3.2.67 \(\int (f x)^m (d+e x^2) (a+b \text {csch}^{-1}(c x)) \, dx\) [167]

Optimal. Leaf size=220 \[ \frac {b e x (f x)^{1+m} \sqrt {-1-c^2 x^2}}{c f \left (6+5 m+m^2\right ) \sqrt {-c^2 x^2}}+\frac {d (f x)^{1+m} \left (a+b \text {csch}^{-1}(c x)\right )}{f (1+m)}+\frac {e (f x)^{3+m} \left (a+b \text {csch}^{-1}(c x)\right )}{f^3 (3+m)}+\frac {b \left (e (1+m)^2-c^2 d (2+m) (3+m)\right ) x (f x)^{1+m} \sqrt {1+c^2 x^2} \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};-c^2 x^2\right )}{c f (1+m)^2 (2+m) (3+m) \sqrt {-c^2 x^2} \sqrt {-1-c^2 x^2}} \]

[Out]

d*(f*x)^(1+m)*(a+b*arccsch(c*x))/f/(1+m)+e*(f*x)^(3+m)*(a+b*arccsch(c*x))/f^3/(3+m)+b*e*x*(f*x)^(1+m)*(-c^2*x^
2-1)^(1/2)/c/f/(m^2+5*m+6)/(-c^2*x^2)^(1/2)+b*(e*(1+m)^2-c^2*d*(2+m)*(3+m))*x*(f*x)^(1+m)*hypergeom([1/2, 1/2+
1/2*m],[3/2+1/2*m],-c^2*x^2)*(c^2*x^2+1)^(1/2)/c/f/(1+m)^2/(2+m)/(3+m)/(-c^2*x^2)^(1/2)/(-c^2*x^2-1)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.15, antiderivative size = 208, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {14, 6437, 12, 470, 372, 371} \begin {gather*} \frac {d (f x)^{m+1} \left (a+b \text {csch}^{-1}(c x)\right )}{f (m+1)}+\frac {e (f x)^{m+3} \left (a+b \text {csch}^{-1}(c x)\right )}{f^3 (m+3)}-\frac {b c x \sqrt {c^2 x^2+1} (f x)^{m+1} \left (\frac {d}{(m+1)^2}-\frac {e}{c^2 (m+2) (m+3)}\right ) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};-c^2 x^2\right )}{f \sqrt {-c^2 x^2} \sqrt {-c^2 x^2-1}}+\frac {b e x \sqrt {-c^2 x^2-1} (f x)^{m+1}}{c f \left (m^2+5 m+6\right ) \sqrt {-c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*e*x*(f*x)^(1 + m)*Sqrt[-1 - c^2*x^2])/(c*f*(6 + 5*m + m^2)*Sqrt[-(c^2*x^2)]) + (d*(f*x)^(1 + m)*(a + b*ArcC
sch[c*x]))/(f*(1 + m)) + (e*(f*x)^(3 + m)*(a + b*ArcCsch[c*x]))/(f^3*(3 + m)) - (b*c*(d/(1 + m)^2 - e/(c^2*(2
+ m)*(3 + m)))*x*(f*x)^(1 + m)*Sqrt[1 + c^2*x^2]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, -(c^2*x^2)])/(f*
Sqrt[-(c^2*x^2)]*Sqrt[-1 - c^2*x^2])

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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

Rule 470

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 6437

Int[((a_.) + ArcCsch[(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*ArcCsch[c*x], u, x] - Dist[b*c*(x/Sqrt[(-c^2)*x^2]), Int[Simp
lifyIntegrand[u/(x*Sqrt[-1 - c^2*x^2]), 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 {csch}^{-1}(c x)\right ) \, dx &=\frac {d (f x)^{1+m} \left (a+b \text {csch}^{-1}(c x)\right )}{f (1+m)}+\frac {e (f x)^{3+m} \left (a+b \text {csch}^{-1}(c x)\right )}{f^3 (3+m)}-\frac {(b c x) \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}{\sqrt {-c^2 x^2}}\\ &=\frac {d (f x)^{1+m} \left (a+b \text {csch}^{-1}(c x)\right )}{f (1+m)}+\frac {e (f x)^{3+m} \left (a+b \text {csch}^{-1}(c x)\right )}{f^3 (3+m)}-\frac {(b c x) \int \frac {(f x)^m \left (d (3+m)+e (1+m) x^2\right )}{\sqrt {-1-c^2 x^2}} \, dx}{\left (3+4 m+m^2\right ) \sqrt {-c^2 x^2}}\\ &=\frac {b e x (f x)^{1+m} \sqrt {-1-c^2 x^2}}{c f \left (6+5 m+m^2\right ) \sqrt {-c^2 x^2}}+\frac {d (f x)^{1+m} \left (a+b \text {csch}^{-1}(c x)\right )}{f (1+m)}+\frac {e (f x)^{3+m} \left (a+b \text {csch}^{-1}(c x)\right )}{f^3 (3+m)}-\frac {\left (b c \left (-\frac {e (1+m)^2}{c^2 (2+m)}+d (3+m)\right ) x\right ) \int \frac {(f x)^m}{\sqrt {-1-c^2 x^2}} \, dx}{\left (3+4 m+m^2\right ) \sqrt {-c^2 x^2}}\\ &=\frac {b e x (f x)^{1+m} \sqrt {-1-c^2 x^2}}{c f \left (6+5 m+m^2\right ) \sqrt {-c^2 x^2}}+\frac {d (f x)^{1+m} \left (a+b \text {csch}^{-1}(c x)\right )}{f (1+m)}+\frac {e (f x)^{3+m} \left (a+b \text {csch}^{-1}(c x)\right )}{f^3 (3+m)}-\frac {\left (b c \left (-\frac {e (1+m)^2}{c^2 (2+m)}+d (3+m)\right ) x \sqrt {1+c^2 x^2}\right ) \int \frac {(f x)^m}{\sqrt {1+c^2 x^2}} \, dx}{\left (3+4 m+m^2\right ) \sqrt {-c^2 x^2} \sqrt {-1-c^2 x^2}}\\ &=\frac {b e x (f x)^{1+m} \sqrt {-1-c^2 x^2}}{c f \left (6+5 m+m^2\right ) \sqrt {-c^2 x^2}}+\frac {d (f x)^{1+m} \left (a+b \text {csch}^{-1}(c x)\right )}{f (1+m)}+\frac {e (f x)^{3+m} \left (a+b \text {csch}^{-1}(c x)\right )}{f^3 (3+m)}+\frac {b c \left (\frac {e (1+m)^2}{c^2 (2+m)}-d (3+m)\right ) x (f x)^{1+m} \sqrt {1+c^2 x^2} \, _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 ) \sqrt {-c^2 x^2} \sqrt {-1-c^2 x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]
time = 0.06, size = 0, normalized size = 0.00 \begin {gather*} \int (f x)^m \left (d+e x^2\right ) \left (a+b \text {csch}^{-1}(c x)\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \left (f x \right )^{m} \left (e \,x^{2}+d \right ) \left (a +b \,\mathrm {arccsch}\left (c x \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (f x\right )^{m} \left (a + b \operatorname {acsch}{\left (c x \right )}\right ) \left (d + e x^{2}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________