3.6.6 \(\int (g+h x)^m (a+b \log (c (d (e+f x)^p)^q)) \, dx\) [506]

Optimal. Leaf size=99 \[ \frac {b f p q (g+h x)^{2+m} \, _2F_1\left (1,2+m;3+m;\frac {f (g+h x)}{f g-e h}\right )}{h (f g-e h) (1+m) (2+m)}+\frac {(g+h x)^{1+m} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h (1+m)} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {2442, 70, 2495} \begin {gather*} \frac {(g+h x)^{m+1} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h (m+1)}+\frac {b f p q (g+h x)^{m+2} \, _2F_1\left (1,m+2;m+3;\frac {f (g+h x)}{f g-e h}\right )}{h (m+1) (m+2) (f g-e h)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(g + h*x)^m*(a + b*Log[c*(d*(e + f*x)^p)^q]),x]

[Out]

(b*f*p*q*(g + h*x)^(2 + m)*Hypergeometric2F1[1, 2 + m, 3 + m, (f*(g + h*x))/(f*g - e*h)])/(h*(f*g - e*h)*(1 +
m)*(2 + m)) + ((g + h*x)^(1 + m)*(a + b*Log[c*(d*(e + f*x)^p)^q]))/(h*(1 + m))

Rule 70

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

Rule 2442

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 2495

Int[((a_.) + Log[(c_.)*((d_.)*((e_.) + (f_.)*(x_))^(m_.))^(n_)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Subst[Int[u*(
a + b*Log[c*d^n*(e + f*x)^(m*n)])^p, x], c*d^n*(e + f*x)^(m*n), c*(d*(e + f*x)^m)^n] /; FreeQ[{a, b, c, d, e,
f, m, n, p}, x] &&  !IntegerQ[n] &&  !(EqQ[d, 1] && EqQ[m, 1]) && IntegralFreeQ[IntHide[u*(a + b*Log[c*d^n*(e
+ f*x)^(m*n)])^p, x]]

Rubi steps

\begin {align*} \int (g+h x)^m \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right ) \, dx &=\text {Subst}\left (\int (g+h x)^m \left (a+b \log \left (c d^q (e+f x)^{p q}\right )\right ) \, dx,c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac {(g+h x)^{1+m} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h (1+m)}-\text {Subst}\left (\frac {(b f p q) \int \frac {(g+h x)^{1+m}}{e+f x} \, dx}{h (1+m)},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac {b f p q (g+h x)^{2+m} \, _2F_1\left (1,2+m;3+m;\frac {f (g+h x)}{f g-e h}\right )}{h (f g-e h) (1+m) (2+m)}+\frac {(g+h x)^{1+m} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h (1+m)}\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 76, normalized size = 0.77 \begin {gather*} \frac {(g+h x)^{1+m} \left (a+a m-b p q+b p q \, _2F_1\left (1,1+m;2+m;\frac {f (g+h x)}{f g-e h}\right )+b (1+m) \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h (1+m)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(g + h*x)^m*(a + b*Log[c*(d*(e + f*x)^p)^q]),x]

[Out]

((g + h*x)^(1 + m)*(a + a*m - b*p*q + b*p*q*Hypergeometric2F1[1, 1 + m, 2 + m, (f*(g + h*x))/(f*g - e*h)] + b*
(1 + m)*Log[c*(d*(e + f*x)^p)^q]))/(h*(1 + m)^2)

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Maple [F]
time = 0.23, size = 0, normalized size = 0.00 \[\int \left (h x +g \right )^{m} \left (a +b \ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((h*x+g)^m*(a+b*ln(c*(d*(f*x+e)^p)^q)),x)

[Out]

int((h*x+g)^m*(a+b*ln(c*(d*(f*x+e)^p)^q)),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((h*x+g)^m*(a+b*log(c*(d*(f*x+e)^p)^q)),x, algorithm="maxima")

[Out]

b*((h*x + g)*(h*x + g)^m*log(((f*x + e)^p)^q)/(h*(m + 1)) + integrate(-(f*g*p*q + (f*h*p*q - f*h*(m + 1)*log(c
) - (m*q + q)*f*h*log(d))*x - (h*(m + 1)*log(c) + (m*q + q)*h*log(d))*e)*(h*x + g)^m/(f*h*(m + 1)*x + h*(m + 1
)*e), x)) + (h*x + g)^(m + 1)*a/(h*(m + 1))

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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((h*x+g)^m*(a+b*log(c*(d*(f*x+e)^p)^q)),x, algorithm="fricas")

[Out]

integral((h*x + g)^m*b*log(((f*x + e)^p*d)^q*c) + (h*x + g)^m*a, x)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: HeuristicGCDFailed} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x+g)**m*(a+b*ln(c*(d*(f*x+e)**p)**q)),x)

[Out]

Exception raised: HeuristicGCDFailed >> no luck

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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((h*x+g)^m*(a+b*log(c*(d*(f*x+e)^p)^q)),x, algorithm="giac")

[Out]

integrate((b*log(((f*x + e)^p*d)^q*c) + a)*(h*x + g)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g + h*x)^m*(a + b*log(c*(d*(e + f*x)^p)^q)),x)

[Out]

int((g + h*x)^m*(a + b*log(c*(d*(e + f*x)^p)^q)), x)

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