3.187 \(\int x^m (d+c^2 d x^2) (a+b \sinh ^{-1}(c x)) \, dx\)

Optimal. Leaf size=128 \[ \frac {c^2 d x^{m+3} \left (a+b \sinh ^{-1}(c x)\right )}{m+3}+\frac {d x^{m+1} \left (a+b \sinh ^{-1}(c x)\right )}{m+1}-\frac {b c d (3 m+7) x^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};-c^2 x^2\right )}{(m+1) (m+2) (m+3)^2}-\frac {b c d \sqrt {c^2 x^2+1} x^{m+2}}{(m+3)^2} \]

[Out]

d*x^(1+m)*(a+b*arcsinh(c*x))/(1+m)+c^2*d*x^(3+m)*(a+b*arcsinh(c*x))/(3+m)-b*c*d*(7+3*m)*x^(2+m)*hypergeom([1/2
, 1+1/2*m],[2+1/2*m],-c^2*x^2)/(3+m)^2/(m^2+3*m+2)-b*c*d*x^(2+m)*(c^2*x^2+1)^(1/2)/(3+m)^2

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Rubi [A]  time = 0.13, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {14, 5730, 12, 459, 364} \[ \frac {c^2 d x^{m+3} \left (a+b \sinh ^{-1}(c x)\right )}{m+3}+\frac {d x^{m+1} \left (a+b \sinh ^{-1}(c x)\right )}{m+1}-\frac {b c d (3 m+7) x^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};-c^2 x^2\right )}{(m+1) (m+2) (m+3)^2}-\frac {b c d \sqrt {c^2 x^2+1} x^{m+2}}{(m+3)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*c*d*x^(2 + m)*Sqrt[1 + c^2*x^2])/(3 + m)^2) + (d*x^(1 + m)*(a + b*ArcSinh[c*x]))/(1 + m) + (c^2*d*x^(3 +
m)*(a + b*ArcSinh[c*x]))/(3 + m) - (b*c*d*(7 + 3*m)*x^(2 + m)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, -(c
^2*x^2)])/((1 + m)*(2 + m)*(3 + m)^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 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 5730

Int[((a_.) + ArcSinh[(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*ArcSinh[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1
+ c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int x^m \left (d+c^2 d x^2\right ) \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac {d x^{1+m} \left (a+b \sinh ^{-1}(c x)\right )}{1+m}+\frac {c^2 d x^{3+m} \left (a+b \sinh ^{-1}(c x)\right )}{3+m}-(b c) \int \frac {d x^{1+m} \left (\frac {1}{1+m}+\frac {c^2 x^2}{3+m}\right )}{\sqrt {1+c^2 x^2}} \, dx\\ &=\frac {d x^{1+m} \left (a+b \sinh ^{-1}(c x)\right )}{1+m}+\frac {c^2 d x^{3+m} \left (a+b \sinh ^{-1}(c x)\right )}{3+m}-(b c d) \int \frac {x^{1+m} \left (\frac {1}{1+m}+\frac {c^2 x^2}{3+m}\right )}{\sqrt {1+c^2 x^2}} \, dx\\ &=-\frac {b c d x^{2+m} \sqrt {1+c^2 x^2}}{(3+m)^2}+\frac {d x^{1+m} \left (a+b \sinh ^{-1}(c x)\right )}{1+m}+\frac {c^2 d x^{3+m} \left (a+b \sinh ^{-1}(c x)\right )}{3+m}-\frac {(b c d (7+3 m)) \int \frac {x^{1+m}}{\sqrt {1+c^2 x^2}} \, dx}{(1+m) (3+m)^2}\\ &=-\frac {b c d x^{2+m} \sqrt {1+c^2 x^2}}{(3+m)^2}+\frac {d x^{1+m} \left (a+b \sinh ^{-1}(c x)\right )}{1+m}+\frac {c^2 d x^{3+m} \left (a+b \sinh ^{-1}(c x)\right )}{3+m}-\frac {b c d (7+3 m) x^{2+m} \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};-c^2 x^2\right )}{(1+m) (2+m) (3+m)^2}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 118, normalized size = 0.92 \[ \frac {d x^{m+1} \left ((m+2) \left (c^2 m x^2+c^2 x^2+m+3\right ) \left (a+b \sinh ^{-1}(c x)\right )-b c (m+1) x \, _2F_1\left (-\frac {1}{2},\frac {m}{2}+1;\frac {m}{2}+2;-c^2 x^2\right )-2 b c x \, _2F_1\left (\frac {1}{2},\frac {m}{2}+1;\frac {m}{2}+2;-c^2 x^2\right )\right )}{(m+1) (m+2) (m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a c^{2} d x^{2} + a d + {\left (b c^{2} d x^{2} + b d\right )} \operatorname {arsinh}\left (c x\right )\right )} x^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*c^2*d*x^2 + a*d + (b*c^2*d*x^2 + b*d)*arcsinh(c*x))*x^m, x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(c^2*d*x^2+d)*(a+b*arcsinh(c*x)),x)

[Out]

int(x^m*(c^2*d*x^2+d)*(a+b*arcsinh(c*x)),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^m\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,\left (d\,c^2\,x^2+d\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(a + b*asinh(c*x))*(d + c^2*d*x^2),x)

[Out]

int(x^m*(a + b*asinh(c*x))*(d + c^2*d*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*(c**2*d*x**2+d)*(a+b*asinh(c*x)),x)

[Out]

d*(Integral(a*x**m, x) + Integral(b*x**m*asinh(c*x), x) + Integral(a*c**2*x**2*x**m, x) + Integral(b*c**2*x**2
*x**m*asinh(c*x), x))

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