3.1.61 \(\int (f x)^m \log (c (d+e \sqrt {x})^p) \, dx\) [61]

Optimal. Leaf size=83 \[ -\frac {e p x^{3/2} (f x)^m \, _2F_1\left (1,3+2 m;2 (2+m);-\frac {e \sqrt {x}}{d}\right )}{d \left (3+5 m+2 m^2\right )}+\frac {(f x)^{1+m} \log \left (c \left (d+e \sqrt {x}\right )^p\right )}{f (1+m)} \]

[Out]

-e*p*x^(3/2)*(f*x)^m*hypergeom([1, 3+2*m],[4+2*m],-e*x^(1/2)/d)/d/(1+m)/(3+2*m)+(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 = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2505, 20, 348, 66} \begin {gather*} \frac {(f x)^{m+1} \log \left (c \left (d+e \sqrt {x}\right )^p\right )}{f (m+1)}-\frac {e p x^{3/2} (f x)^m \, _2F_1\left (1,2 m+3;2 (m+2);-\frac {e \sqrt {x}}{d}\right )}{d \left (2 m^2+5 m+3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((e*p*x^(3/2)*(f*x)^m*Hypergeometric2F1[1, 3 + 2*m, 2*(2 + m), -((e*Sqrt[x])/d)])/(d*(3 + 5*m + 2*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 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], 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 348

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 2505

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+e \sqrt {x}\right )^p\right ) \, dx &=\frac {(f x)^{1+m} \log \left (c \left (d+e \sqrt {x}\right )^p\right )}{f (1+m)}-\frac {(e p) \int \frac {(f x)^{1+m}}{\left (d+e \sqrt {x}\right ) \sqrt {x}} \, dx}{2 f (1+m)}\\ &=\frac {(f x)^{1+m} \log \left (c \left (d+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+e \sqrt {x}} \, dx}{2 (1+m)}\\ &=\frac {(f x)^{1+m} \log \left (c \left (d+e \sqrt {x}\right )^p\right )}{f (1+m)}-\frac {\left (e p x^{-m} (f x)^m\right ) \text {Subst}\left (\int \frac {x^{-1+2 \left (\frac {3}{2}+m\right )}}{d+e x} \, dx,x,\sqrt {x}\right )}{1+m}\\ &=-\frac {e p x^{3/2} (f x)^m \, _2F_1\left (1,3+2 m;2 (2+m);-\frac {e \sqrt {x}}{d}\right )}{d \left (3+5 m+2 m^2\right )}+\frac {(f x)^{1+m} \log \left (c \left (d+e \sqrt {x}\right )^p\right )}{f (1+m)}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 76, normalized size = 0.92 \begin {gather*} \frac {x (f x)^m \left (-e p \sqrt {x} \, _2F_1\left (1,3+2 m;4+2 m;-\frac {e \sqrt {x}}{d}\right )+d (3+2 m) \log \left (c \left (d+e \sqrt {x}\right )^p\right )\right )}{d (1+m) (3+2 m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(f*x)^m*(-(e*p*Sqrt[x]*Hypergeometric2F1[1, 3 + 2*m, 4 + 2*m, -((e*Sqrt[x])/d)]) + d*(3 + 2*m)*Log[c*(d + e
*Sqrt[x])^p]))/(d*(1 + m)*(3 + 2*m))

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Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int \left (f x \right )^{m} \ln \left (c \left (d +e \sqrt {x}\right )^{p}\right )\, dx\]

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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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((sqrt(x)*e + d)^p*c), x)

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

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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((sqrt(x)*e + d)^p*c), x)

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

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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