3.210 \(\int (d+e x)^m \log (c (a+\frac{b}{x^2})^p) \, dx\)

Optimal. Leaf size=257 \[ \frac{(d+e x)^{m+1} \log \left (c \left (a+\frac{b}{x^2}\right )^p\right )}{e (m+1)}+\frac{\sqrt{-a} p (d+e x)^{m+2} \, _2F_1\left (1,m+2;m+3;\frac{\sqrt{-a} (d+e x)}{\sqrt{-a} d-\sqrt{b} e}\right )}{e (m+1) (m+2) \left (\sqrt{-a} d-\sqrt{b} e\right )}+\frac{\sqrt{-a} p (d+e x)^{m+2} \, _2F_1\left (1,m+2;m+3;\frac{\sqrt{-a} (d+e x)}{\sqrt{-a} d+\sqrt{b} e}\right )}{e (m+1) (m+2) \left (\sqrt{-a} d+\sqrt{b} e\right )}-\frac{2 p (d+e x)^{m+2} \, _2F_1\left (1,m+2;m+3;\frac{e x}{d}+1\right )}{d e \left (m^2+3 m+2\right )} \]

[Out]

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

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Rubi [A]  time = 0.532742, antiderivative size = 257, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {2463, 1570, 961, 65, 831, 68} \[ \frac{(d+e x)^{m+1} \log \left (c \left (a+\frac{b}{x^2}\right )^p\right )}{e (m+1)}+\frac{\sqrt{-a} p (d+e x)^{m+2} \, _2F_1\left (1,m+2;m+3;\frac{\sqrt{-a} (d+e x)}{\sqrt{-a} d-\sqrt{b} e}\right )}{e (m+1) (m+2) \left (\sqrt{-a} d-\sqrt{b} e\right )}+\frac{\sqrt{-a} p (d+e x)^{m+2} \, _2F_1\left (1,m+2;m+3;\frac{\sqrt{-a} (d+e x)}{\sqrt{-a} d+\sqrt{b} e}\right )}{e (m+1) (m+2) \left (\sqrt{-a} d+\sqrt{b} e\right )}-\frac{2 p (d+e x)^{m+2} \, _2F_1\left (1,m+2;m+3;\frac{e x}{d}+1\right )}{d e \left (m^2+3 m+2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2463

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.) + (g_.)*(x_))^(r_.), x_Symbol] :> Simp[((
f + g*x)^(r + 1)*(a + b*Log[c*(d + e*x^n)^p]))/(g*(r + 1)), x] - Dist[(b*e*n*p)/(g*(r + 1)), Int[(x^(n - 1)*(f
+ g*x)^(r + 1))/(d + e*x^n), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, r}, x] && (IGtQ[r, 0] || RationalQ[n
]) && NeQ[r, -1]

Rule 1570

Int[(x_)^(m_.)*((a_.) + (c_.)*(x_)^(mn2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_.))^(q_.), x_Symbol] :> Int[x^(m - 2*n
*p)*(d + e*x^n)^q*(c + a*x^(2*n))^p, x] /; FreeQ[{a, c, d, e, m, n, q}, x] && EqQ[mn2, -2*n] && IntegerQ[p]

Rule 961

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (ILtQ[m, 0] && ILtQ[n, 0])) &&  !(IGtQ[m, 0] || IGtQ[n, 0])

Rule 65

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

Rule 831

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m, (f + g*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !Ration
alQ[m]

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

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

Mathematica [A]  time = 0.507459, size = 211, normalized size = 0.82 \[ \frac{(d+e x)^{m+1} \left (\log \left (c \left (a+\frac{b}{x^2}\right )^p\right )+\frac{p (d+e x) \left (-2 \left (a d^2+b e^2\right ) \, _2F_1\left (1,m+2;m+3;\frac{e x}{d}+1\right )+d \left (a d-\sqrt{-a} \sqrt{b} e\right ) \, _2F_1\left (1,m+2;m+3;\frac{\sqrt{-a} (d+e x)}{\sqrt{-a} d-\sqrt{b} e}\right )+d \left (\sqrt{-a} \sqrt{b} e+a d\right ) \, _2F_1\left (1,m+2;m+3;\frac{\sqrt{-a} (d+e x)}{\sqrt{-a} d+\sqrt{b} e}\right )\right )}{d (m+2) \left (a d^2+b e^2\right )}\right )}{e (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 3.108, size = 0, normalized size = 0. \begin{align*} \int \left ( ex+d \right ) ^{m}\ln \left ( c \left ( a+{\frac{b}{{x}^{2}}} \right ) ^{p} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((e*x+d)^m*ln(c*(a+b/x^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((e*x+d)^m*log(c*(a+b/x^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 (e x + d\right )}^{m} \log \left (c \left (\frac{a x^{2} + b}{x^{2}}\right )^{p}\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e*x + d)^m*log(c*((a*x^2 + b)/x^2)^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((e*x+d)**m*ln(c*(a+b/x**2)**p),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x + d\right )}^{m} \log \left ({\left (a + \frac{b}{x^{2}}\right )}^{p} c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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