3.6.19 \(\int (d+c^2 d x^2)^{3/2} (a+b \sinh ^{-1}(c x))^n \, dx\) [519]

Optimal. Leaf size=420 \[ \frac {3 d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^{1+n}}{8 b c (1+n) \sqrt {1+c^2 x^2}}+\frac {2^{-2 (3+n)} d e^{-\frac {4 a}{b}} \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \Gamma \left (1+n,-\frac {4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{c \sqrt {1+c^2 x^2}}+\frac {2^{-3-n} d e^{-\frac {2 a}{b}} \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \Gamma \left (1+n,-\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{c \sqrt {1+c^2 x^2}}-\frac {2^{-3-n} d e^{\frac {2 a}{b}} \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^n \left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \Gamma \left (1+n,\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{c \sqrt {1+c^2 x^2}}-\frac {2^{-2 (3+n)} d e^{\frac {4 a}{b}} \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^n \left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \Gamma \left (1+n,\frac {4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{c \sqrt {1+c^2 x^2}} \]

[Out]

3/8*d*(a+b*arcsinh(c*x))^(1+n)*(c^2*d*x^2+d)^(1/2)/b/c/(1+n)/(c^2*x^2+1)^(1/2)+d*(a+b*arcsinh(c*x))^n*GAMMA(1+
n,-4*(a+b*arcsinh(c*x))/b)*(c^2*d*x^2+d)^(1/2)/(2^(6+2*n))/c/exp(4*a/b)/(((-a-b*arcsinh(c*x))/b)^n)/(c^2*x^2+1
)^(1/2)+2^(-3-n)*d*(a+b*arcsinh(c*x))^n*GAMMA(1+n,-2*(a+b*arcsinh(c*x))/b)*(c^2*d*x^2+d)^(1/2)/c/exp(2*a/b)/((
(-a-b*arcsinh(c*x))/b)^n)/(c^2*x^2+1)^(1/2)-2^(-3-n)*d*exp(2*a/b)*(a+b*arcsinh(c*x))^n*GAMMA(1+n,2*(a+b*arcsin
h(c*x))/b)*(c^2*d*x^2+d)^(1/2)/c/(((a+b*arcsinh(c*x))/b)^n)/(c^2*x^2+1)^(1/2)-d*exp(4*a/b)*(a+b*arcsinh(c*x))^
n*GAMMA(1+n,4*(a+b*arcsinh(c*x))/b)*(c^2*d*x^2+d)^(1/2)/(2^(6+2*n))/c/(((a+b*arcsinh(c*x))/b)^n)/(c^2*x^2+1)^(
1/2)

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Rubi [A]
time = 0.26, antiderivative size = 420, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {5791, 3393, 3388, 2212} \begin {gather*} \frac {d 2^{-2 (n+3)} e^{-\frac {4 a}{b}} \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \text {Gamma}\left (n+1,-\frac {4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{c \sqrt {c^2 x^2+1}}+\frac {d 2^{-n-3} e^{-\frac {2 a}{b}} \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \text {Gamma}\left (n+1,-\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{c \sqrt {c^2 x^2+1}}-\frac {d 2^{-n-3} e^{\frac {2 a}{b}} \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^n \left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \text {Gamma}\left (n+1,\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{c \sqrt {c^2 x^2+1}}-\frac {d 2^{-2 (n+3)} e^{\frac {4 a}{b}} \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^n \left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \text {Gamma}\left (n+1,\frac {4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{c \sqrt {c^2 x^2+1}}+\frac {3 d \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^{n+1}}{8 b c (n+1) \sqrt {c^2 x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x])^n,x]

[Out]

(3*d*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^(1 + n))/(8*b*c*(1 + n)*Sqrt[1 + c^2*x^2]) + (d*Sqrt[d + c^2*d*x
^2]*(a + b*ArcSinh[c*x])^n*Gamma[1 + n, (-4*(a + b*ArcSinh[c*x]))/b])/(2^(2*(3 + n))*c*E^((4*a)/b)*Sqrt[1 + c^
2*x^2]*(-((a + b*ArcSinh[c*x])/b))^n) + (2^(-3 - n)*d*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^n*Gamma[1 + n,
(-2*(a + b*ArcSinh[c*x]))/b])/(c*E^((2*a)/b)*Sqrt[1 + c^2*x^2]*(-((a + b*ArcSinh[c*x])/b))^n) - (2^(-3 - n)*d*
E^((2*a)/b)*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^n*Gamma[1 + n, (2*(a + b*ArcSinh[c*x]))/b])/(c*Sqrt[1 + c
^2*x^2]*((a + b*ArcSinh[c*x])/b)^n) - (d*E^((4*a)/b)*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^n*Gamma[1 + n, (
4*(a + b*ArcSinh[c*x]))/b])/(2^(2*(3 + n))*c*Sqrt[1 + c^2*x^2]*((a + b*ArcSinh[c*x])/b)^n)

Rule 2212

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-F^(g*(e - c*(f/d))))*((c
+ d*x)^FracPart[m]/(d*((-f)*g*(Log[F]/d))^(IntPart[m] + 1)*((-f)*g*Log[F]*((c + d*x)/d))^FracPart[m]))*Gamma[m
+ 1, ((-f)*g*(Log[F]/d))*(c + d*x)], x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rule 3388

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/(E^(
I*k*Pi)*E^(I*(e + f*x))), x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*k*Pi)*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d
, e, f, m}, x] && IntegerQ[2*k]

Rule 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 5791

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[(1/(b*c))*Simp[(d
+ e*x^2)^p/(1 + c^2*x^2)^p], Subst[Int[x^n*Cosh[-a/b + x/b]^(2*p + 1), x], x, a + b*ArcSinh[c*x]], x] /; Free
Q[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && IGtQ[2*p, 0]

Rubi steps

\begin {align*} \int \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^n \, dx &=\frac {\left (d \sqrt {d+c^2 d x^2}\right ) \int \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^n \, dx}{\sqrt {1+c^2 x^2}}\\ &=\frac {\left (d \sqrt {d+c^2 d x^2}\right ) \text {Subst}\left (\int (a+b x)^n \cosh ^4(x) \, dx,x,\sinh ^{-1}(c x)\right )}{c \sqrt {1+c^2 x^2}}\\ &=\frac {\left (d \sqrt {d+c^2 d x^2}\right ) \text {Subst}\left (\int \left (\frac {3}{8} (a+b x)^n+\frac {1}{2} (a+b x)^n \cosh (2 x)+\frac {1}{8} (a+b x)^n \cosh (4 x)\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c \sqrt {1+c^2 x^2}}\\ &=\frac {3 d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^{1+n}}{8 b c (1+n) \sqrt {1+c^2 x^2}}+\frac {\left (d \sqrt {d+c^2 d x^2}\right ) \text {Subst}\left (\int (a+b x)^n \cosh (4 x) \, dx,x,\sinh ^{-1}(c x)\right )}{8 c \sqrt {1+c^2 x^2}}+\frac {\left (d \sqrt {d+c^2 d x^2}\right ) \text {Subst}\left (\int (a+b x)^n \cosh (2 x) \, dx,x,\sinh ^{-1}(c x)\right )}{2 c \sqrt {1+c^2 x^2}}\\ &=\frac {3 d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^{1+n}}{8 b c (1+n) \sqrt {1+c^2 x^2}}+\frac {\left (d \sqrt {d+c^2 d x^2}\right ) \text {Subst}\left (\int e^{-4 x} (a+b x)^n \, dx,x,\sinh ^{-1}(c x)\right )}{16 c \sqrt {1+c^2 x^2}}+\frac {\left (d \sqrt {d+c^2 d x^2}\right ) \text {Subst}\left (\int e^{4 x} (a+b x)^n \, dx,x,\sinh ^{-1}(c x)\right )}{16 c \sqrt {1+c^2 x^2}}+\frac {\left (d \sqrt {d+c^2 d x^2}\right ) \text {Subst}\left (\int e^{-2 x} (a+b x)^n \, dx,x,\sinh ^{-1}(c x)\right )}{4 c \sqrt {1+c^2 x^2}}+\frac {\left (d \sqrt {d+c^2 d x^2}\right ) \text {Subst}\left (\int e^{2 x} (a+b x)^n \, dx,x,\sinh ^{-1}(c x)\right )}{4 c \sqrt {1+c^2 x^2}}\\ &=\frac {3 d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^{1+n}}{8 b c (1+n) \sqrt {1+c^2 x^2}}+\frac {4^{-3-n} d e^{-\frac {4 a}{b}} \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \Gamma \left (1+n,-\frac {4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{c \sqrt {1+c^2 x^2}}+\frac {2^{-3-n} d e^{-\frac {2 a}{b}} \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^n \left (-\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \Gamma \left (1+n,-\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{c \sqrt {1+c^2 x^2}}-\frac {2^{-3-n} d e^{\frac {2 a}{b}} \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^n \left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \Gamma \left (1+n,\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{c \sqrt {1+c^2 x^2}}-\frac {4^{-3-n} d e^{\frac {4 a}{b}} \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^n \left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \Gamma \left (1+n,\frac {4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{c \sqrt {1+c^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 1.00, size = 287, normalized size = 0.68 \begin {gather*} \frac {d^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^n \left (-\frac {8 \left (a+b \sinh ^{-1}(c x)\right )}{b+b n}+8 \left (\frac {4 a+4 b \sinh ^{-1}(c x)}{b+b n}+2^{-n} e^{-\frac {2 a}{b}} \left (-\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \Gamma \left (1+n,-\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )-2^{-n} e^{\frac {2 a}{b}} \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )^{-n} \Gamma \left (1+n,\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )\right )+4^{-n} e^{-\frac {4 a}{b}} \left (-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{b^2}\right )^{-n} \left (\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )^n \Gamma \left (1+n,-\frac {4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )-e^{\frac {8 a}{b}} \left (-\frac {a+b \sinh ^{-1}(c x)}{b}\right )^n \Gamma \left (1+n,\frac {4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )\right )\right )}{64 c \sqrt {d+c^2 d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x])^n,x]

[Out]

(d^2*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^n*((-8*(a + b*ArcSinh[c*x]))/(b + b*n) + 8*((4*a + 4*b*ArcSinh[c*x
])/(b + b*n) + Gamma[1 + n, (-2*(a + b*ArcSinh[c*x]))/b]/(2^n*E^((2*a)/b)*(-((a + b*ArcSinh[c*x])/b))^n) - (E^
((2*a)/b)*Gamma[1 + n, (2*(a + b*ArcSinh[c*x]))/b])/(2^n*(a/b + ArcSinh[c*x])^n)) + ((a/b + ArcSinh[c*x])^n*Ga
mma[1 + n, (-4*(a + b*ArcSinh[c*x]))/b] - E^((8*a)/b)*(-((a + b*ArcSinh[c*x])/b))^n*Gamma[1 + n, (4*(a + b*Arc
Sinh[c*x]))/b])/(4^n*E^((4*a)/b)*(-((a + b*ArcSinh[c*x])^2/b^2))^n)))/(64*c*Sqrt[d + c^2*d*x^2])

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \arcsinh \left (c x \right )\right )^{n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))^n,x)

[Out]

int((c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))^n,x)

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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((c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))^n,x, algorithm="maxima")

[Out]

integrate((c^2*d*x^2 + d)^(3/2)*(b*arcsinh(c*x) + a)^n, x)

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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((c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))^n,x, algorithm="fricas")

[Out]

integral((c^2*d*x^2 + d)^(3/2)*(b*arcsinh(c*x) + a)^n, x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c**2*d*x**2+d)**(3/2)*(a+b*asinh(c*x))**n,x)

[Out]

Timed out

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))^n,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^n\,{\left (d\,c^2\,x^2+d\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c*x))^n*(d + c^2*d*x^2)^(3/2),x)

[Out]

int((a + b*asinh(c*x))^n*(d + c^2*d*x^2)^(3/2), x)

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