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

Optimal. Leaf size=372 \[ \frac {d^2 (f x)^{m+1} \left (a+b \text {sech}^{-1}(c x)\right )}{f (m+1)}+\frac {2 d e (f x)^{m+3} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (m+3)}+\frac {e^2 (f x)^{m+5} \left (a+b \text {sech}^{-1}(c x)\right )}{f^5 (m+5)}-\frac {b e^2 \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} (f x)^{m+3}}{c^2 f^3 (m+4) (m+5)}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} (f x)^{m+1} \left (c^4 d^2 (m+2) (m+3) (m+4) (m+5)+e (m+1)^2 \left (2 c^2 d \left (m^2+9 m+20\right )+e (m+3)^2\right )\right ) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};c^2 x^2\right )}{c^4 f (m+1)^2 (m+2) (m+3) (m+4) (m+5)}-\frac {b e \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} (f x)^{m+1} \left (2 c^2 d \left (m^2+9 m+20\right )+e (m+3)^2\right )}{c^4 f (m+2) (m+3) (m+4) (m+5)} \]

[Out]

d^2*(f*x)^(1+m)*(a+b*arcsech(c*x))/f/(1+m)+2*d*e*(f*x)^(3+m)*(a+b*arcsech(c*x))/f^3/(3+m)+e^2*(f*x)^(5+m)*(a+b
*arcsech(c*x))/f^5/(5+m)+b*(c^4*d^2*(2+m)*(3+m)*(4+m)*(5+m)+e*(1+m)^2*(e*(3+m)^2+2*c^2*d*(m^2+9*m+20)))*(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^4/f/(1+m)^2/(2+m)/(3+m
)/(4+m)/(5+m)-b*e*(e*(3+m)^2+2*c^2*d*(m^2+9*m+20))*(f*x)^(1+m)*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)*(-c^2*x^2+1)^(1
/2)/c^4/f/(4+m)/(5+m)/(m^2+5*m+6)-b*e^2*(f*x)^(3+m)*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)*(-c^2*x^2+1)^(1/2)/c^2/f^3
/(4+m)/(5+m)

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Rubi [A]  time = 0.43, antiderivative size = 352, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {270, 6301, 12, 1267, 459, 364} \[ \frac {d^2 (f x)^{m+1} \left (a+b \text {sech}^{-1}(c x)\right )}{f (m+1)}+\frac {2 d e (f x)^{m+3} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (m+3)}+\frac {e^2 (f x)^{m+5} \left (a+b \text {sech}^{-1}(c x)\right )}{f^5 (m+5)}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} (f x)^{m+1} \left (\frac {e \left (2 c^2 d \left (m^2+9 m+20\right )+e (m+3)^2\right )}{c^4 (m+2) (m+3) (m+4) (m+5)}+\frac {d^2}{(m+1)^2}\right ) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};c^2 x^2\right )}{f}-\frac {b e \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} (f x)^{m+1} \left (2 c^2 d \left (m^2+9 m+20\right )+e (m+3)^2\right )}{c^4 f (m+2) (m+3) (m+4) (m+5)}-\frac {b e^2 \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} (f x)^{m+3}}{c^2 f^3 (m+4) (m+5)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*e*(e*(3 + m)^2 + 2*c^2*d*(20 + 9*m + m^2))*(f*x)^(1 + m)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*
x^2])/(c^4*f*(2 + m)*(3 + m)*(4 + m)*(5 + m))) - (b*e^2*(f*x)^(3 + m)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[
1 - c^2*x^2])/(c^2*f^3*(4 + m)*(5 + m)) + (d^2*(f*x)^(1 + m)*(a + b*ArcSech[c*x]))/(f*(1 + m)) + (2*d*e*(f*x)^
(3 + m)*(a + b*ArcSech[c*x]))/(f^3*(3 + m)) + (e^2*(f*x)^(5 + m)*(a + b*ArcSech[c*x]))/(f^5*(5 + m)) + (b*(d^2
/(1 + m)^2 + (e*(e*(3 + m)^2 + 2*c^2*d*(20 + 9*m + m^2)))/(c^4*(2 + m)*(3 + m)*(4 + m)*(5 + 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 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

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 1267

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Si
mp[(c^p*(f*x)^(m + 4*p - 1)*(d + e*x^2)^(q + 1))/(e*f^(4*p - 1)*(m + 4*p + 2*q + 1)), x] + Dist[1/(e*(m + 4*p
+ 2*q + 1)), Int[(f*x)^m*(d + e*x^2)^q*ExpandToSum[e*(m + 4*p + 2*q + 1)*((a + b*x^2 + c*x^4)^p - c^p*x^(4*p))
- d*c^p*(m + 4*p - 1)*x^(4*p - 2), x], x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[b^2 - 4*a*c, 0] &&
IGtQ[p, 0] &&  !IntegerQ[q] && NeQ[m + 4*p + 2*q + 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 )^2 \left (a+b \text {sech}^{-1}(c x)\right ) \, dx &=\frac {d^2 (f x)^{1+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f (1+m)}+\frac {2 d e (f x)^{3+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f^5 (5+m)}+\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {(f x)^m \left (d^2 \left (15+8 m+m^2\right )+2 d e \left (5+6 m+m^2\right ) x^2+e^2 \left (3+4 m+m^2\right ) x^4\right )}{(1+m) (3+m) (5+m) \sqrt {1-c^2 x^2}} \, dx\\ &=\frac {d^2 (f x)^{1+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f (1+m)}+\frac {2 d e (f x)^{3+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f^5 (5+m)}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {(f x)^m \left (d^2 \left (15+8 m+m^2\right )+2 d e \left (5+6 m+m^2\right ) x^2+e^2 \left (3+4 m+m^2\right ) x^4\right )}{\sqrt {1-c^2 x^2}} \, dx}{15+23 m+9 m^2+m^3}\\ &=-\frac {b e^2 (f x)^{3+m} \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{c^2 f^3 (4+m) (5+m)}+\frac {d^2 (f x)^{1+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f (1+m)}+\frac {2 d e (f x)^{3+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f^5 (5+m)}-\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {(f x)^m \left (-c^2 d^2 (3+m) (4+m) (5+m)-e (1+m) \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right ) x^2\right )}{\sqrt {1-c^2 x^2}} \, dx}{c^2 (4+m) \left (15+23 m+9 m^2+m^3\right )}\\ &=-\frac {b e \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right ) (f x)^{1+m} \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{c^4 f (2+m) (4+m) \left (15+8 m+m^2\right )}-\frac {b e^2 (f x)^{3+m} \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{c^2 f^3 (4+m) (5+m)}+\frac {d^2 (f x)^{1+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f (1+m)}+\frac {2 d e (f x)^{3+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f^5 (5+m)}+-\frac {\left (b \left (-c^4 d^2 (2+m) (3+m) (4+m) (5+m)-e (1+m)^2 \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right )\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {(f x)^m}{\sqrt {1-c^2 x^2}} \, dx}{c^4 (2+m) (4+m) \left (15+23 m+9 m^2+m^3\right )}\\ &=-\frac {b e \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right ) (f x)^{1+m} \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{c^4 f (2+m) (4+m) \left (15+8 m+m^2\right )}-\frac {b e^2 (f x)^{3+m} \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{c^2 f^3 (4+m) (5+m)}+\frac {d^2 (f x)^{1+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f (1+m)}+\frac {2 d e (f x)^{3+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f^5 (5+m)}+\frac {b \left (c^4 d^2 (2+m) (3+m) (4+m) (5+m)+e (1+m)^2 \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right )\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 )}{c^4 f (1+m) (2+m) (4+m) \left (15+23 m+9 m^2+m^3\right )}\\ \end {align*}

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Mathematica [F]  time = 0.16, size = 0, normalized size = 0.00 \[ \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \text {sech}^{-1}(c x)\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a e^{2} x^{4} + 2 \, a d e x^{2} + a d^{2} + {\left (b e^{2} x^{4} + 2 \, b d e x^{2} + b d^{2}\right )} \operatorname {arsech}\left (c x\right )\right )} \left (f x\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*e^2*x^4 + 2*a*d*e*x^2 + a*d^2 + (b*e^2*x^4 + 2*b*d*e*x^2 + b*d^2)*arcsech(c*x))*(f*x)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x^2 + d)^2*(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 \[ \int \left (f x \right )^{m} \left (e \,x^{2}+d \right )^{2} \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {a e^{2} f^{m} x^{5} x^{m}}{m + 5} + \frac {2 \, a d e f^{m} x^{3} x^{m}}{m + 3} + \frac {\left (f x\right )^{m + 1} a d^{2}}{f {\left (m + 1\right )}} + \frac {{\left ({\left (m^{2} + 4 \, m + 3\right )} b e^{2} f^{m} x^{5} x^{m} + 2 \, {\left (m^{2} + 6 \, m + 5\right )} b d e f^{m} x^{3} x^{m} + {\left (m^{2} + 8 \, m + 15\right )} b d^{2} f^{m} x x^{m}\right )} \log \left (\sqrt {c x + 1} \sqrt {-c x + 1} + 1\right ) - {\left ({\left (m^{2} + 4 \, m + 3\right )} b e^{2} f^{m} x^{5} x^{m} + 2 \, {\left (m^{2} + 6 \, m + 5\right )} b d e f^{m} x^{3} x^{m} + {\left (m^{2} + 8 \, m + 15\right )} b d^{2} f^{m} x x^{m}\right )} \log \relax (x)}{m^{3} + 9 \, m^{2} + 23 \, m + 15} - \int \frac {{\left (b c^{2} e^{2} f^{m} {\left (m + 5\right )} x^{2} \log \relax (c) - {\left (e^{2} f^{m} {\left (m + 5\right )} \log \relax (c) - e^{2} f^{m}\right )} b\right )} x^{4} x^{m}}{c^{2} {\left (m + 5\right )} x^{2} - m - 5}\,{d x} - \int \frac {2 \, {\left (b c^{2} d e f^{m} {\left (m + 3\right )} x^{2} \log \relax (c) - {\left (d e f^{m} {\left (m + 3\right )} \log \relax (c) - d e f^{m}\right )} b\right )} x^{2} x^{m}}{c^{2} {\left (m + 3\right )} x^{2} - m - 3}\,{d x} - \int \frac {{\left (b c^{2} d^{2} f^{m} {\left (m + 1\right )} x^{2} \log \relax (c) - {\left (d^{2} f^{m} {\left (m + 1\right )} \log \relax (c) - d^{2} f^{m}\right )} b\right )} x^{m}}{c^{2} {\left (m + 1\right )} x^{2} - m - 1}\,{d x} + \int \frac {{\left (m^{2} + 4 \, m + 3\right )} b c^{2} e^{2} f^{m} x^{6} x^{m} + 2 \, {\left (m^{2} + 6 \, m + 5\right )} b c^{2} d e f^{m} x^{4} x^{m} + {\left (m^{2} + 8 \, m + 15\right )} b c^{2} d^{2} f^{m} x^{2} x^{m}}{{\left (m^{3} + 9 \, m^{2} + 23 \, m + 15\right )} c^{2} x^{2} - m^{3} + {\left ({\left (m^{3} + 9 \, m^{2} + 23 \, m + 15\right )} c^{2} x^{2} - m^{3} - 9 \, m^{2} - 23 \, m - 15\right )} \sqrt {c x + 1} \sqrt {-c x + 1} - 9 \, m^{2} - 23 \, m - 15}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*e^2*f^m*x^5*x^m/(m + 5) + 2*a*d*e*f^m*x^3*x^m/(m + 3) + (f*x)^(m + 1)*a*d^2/(f*(m + 1)) + (((m^2 + 4*m + 3)*
b*e^2*f^m*x^5*x^m + 2*(m^2 + 6*m + 5)*b*d*e*f^m*x^3*x^m + (m^2 + 8*m + 15)*b*d^2*f^m*x*x^m)*log(sqrt(c*x + 1)*
sqrt(-c*x + 1) + 1) - ((m^2 + 4*m + 3)*b*e^2*f^m*x^5*x^m + 2*(m^2 + 6*m + 5)*b*d*e*f^m*x^3*x^m + (m^2 + 8*m +
15)*b*d^2*f^m*x*x^m)*log(x))/(m^3 + 9*m^2 + 23*m + 15) - integrate((b*c^2*e^2*f^m*(m + 5)*x^2*log(c) - (e^2*f^
m*(m + 5)*log(c) - e^2*f^m)*b)*x^4*x^m/(c^2*(m + 5)*x^2 - m - 5), x) - integrate(2*(b*c^2*d*e*f^m*(m + 3)*x^2*
log(c) - (d*e*f^m*(m + 3)*log(c) - d*e*f^m)*b)*x^2*x^m/(c^2*(m + 3)*x^2 - m - 3), x) - integrate((b*c^2*d^2*f^
m*(m + 1)*x^2*log(c) - (d^2*f^m*(m + 1)*log(c) - d^2*f^m)*b)*x^m/(c^2*(m + 1)*x^2 - m - 1), x) + integrate(((m
^2 + 4*m + 3)*b*c^2*e^2*f^m*x^6*x^m + 2*(m^2 + 6*m + 5)*b*c^2*d*e*f^m*x^4*x^m + (m^2 + 8*m + 15)*b*c^2*d^2*f^m
*x^2*x^m)/((m^3 + 9*m^2 + 23*m + 15)*c^2*x^2 - m^3 + ((m^3 + 9*m^2 + 23*m + 15)*c^2*x^2 - m^3 - 9*m^2 - 23*m -
15)*sqrt(c*x + 1)*sqrt(-c*x + 1) - 9*m^2 - 23*m - 15), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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