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]

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

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Rubi [A]  time = 0.04, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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 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])))

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 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]

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*}

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Mathematica [A]  time = 0.04, size = 77, normalized size = 1.10 \[ \frac {\sqrt {x} (f x)^m \left (d (2 m+1) \sqrt {x} \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^p\right )+e p \, _2F_1\left (1,-2 m-1;-2 m;-\frac {e}{d \sqrt {x}}\right )\right )}{d (m+1) (2 m+1)} \]

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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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (f x\right )^{m} \log \left (c \left (\frac {d x + e \sqrt {x}}{x}\right )^{p}\right ), x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (f x\right )^{m} \log \left (c {\left (d + \frac {e}{\sqrt {x}}\right )}^{p}\right )\,{d x} \]

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)

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maple [F]  time = 0.17, size = 0, normalized size = 0.00 \[ \int \left (f x \right )^{m} \ln \left (c \left (d +\frac {e}{\sqrt {x}}\right )^{p}\right )\, dx \]

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]  time = 0.00, size = 0, normalized size = 0.00 \[ d^{2} f^{m} p \int \frac {x x^{m}}{2 \, {\left (d e {\left (m + 1\right )} \sqrt {x} + e^{2} {\left (m + 1\right )}\right )}}\,{d x} + \frac {2 \, {\left (2 \, m^{2} + 5 \, m + 3\right )} e f^{m} p x x^{m} \log \left (d \sqrt {x} + e\right ) - 2 \, {\left (2 \, m^{2} + 5 \, m + 3\right )} e f^{m} x x^{m} \log \left (x^{\frac {1}{2} \, p}\right ) - 2 \, {\left (m p + p\right )} d f^{m} x^{\frac {3}{2}} x^{m} + {\left (2 \, {\left (2 \, m^{2} + 5 \, m + 3\right )} e f^{m} \log \relax (c) + {\left (2 \, m p + 3 \, p\right )} e f^{m}\right )} x x^{m}}{2 \, {\left (2 \, m^{3} + 7 \, m^{2} + 8 \, m + 3\right )} e} \]

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]

d^2*f^m*p*integrate(1/2*x*x^m/(d*e*(m + 1)*sqrt(x) + e^2*(m + 1)), x) + 1/2*(2*(2*m^2 + 5*m + 3)*e*f^m*p*x*x^m
*log(d*sqrt(x) + e) - 2*(2*m^2 + 5*m + 3)*e*f^m*x*x^m*log(x^(1/2*p)) - 2*(m*p + p)*d*f^m*x^(3/2)*x^m + (2*(2*m
^2 + 5*m + 3)*e*f^m*log(c) + (2*m*p + 3*p)*e*f^m)*x*x^m)/((2*m^3 + 7*m^2 + 8*m + 3)*e)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \ln \left (c\,{\left (d+\frac {e}{\sqrt {x}}\right )}^p\right )\,{\left (f\,x\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (f x\right )^{m} \log {\left (c \left (d + \frac {e}{\sqrt {x}}\right )^{p} \right )}\, dx \]

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]

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

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