, -((d*Sqrt[x])/e)])/(2*(1 + m)^2) + ((f*x)^(1 + m)*Log[c*
(d + e/Sqrt[x])^p])/(f*(1 + m))

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Rubi [A]  time = 0.0422632, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2455, 20, 263, 341, 64} \[ \frac{(f x)^{m+1} \log \left (c \left (d+\frac{e}{\sqrt{x}}\right )^p\right )}{f (m+1)}+\frac{p x (f x)^m \, _2F_1\left (1,2 (m+1);2 m+3;-\frac{d \sqrt{x}}{e}\right )}{2 (m+1)^2} \]

Antiderivative was successfully verified.

[In]

Int[(f*x)^m*Log[c*(d + e/Sqrt[x])^p],x]

[Out]

(p*x*(f*x)^m*Hypergeometric2F1[1, 2*(1 + m), 3 + 2*m, -((d*Sqrt[x])/e)])/(2*(1 + m)^2) + ((f*x)^(1 + m)*Log[c*
(d + e/Sqrt[x])^p])/(f*(1 + m))

Rule 2455

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

Rule 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(b^IntPart[n]*(b*v)^FracPart[n])/(a^IntPart[n
]*(a*v)^FracPart[n]), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + n]

Rule 263

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rule 341

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[Int[x^(k*(
m + 1) - 1)*(a + b*x^(k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, m, p}, x] && FractionQ[n]

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

Rubi steps

\begin{align*} \int (f x)^m \log \left (c \left (d+\frac{e}{\sqrt{x}}\right )^p\right ) \, dx &=\frac{(f x)^{1+m} \log \left (c \left (d+\frac{e}{\sqrt{x}}\right )^p\right )}{f (1+m)}+\frac{(e p) \int \frac{(f x)^{1+m}}{\left (d+\frac{e}{\sqrt{x}}\right ) x^{3/2}} \, dx}{2 f (1+m)}\\ &=\frac{(f x)^{1+m} \log \left (c \left (d+\frac{e}{\sqrt{x}}\right )^p\right )}{f (1+m)}+\frac{\left (e p x^{-m} (f x)^m\right ) \int \frac{x^{-\frac{1}{2}+m}}{d+\frac{e}{\sqrt{x}}} \, dx}{2 (1+m)}\\ &=\frac{(f x)^{1+m} \log \left (c \left (d+\frac{e}{\sqrt{x}}\right )^p\right )}{f (1+m)}+\frac{\left (e p x^{-m} (f x)^m\right ) \int \frac{x^m}{e+d \sqrt{x}} \, dx}{2 (1+m)}\\ &=\frac{(f x)^{1+m} \log \left (c \left (d+\frac{e}{\sqrt{x}}\right )^p\right )}{f (1+m)}+\frac{\left (e p x^{-m} (f x)^m\right ) \operatorname{Subst}\left (\int \frac{x^{-1+2 (1+m)}}{e+d x} \, dx,x,\sqrt{x}\right )}{1+m}\\ &=\frac{p x (f x)^m \, _2F_1\left (1,2 (1+m);3+2 m;-\frac{d \sqrt{x}}{e}\right )}{2 (1+m)^2}+\frac{(f x)^{1+m} \log \left (c \left (d+\frac{e}{\sqrt{x}}\right )^p\right )}{f (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.0372645, size = 77, normalized size = 1.1 \[ \frac{\sqrt{x} (f x)^m \left (d (2 m+1) \sqrt{x} \log \left (c \left (d+\frac{e}{\sqrt{x}}\right )^p\right )+e p \, _2F_1\left (1,-2 m-1;-2 m;-\frac{e}{d \sqrt{x}}\right )\right )}{d (m+1) (2 m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(f*x)^m*Log[c*(d + e/Sqrt[x])^p],x]

[Out]

(Sqrt[x]*(f*x)^m*(e*p*Hypergeometric2F1[1, -1 - 2*m, -2*m, -(e/(d*Sqrt[x]))] + d*(1 + 2*m)*Sqrt[x]*Log[c*(d +
e/Sqrt[x])^p]))/(d*(1 + m)*(1 + 2*m))

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Maple [F]  time = 0.389, size = 0, normalized size = 0. \begin{align*} \int \left ( fx \right ) ^{m}\ln \left ( c \left ( d+{e{\frac{1}{\sqrt{x}}}} \right ) ^{p} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x)^m*ln(c*(d+e/x^(1/2))^p),x)

[Out]

int((f*x)^m*ln(c*(d+e/x^(1/2))^p),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((f*x)^m*log(c*(d+e/x^(1/2))^p),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (f x\right )^{m} \log \left (c \left (\frac{d x + e \sqrt{x}}{x}\right )^{p}\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*log(c*(d+e/x^(1/2))^p),x, algorithm="fricas")

[Out]

integral((f*x)^m*log(c*((d*x + e*sqrt(x))/x)^p), 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((f*x)**m*ln(c*(d+e/x**(1/2))**p),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (f x\right )^{m} \log \left (c{\left (d + \frac{e}{\sqrt{x}}\right )}^{p}\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*log(c*(d+e/x^(1/2))^p),x, algorithm="giac")

[Out]

integrate((f*x)^m*log(c*(d + e/sqrt(x))^p), x)