3.179 \(\int (f x)^m (d+e x^2) (a+b \text {sech}^{-1}(c x)) \, dx\)

Optimal. Leaf size=206 \[ \frac {d (f x)^{m+1} \left (a+b \text {sech}^{-1}(c x)\right )}{f (m+1)}+\frac {e (f x)^{m+3} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (m+3)}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} (f x)^{m+1} \left (c^2 d (m+2) (m+3)+e (m+1)^2\right ) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};c^2 x^2\right )}{c^2 f (m+1)^2 (m+2) (m+3)}-\frac {b e \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} (f x)^{m+1}}{c^2 f (m+2) (m+3)} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 192, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {14, 6301, 12, 459, 364} \[ \frac {d (f x)^{m+1} \left (a+b \text {sech}^{-1}(c x)\right )}{f (m+1)}+\frac {e (f x)^{m+3} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (m+3)}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} (f x)^{m+1} \left (\frac {e}{c^2 (m+2) (m+3)}+\frac {d}{(m+1)^2}\right ) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};c^2 x^2\right )}{f}-\frac {b e \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} (f x)^{m+1}}{c^2 f (m+2) (m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 459

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

Rule 6301

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u
= IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSech[c*x], u, x] + Dist[b*Sqrt[1 + c*x]*Sqrt[1/(1 + c*x)],
Int[SimplifyIntegrand[u/(x*Sqrt[1 - c*x]*Sqrt[1 + c*x]), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, p}, x] &&
((IGtQ[p, 0] &&  !(ILtQ[(m - 1)/2, 0] && GtQ[m + 2*p + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[p, 0] && GtQ
[m + 2*p + 3, 0])) || (ILtQ[(m + 2*p + 1)/2, 0] &&  !ILtQ[(m - 1)/2, 0]))

Rubi steps

\begin {align*} \int (f x)^m \left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right ) \, dx &=\frac {d (f x)^{1+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f (1+m)}+\frac {e (f x)^{3+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (3+m)}+\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {(f x)^m \left (d (3+m)+e (1+m) x^2\right )}{(1+m) (3+m) \sqrt {1-c^2 x^2}} \, dx\\ &=\frac {d (f x)^{1+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f (1+m)}+\frac {e (f x)^{3+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (3+m)}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {(f x)^m \left (d (3+m)+e (1+m) x^2\right )}{\sqrt {1-c^2 x^2}} \, dx}{3+4 m+m^2}\\ &=-\frac {b e (f x)^{1+m} \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{c^2 f (2+m) (3+m)}+\frac {d (f x)^{1+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f (1+m)}+\frac {e (f x)^{3+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (3+m)}+\frac {\left (b \left (\frac {e (1+m)^2}{c^2 (2+m)}+d (3+m)\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {(f x)^m}{\sqrt {1-c^2 x^2}} \, dx}{3+4 m+m^2}\\ &=-\frac {b e (f x)^{1+m} \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{c^2 f (2+m) (3+m)}+\frac {d (f x)^{1+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f (1+m)}+\frac {e (f x)^{3+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (3+m)}+\frac {b \left (\frac {e (1+m)^2}{c^2 (2+m)}+d (3+m)\right ) (f x)^{1+m} \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};c^2 x^2\right )}{f (1+m) \left (3+4 m+m^2\right )}\\ \end {align*}

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Mathematica [F]  time = 0.13, size = 0, normalized size = 0.00 \[ \int (f x)^m \left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a e x^{2} + a d + {\left (b e x^{2} + b d\right )} \operatorname {arsech}\left (c x\right )\right )} \left (f x\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*e*x^2 + a*d + (b*e*x^2 + b*d)*arcsech(c*x))*(f*x)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x^2 + d)*(b*arcsech(c*x) + a)*(f*x)^m, x)

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \left (f x \right )^{m} \left (e \,x^{2}+d \right ) \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x)^m*(e*x^2+d)*(a+b*arcsech(c*x)),x)

[Out]

int((f*x)^m*(e*x^2+d)*(a+b*arcsech(c*x)),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {a e f^{m} x^{3} x^{m}}{m + 3} + \frac {\left (f x\right )^{m + 1} a d}{f {\left (m + 1\right )}} + \frac {{\left (b e f^{m} {\left (m + 1\right )} x^{3} x^{m} + b d f^{m} {\left (m + 3\right )} x x^{m}\right )} \log \left (\sqrt {c x + 1} \sqrt {-c x + 1} + 1\right ) - {\left (b e f^{m} {\left (m + 1\right )} x^{3} x^{m} + b d f^{m} {\left (m + 3\right )} x x^{m}\right )} \log \relax (x)}{m^{2} + 4 \, m + 3} - \int \frac {{\left (b c^{2} e f^{m} {\left (m + 3\right )} x^{2} \log \relax (c) - {\left (e f^{m} {\left (m + 3\right )} \log \relax (c) - e f^{m}\right )} b\right )} x^{2} x^{m}}{c^{2} {\left (m + 3\right )} x^{2} - m - 3}\,{d x} - \int \frac {{\left (b c^{2} d f^{m} {\left (m + 1\right )} x^{2} \log \relax (c) - {\left (d f^{m} {\left (m + 1\right )} \log \relax (c) - d f^{m}\right )} b\right )} x^{m}}{c^{2} {\left (m + 1\right )} x^{2} - m - 1}\,{d x} + \int \frac {b c^{2} e f^{m} {\left (m + 1\right )} x^{4} x^{m} + b c^{2} d f^{m} {\left (m + 3\right )} x^{2} x^{m}}{{\left (m^{2} + 4 \, m + 3\right )} c^{2} x^{2} + {\left ({\left (m^{2} + 4 \, m + 3\right )} c^{2} x^{2} - m^{2} - 4 \, m - 3\right )} \sqrt {c x + 1} \sqrt {-c x + 1} - m^{2} - 4 \, m - 3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*e*f^m*x^3*x^m/(m + 3) + (f*x)^(m + 1)*a*d/(f*(m + 1)) + ((b*e*f^m*(m + 1)*x^3*x^m + b*d*f^m*(m + 3)*x*x^m)*l
og(sqrt(c*x + 1)*sqrt(-c*x + 1) + 1) - (b*e*f^m*(m + 1)*x^3*x^m + b*d*f^m*(m + 3)*x*x^m)*log(x))/(m^2 + 4*m +
3) - integrate((b*c^2*e*f^m*(m + 3)*x^2*log(c) - (e*f^m*(m + 3)*log(c) - e*f^m)*b)*x^2*x^m/(c^2*(m + 3)*x^2 -
m - 3), x) - integrate((b*c^2*d*f^m*(m + 1)*x^2*log(c) - (d*f^m*(m + 1)*log(c) - d*f^m)*b)*x^m/(c^2*(m + 1)*x^
2 - m - 1), x) + integrate((b*c^2*e*f^m*(m + 1)*x^4*x^m + b*c^2*d*f^m*(m + 3)*x^2*x^m)/((m^2 + 4*m + 3)*c^2*x^
2 + ((m^2 + 4*m + 3)*c^2*x^2 - m^2 - 4*m - 3)*sqrt(c*x + 1)*sqrt(-c*x + 1) - m^2 - 4*m - 3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x)^m*(d + e*x^2)*(a + b*acosh(1/(c*x))),x)

[Out]

int((f*x)^m*(d + e*x^2)*(a + b*acosh(1/(c*x))), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (f x\right )^{m} \left (a + b \operatorname {asech}{\left (c x \right )}\right ) \left (d + e x^{2}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)**m*(e*x**2+d)*(a+b*asech(c*x)),x)

[Out]

Integral((f*x)**m*(a + b*asech(c*x))*(d + e*x**2), x)

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