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

Optimal. Leaf size=379 \[ -\frac{b x \sqrt{c^2 x^2+1} (f x)^{m+1} \left (c^4 d^2 (m+2) (m+3) (m+4) (m+5)+e (m+1)^2 \left (e (m+3)^2-2 c^2 d \left (m^2+9 m+20\right )\right )\right ) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},-c^2 x^2\right )}{c^3 f (m+1)^2 (m+2) (m+3) (m+4) (m+5) \sqrt{-c^2 x^2} \sqrt{-c^2 x^2-1}}+\frac{d^2 (f x)^{m+1} \left (a+b \text{csch}^{-1}(c x)\right )}{f (m+1)}+\frac{2 d e (f x)^{m+3} \left (a+b \text{csch}^{-1}(c x)\right )}{f^3 (m+3)}+\frac{e^2 (f x)^{m+5} \left (a+b \text{csch}^{-1}(c x)\right )}{f^5 (m+5)}-\frac{b e x \sqrt{-c^2 x^2-1} (f x)^{m+1} \left (e (m+3)^2-2 c^2 d \left (m^2+9 m+20\right )\right )}{c^3 f (m+2) (m+3) (m+4) (m+5) \sqrt{-c^2 x^2}}+\frac{b e^2 x \sqrt{-c^2 x^2-1} (f x)^{m+3}}{c f^3 (m+4) (m+5) \sqrt{-c^2 x^2}} \]

[Out]

-((b*e*(e*(3 + m)^2 - 2*c^2*d*(20 + 9*m + m^2))*x*(f*x)^(1 + m)*Sqrt[-1 - c^2*x^2])/(c^3*f*(2 + m)*(3 + m)*(4
+ m)*(5 + m)*Sqrt[-(c^2*x^2)])) + (b*e^2*x*(f*x)^(3 + m)*Sqrt[-1 - c^2*x^2])/(c*f^3*(4 + m)*(5 + m)*Sqrt[-(c^2
*x^2)]) + (d^2*(f*x)^(1 + m)*(a + b*ArcCsch[c*x]))/(f*(1 + m)) + (2*d*e*(f*x)^(3 + m)*(a + b*ArcCsch[c*x]))/(f
^3*(3 + m)) + (e^2*(f*x)^(5 + m)*(a + b*ArcCsch[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*(20 + 9*m + m^2)))*x*(f*x)^(1 + m)*Sqrt[1 + c^2*x^2]*Hypergeometric2
F1[1/2, (1 + m)/2, (3 + m)/2, -(c^2*x^2)])/(c^3*f*(1 + m)^2*(2 + m)*(3 + m)*(4 + m)*(5 + m)*Sqrt[-(c^2*x^2)]*S
qrt[-1 - c^2*x^2])

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Rubi [A]  time = 0.491282, antiderivative size = 360, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {270, 6302, 12, 1267, 459, 365, 364} \[ \frac{d^2 (f x)^{m+1} \left (a+b \text{csch}^{-1}(c x)\right )}{f (m+1)}+\frac{2 d e (f x)^{m+3} \left (a+b \text{csch}^{-1}(c x)\right )}{f^3 (m+3)}+\frac{e^2 (f x)^{m+5} \left (a+b \text{csch}^{-1}(c x)\right )}{f^5 (m+5)}-\frac{b c x \sqrt{c^2 x^2+1} (f x)^{m+1} \left (\frac{e \left (e (m+3)^2-2 c^2 d \left (m^2+9 m+20\right )\right )}{c^4 (m+2) (m+3) (m+4) (m+5)}+\frac{d^2}{(m+1)^2}\right ) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};-c^2 x^2\right )}{f \sqrt{-c^2 x^2} \sqrt{-c^2 x^2-1}}-\frac{b e x \sqrt{-c^2 x^2-1} (f x)^{m+1} \left (e (m+3)^2-2 c^2 d \left (m^2+9 m+20\right )\right )}{c^3 f (m+2) (m+3) (m+4) (m+5) \sqrt{-c^2 x^2}}+\frac{b e^2 x \sqrt{-c^2 x^2-1} (f x)^{m+3}}{c f^3 (m+4) (m+5) \sqrt{-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*e*(e*(3 + m)^2 - 2*c^2*d*(20 + 9*m + m^2))*x*(f*x)^(1 + m)*Sqrt[-1 - c^2*x^2])/(c^3*f*(2 + m)*(3 + m)*(4
+ m)*(5 + m)*Sqrt[-(c^2*x^2)])) + (b*e^2*x*(f*x)^(3 + m)*Sqrt[-1 - c^2*x^2])/(c*f^3*(4 + m)*(5 + m)*Sqrt[-(c^2
*x^2)]) + (d^2*(f*x)^(1 + m)*(a + b*ArcCsch[c*x]))/(f*(1 + m)) + (2*d*e*(f*x)^(3 + m)*(a + b*ArcCsch[c*x]))/(f
^3*(3 + m)) + (e^2*(f*x)^(5 + m)*(a + b*ArcCsch[c*x]))/(f^5*(5 + m)) - (b*c*(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)))*x*(f*x)^(1 + m)*Sqrt[1 + c^2*x^2]*Hypergeom
etric2F1[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 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 6302

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]))

Rule 12

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

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 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 365

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 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])

Rubi steps

\begin{align*} \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \text{csch}^{-1}(c x)\right ) \, dx &=\frac{d^2 (f x)^{1+m} \left (a+b \text{csch}^{-1}(c x)\right )}{f (1+m)}+\frac{2 d e (f x)^{3+m} \left (a+b \text{csch}^{-1}(c x)\right )}{f^3 (3+m)}+\frac{e^2 (f x)^{5+m} \left (a+b \text{csch}^{-1}(c x)\right )}{f^5 (5+m)}-\frac{(b c x) \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}{\sqrt{-c^2 x^2}}\\ &=\frac{d^2 (f x)^{1+m} \left (a+b \text{csch}^{-1}(c x)\right )}{f (1+m)}+\frac{2 d e (f x)^{3+m} \left (a+b \text{csch}^{-1}(c x)\right )}{f^3 (3+m)}+\frac{e^2 (f x)^{5+m} \left (a+b \text{csch}^{-1}(c x)\right )}{f^5 (5+m)}-\frac{(b c x) \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}{\left (15+23 m+9 m^2+m^3\right ) \sqrt{-c^2 x^2}}\\ &=\frac{b e^2 x (f x)^{3+m} \sqrt{-1-c^2 x^2}}{c f^3 (4+m) (5+m) \sqrt{-c^2 x^2}}+\frac{d^2 (f x)^{1+m} \left (a+b \text{csch}^{-1}(c x)\right )}{f (1+m)}+\frac{2 d e (f x)^{3+m} \left (a+b \text{csch}^{-1}(c x)\right )}{f^3 (3+m)}+\frac{e^2 (f x)^{5+m} \left (a+b \text{csch}^{-1}(c x)\right )}{f^5 (5+m)}+\frac{(b x) \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 (4+m) \left (15+23 m+9 m^2+m^3\right ) \sqrt{-c^2 x^2}}\\ &=-\frac{b e \left (e (3+m)^2-2 c^2 d \left (20+9 m+m^2\right )\right ) x (f x)^{1+m} \sqrt{-1-c^2 x^2}}{c^3 f (2+m) (4+m) \left (15+8 m+m^2\right ) \sqrt{-c^2 x^2}}+\frac{b e^2 x (f x)^{3+m} \sqrt{-1-c^2 x^2}}{c f^3 (4+m) (5+m) \sqrt{-c^2 x^2}}+\frac{d^2 (f x)^{1+m} \left (a+b \text{csch}^{-1}(c x)\right )}{f (1+m)}+\frac{2 d e (f x)^{3+m} \left (a+b \text{csch}^{-1}(c x)\right )}{f^3 (3+m)}+\frac{e^2 (f x)^{5+m} \left (a+b \text{csch}^{-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 ) x\right ) \int \frac{(f x)^m}{\sqrt{-1-c^2 x^2}} \, dx}{c^3 (2+m) (4+m) \left (15+23 m+9 m^2+m^3\right ) \sqrt{-c^2 x^2}}\\ &=-\frac{b e \left (e (3+m)^2-2 c^2 d \left (20+9 m+m^2\right )\right ) x (f x)^{1+m} \sqrt{-1-c^2 x^2}}{c^3 f (2+m) (4+m) \left (15+8 m+m^2\right ) \sqrt{-c^2 x^2}}+\frac{b e^2 x (f x)^{3+m} \sqrt{-1-c^2 x^2}}{c f^3 (4+m) (5+m) \sqrt{-c^2 x^2}}+\frac{d^2 (f x)^{1+m} \left (a+b \text{csch}^{-1}(c x)\right )}{f (1+m)}+\frac{2 d e (f x)^{3+m} \left (a+b \text{csch}^{-1}(c x)\right )}{f^3 (3+m)}+\frac{e^2 (f x)^{5+m} \left (a+b \text{csch}^{-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 ) x \sqrt{1+c^2 x^2}\right ) \int \frac{(f x)^m}{\sqrt{1+c^2 x^2}} \, dx}{c^3 (2+m) (4+m) \left (15+23 m+9 m^2+m^3\right ) \sqrt{-c^2 x^2} \sqrt{-1-c^2 x^2}}\\ &=-\frac{b e \left (e (3+m)^2-2 c^2 d \left (20+9 m+m^2\right )\right ) x (f x)^{1+m} \sqrt{-1-c^2 x^2}}{c^3 f (2+m) (4+m) \left (15+8 m+m^2\right ) \sqrt{-c^2 x^2}}+\frac{b e^2 x (f x)^{3+m} \sqrt{-1-c^2 x^2}}{c f^3 (4+m) (5+m) \sqrt{-c^2 x^2}}+\frac{d^2 (f x)^{1+m} \left (a+b \text{csch}^{-1}(c x)\right )}{f (1+m)}+\frac{2 d e (f x)^{3+m} \left (a+b \text{csch}^{-1}(c x)\right )}{f^3 (3+m)}+\frac{e^2 (f x)^{5+m} \left (a+b \text{csch}^{-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 ) 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^3 f (1+m) (2+m) (4+m) \left (15+23 m+9 m^2+m^3\right ) \sqrt{-c^2 x^2} \sqrt{-1-c^2 x^2}}\\ \end{align*}

Mathematica [F]  time = 0.139643, size = 0, normalized size = 0. \[ \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \text{csch}^{-1}(c x)\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.374, size = 0, normalized size = 0. \begin{align*} \int \left ( fx \right ) ^{m} \left ( e{x}^{2}+d \right ) ^{2} \left ( a+b{\rm arccsch} \left (cx\right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((f*x)^m*(e*x^2+d)^2*(a+b*arccsch(c*x)),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((f*x)^m*(e*x^2+d)^2*(a+b*arccsch(c*x)),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 (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{arcsch}\left (c x\right )\right )} \left (f x\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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