3.207 \(\int (d+e x)^m \log (c (a+b x^2)^p) \, dx\)

Optimal. Leaf size=205 \[ \frac {(d+e x)^{m+1} \log \left (c \left (a+b x^2\right )^p\right )}{e (m+1)}+\frac {\sqrt {b} p (d+e x)^{m+2} \, _2F_1\left (1,m+2;m+3;\frac {\sqrt {b} (d+e x)}{\sqrt {b} d-\sqrt {-a} e}\right )}{e (m+1) (m+2) \left (\sqrt {b} d-\sqrt {-a} e\right )}+\frac {\sqrt {b} p (d+e x)^{m+2} \, _2F_1\left (1,m+2;m+3;\frac {\sqrt {b} (d+e x)}{\sqrt {b} d+\sqrt {-a} e}\right )}{e (m+1) (m+2) \left (\sqrt {-a} e+\sqrt {b} d\right )} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2463, 831, 68} \[ \frac {(d+e x)^{m+1} \log \left (c \left (a+b x^2\right )^p\right )}{e (m+1)}+\frac {\sqrt {b} p (d+e x)^{m+2} \, _2F_1\left (1,m+2;m+3;\frac {\sqrt {b} (d+e x)}{\sqrt {b} d-\sqrt {-a} e}\right )}{e (m+1) (m+2) \left (\sqrt {b} d-\sqrt {-a} e\right )}+\frac {\sqrt {b} p (d+e x)^{m+2} \, _2F_1\left (1,m+2;m+3;\frac {\sqrt {b} (d+e x)}{\sqrt {b} d+\sqrt {-a} e}\right )}{e (m+1) (m+2) \left (\sqrt {-a} e+\sqrt {b} d\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.25, size = 176, normalized size = 0.86 \[ \frac {(d+e x)^{m+1} \left (\log \left (c \left (a+b x^2\right )^p\right )+\frac {\sqrt {b} p (d+e x) \left (\left (\sqrt {-a} e+\sqrt {b} d\right ) \, _2F_1\left (1,m+2;m+3;\frac {\sqrt {b} (d+e x)}{\sqrt {b} d-\sqrt {-a} e}\right )+\left (\sqrt {b} d-\sqrt {-a} e\right ) \, _2F_1\left (1,m+2;m+3;\frac {\sqrt {b} (d+e x)}{\sqrt {b} d+\sqrt {-a} e}\right )\right )}{(m+2) \left (a e^2+b d^2\right )}\right )}{e (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (e x + d\right )}^{m} \log \left ({\left (b x^{2} + a\right )}^{p} c\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x + d\right )}^{m} \log \left ({\left (b x^{2} + a\right )}^{p} c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.36, size = 0, normalized size = 0.00 \[ \int \left (e x +d \right )^{m} \ln \left (c \left (b \,x^{2}+a \right )^{p}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left (e p x + d p\right )} {\left (e x + d\right )}^{m} \log \left (b x^{2} + a\right )}{e {\left (m + 1\right )}} + \int -\frac {{\left (2 \, b d p x - {\left (e {\left (m + 1\right )} \log \relax (c) - 2 \, e p\right )} b x^{2} - a e {\left (m + 1\right )} \log \relax (c)\right )} {\left (e x + d\right )}^{m}}{b e {\left (m + 1\right )} x^{2} + a e {\left (m + 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e*p*x + d*p)*(e*x + d)^m*log(b*x^2 + a)/(e*(m + 1)) + integrate(-(2*b*d*p*x - (e*(m + 1)*log(c) - 2*e*p)*b*x^
2 - a*e*(m + 1)*log(c))*(e*x + d)^m/(b*e*(m + 1)*x^2 + a*e*(m + 1)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \ln \left (c\,{\left (b\,x^2+a\right )}^p\right )\,{\left (d+e\,x\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**m*ln(c*(b*x**2+a)**p),x)

[Out]

Timed out

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