\(\int \frac {a+b \text {arcsinh}(c x)}{(d x)^{5/2}} \, dx\) [82]

Optimal result

Integrand size = 16, antiderivative size = 248 \[ \int \frac {a+b \text {arcsinh}(c x)}{(d x)^{5/2}} \, dx=-\frac {4 b c \sqrt {1+c^2 x^2}}{3 d^2 \sqrt {d x}}+\frac {4 b c^2 \sqrt {d x} \sqrt {1+c^2 x^2}}{3 d^3 (1+c x)}-\frac {2 (a+b \text {arcsinh}(c x))}{3 d (d x)^{3/2}}-\frac {4 b c^{3/2} (1+c x) \sqrt {\frac {1+c^2 x^2}{(1+c x)^2}} E\left (2 \arctan \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )|\frac {1}{2}\right )}{3 d^{5/2} \sqrt {1+c^2 x^2}}+\frac {2 b c^{3/2} (1+c x) \sqrt {\frac {1+c^2 x^2}{(1+c x)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right ),\frac {1}{2}\right )}{3 d^{5/2} \sqrt {1+c^2 x^2}} \] Output:

-4/3*b*c*(c^2*x^2+1)^(1/2)/d^2/(d*x)^(1/2)+4/3*b*c^2*(d*x)^(1/2)*(c^2*x^2+ 
1)^(1/2)/d^3/(c*x+1)-2/3*(a+b*arcsinh(c*x))/d/(d*x)^(3/2)-4/3*b*c^(3/2)*(c 
*x+1)*((c^2*x^2+1)/(c*x+1)^2)^(1/2)*EllipticE(sin(2*arctan(c^(1/2)*(d*x)^( 
1/2)/d^(1/2))),1/2*2^(1/2))/d^(5/2)/(c^2*x^2+1)^(1/2)+2/3*b*c^(3/2)*(c*x+1 
)*((c^2*x^2+1)/(c*x+1)^2)^(1/2)*InverseJacobiAM(2*arctan(c^(1/2)*(d*x)^(1/ 
2)/d^(1/2)),1/2*2^(1/2))/d^(5/2)/(c^2*x^2+1)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.17 \[ \int \frac {a+b \text {arcsinh}(c x)}{(d x)^{5/2}} \, dx=-\frac {2 x \left (a+b \text {arcsinh}(c x)+2 b c x \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-c^2 x^2\right )\right )}{3 (d x)^{5/2}} \] Input:

Integrate[(a + b*ArcSinh[c*x])/(d*x)^(5/2),x]
 

Output:

(-2*x*(a + b*ArcSinh[c*x] + 2*b*c*x*Hypergeometric2F1[-1/4, 1/2, 3/4, -(c^ 
2*x^2)]))/(3*(d*x)^(5/2))
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {6191, 264, 266, 834, 27, 761, 1510}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {a+b \text {arcsinh}(c x)}{(d x)^{5/2}} \, dx\)

\(\Big \downarrow \) 6191

\(\displaystyle \frac {2 b c \int \frac {1}{(d x)^{3/2} \sqrt {c^2 x^2+1}}dx}{3 d}-\frac {2 (a+b \text {arcsinh}(c x))}{3 d (d x)^{3/2}}\)

\(\Big \downarrow \) 264

\(\displaystyle \frac {2 b c \left (\frac {c^2 \int \frac {\sqrt {d x}}{\sqrt {c^2 x^2+1}}dx}{d^2}-\frac {2 \sqrt {c^2 x^2+1}}{d \sqrt {d x}}\right )}{3 d}-\frac {2 (a+b \text {arcsinh}(c x))}{3 d (d x)^{3/2}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2 b c \left (\frac {2 c^2 \int \frac {d x}{\sqrt {c^2 x^2+1}}d\sqrt {d x}}{d^3}-\frac {2 \sqrt {c^2 x^2+1}}{d \sqrt {d x}}\right )}{3 d}-\frac {2 (a+b \text {arcsinh}(c x))}{3 d (d x)^{3/2}}\)

\(\Big \downarrow \) 834

\(\displaystyle \frac {2 b c \left (\frac {2 c^2 \left (\frac {d \int \frac {1}{\sqrt {c^2 x^2+1}}d\sqrt {d x}}{c}-\frac {d \int \frac {d-c d x}{d \sqrt {c^2 x^2+1}}d\sqrt {d x}}{c}\right )}{d^3}-\frac {2 \sqrt {c^2 x^2+1}}{d \sqrt {d x}}\right )}{3 d}-\frac {2 (a+b \text {arcsinh}(c x))}{3 d (d x)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 b c \left (\frac {2 c^2 \left (\frac {d \int \frac {1}{\sqrt {c^2 x^2+1}}d\sqrt {d x}}{c}-\frac {\int \frac {d-c d x}{\sqrt {c^2 x^2+1}}d\sqrt {d x}}{c}\right )}{d^3}-\frac {2 \sqrt {c^2 x^2+1}}{d \sqrt {d x}}\right )}{3 d}-\frac {2 (a+b \text {arcsinh}(c x))}{3 d (d x)^{3/2}}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {2 b c \left (\frac {2 c^2 \left (\frac {\sqrt {d} (c d x+d) \sqrt {\frac {c^2 d^2 x^2+d^2}{(c d x+d)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right ),\frac {1}{2}\right )}{2 c^{3/2} \sqrt {c^2 x^2+1}}-\frac {\int \frac {d-c d x}{\sqrt {c^2 x^2+1}}d\sqrt {d x}}{c}\right )}{d^3}-\frac {2 \sqrt {c^2 x^2+1}}{d \sqrt {d x}}\right )}{3 d}-\frac {2 (a+b \text {arcsinh}(c x))}{3 d (d x)^{3/2}}\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {2 b c \left (\frac {2 c^2 \left (\frac {\sqrt {d} (c d x+d) \sqrt {\frac {c^2 d^2 x^2+d^2}{(c d x+d)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right ),\frac {1}{2}\right )}{2 c^{3/2} \sqrt {c^2 x^2+1}}-\frac {\frac {\sqrt {d} (c d x+d) \sqrt {\frac {c^2 d^2 x^2+d^2}{(c d x+d)^2}} E\left (2 \arctan \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )|\frac {1}{2}\right )}{\sqrt {c} \sqrt {c^2 x^2+1}}-\frac {d^2 \sqrt {c^2 x^2+1} \sqrt {d x}}{c d x+d}}{c}\right )}{d^3}-\frac {2 \sqrt {c^2 x^2+1}}{d \sqrt {d x}}\right )}{3 d}-\frac {2 (a+b \text {arcsinh}(c x))}{3 d (d x)^{3/2}}\)

Input:

Int[(a + b*ArcSinh[c*x])/(d*x)^(5/2),x]
 

Output:

(-2*(a + b*ArcSinh[c*x]))/(3*d*(d*x)^(3/2)) + (2*b*c*((-2*Sqrt[1 + c^2*x^2 
])/(d*Sqrt[d*x]) + (2*c^2*(-((-((d^2*Sqrt[d*x]*Sqrt[1 + c^2*x^2])/(d + c*d 
*x)) + (Sqrt[d]*(d + c*d*x)*Sqrt[(d^2 + c^2*d^2*x^2)/(d + c*d*x)^2]*Ellipt 
icE[2*ArcTan[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]], 1/2])/(Sqrt[c]*Sqrt[1 + c^2*x^2 
]))/c) + (Sqrt[d]*(d + c*d*x)*Sqrt[(d^2 + c^2*d^2*x^2)/(d + c*d*x)^2]*Elli 
pticF[2*ArcTan[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]], 1/2])/(2*c^(3/2)*Sqrt[1 + c^2 
*x^2])))/d^3))/(3*d)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 761

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 834

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S 
imp[1/q   Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q   Int[(1 - q*x^2)/Sqrt[a 
 + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 1510

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* 
(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/a]
 

rule 6191

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] 
 :> Simp[(d*x)^(m + 1)*((a + b*ArcSinh[c*x])^n/(d*(m + 1))), x] - Simp[b*c* 
(n/(d*(m + 1)))   Int[(d*x)^(m + 1)*((a + b*ArcSinh[c*x])^(n - 1)/Sqrt[1 + 
c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.41 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.56

method	result	size
derivativedivides	\(\frac {-\frac {2 a}{3 \left (d x \right )^{\frac {3}{2}}}+2 b \left (-\frac {\operatorname {arcsinh}\left (x c \right )}{3 \left (d x \right )^{\frac {3}{2}}}+\frac {2 c \left (-\frac {\sqrt {c^{2} x^{2}+1}}{\sqrt {d x}}+\frac {i c \sqrt {-i x c +1}\, \sqrt {i x c +1}\, \left (\operatorname {EllipticF}\left (\sqrt {d x}\, \sqrt {\frac {i c}{d}}, i\right )-\operatorname {EllipticE}\left (\sqrt {d x}\, \sqrt {\frac {i c}{d}}, i\right )\right )}{d \sqrt {\frac {i c}{d}}\, \sqrt {c^{2} x^{2}+1}}\right )}{3 d}\right )}{d}\)	\(138\)
default	\(\frac {-\frac {2 a}{3 \left (d x \right )^{\frac {3}{2}}}+2 b \left (-\frac {\operatorname {arcsinh}\left (x c \right )}{3 \left (d x \right )^{\frac {3}{2}}}+\frac {2 c \left (-\frac {\sqrt {c^{2} x^{2}+1}}{\sqrt {d x}}+\frac {i c \sqrt {-i x c +1}\, \sqrt {i x c +1}\, \left (\operatorname {EllipticF}\left (\sqrt {d x}\, \sqrt {\frac {i c}{d}}, i\right )-\operatorname {EllipticE}\left (\sqrt {d x}\, \sqrt {\frac {i c}{d}}, i\right )\right )}{d \sqrt {\frac {i c}{d}}\, \sqrt {c^{2} x^{2}+1}}\right )}{3 d}\right )}{d}\)	\(138\)
parts	\(-\frac {2 a}{3 \left (d x \right )^{\frac {3}{2}} d}+\frac {2 b \left (-\frac {\operatorname {arcsinh}\left (x c \right )}{3 \left (d x \right )^{\frac {3}{2}}}+\frac {2 c \left (-\frac {\sqrt {c^{2} x^{2}+1}}{\sqrt {d x}}+\frac {i c \sqrt {-i x c +1}\, \sqrt {i x c +1}\, \left (\operatorname {EllipticF}\left (\sqrt {d x}\, \sqrt {\frac {i c}{d}}, i\right )-\operatorname {EllipticE}\left (\sqrt {d x}\, \sqrt {\frac {i c}{d}}, i\right )\right )}{d \sqrt {\frac {i c}{d}}\, \sqrt {c^{2} x^{2}+1}}\right )}{3 d}\right )}{d}\)	\(140\)

Input:

int((a+b*arcsinh(x*c))/(d*x)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

2/d*(-1/3*a/(d*x)^(3/2)+b*(-1/3/(d*x)^(3/2)*arcsinh(x*c)+2/3/d*c*(-(c^2*x^ 
2+1)^(1/2)/(d*x)^(1/2)+I*c/d/(I/d*c)^(1/2)*(1-I*x*c)^(1/2)*(1+I*x*c)^(1/2) 
/(c^2*x^2+1)^(1/2)*(EllipticF((d*x)^(1/2)*(I/d*c)^(1/2),I)-EllipticE((d*x) 
^(1/2)*(I/d*c)^(1/2),I)))))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.36 \[ \int \frac {a+b \text {arcsinh}(c x)}{(d x)^{5/2}} \, dx=-\frac {2 \, {\left (2 \, \sqrt {c^{2} d} b c x^{2} {\rm weierstrassZeta}\left (-\frac {4}{c^{2}}, 0, {\rm weierstrassPInverse}\left (-\frac {4}{c^{2}}, 0, x\right )\right ) + 2 \, \sqrt {c^{2} x^{2} + 1} \sqrt {d x} b c x + \sqrt {d x} b \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + \sqrt {d x} a\right )}}{3 \, d^{3} x^{2}} \] Input:

integrate((a+b*arcsinh(c*x))/(d*x)^(5/2),x, algorithm="fricas")
 

Output:

-2/3*(2*sqrt(c^2*d)*b*c*x^2*weierstrassZeta(-4/c^2, 0, weierstrassPInverse 
(-4/c^2, 0, x)) + 2*sqrt(c^2*x^2 + 1)*sqrt(d*x)*b*c*x + sqrt(d*x)*b*log(c* 
x + sqrt(c^2*x^2 + 1)) + sqrt(d*x)*a)/(d^3*x^2)
 

Sympy [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{(d x)^{5/2}} \, dx=\int \frac {a + b \operatorname {asinh}{\left (c x \right )}}{\left (d x\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((a+b*asinh(c*x))/(d*x)**(5/2),x)
 

Output:

Integral((a + b*asinh(c*x))/(d*x)**(5/2), x)
 

Maxima [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{(d x)^{5/2}} \, dx=\int { \frac {b \operatorname {arsinh}\left (c x\right ) + a}{\left (d x\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((a+b*arcsinh(c*x))/(d*x)^(5/2),x, algorithm="maxima")
 

Output:

-1/6*(c^2*(I*sqrt(2)*(log(1/2*I*sqrt(2)*(2*c*sqrt(x) + sqrt(2)*sqrt(c))/sq 
rt(c) + 1) - log(-1/2*I*sqrt(2)*(2*c*sqrt(x) + sqrt(2)*sqrt(c))/sqrt(c) + 
1))/(sqrt(c)*d^(5/2)) + I*sqrt(2)*(log(1/2*I*sqrt(2)*(2*c*sqrt(x) - sqrt(2 
)*sqrt(c))/sqrt(c) + 1) - log(-1/2*I*sqrt(2)*(2*c*sqrt(x) - sqrt(2)*sqrt(c 
))/sqrt(c) + 1))/(sqrt(c)*d^(5/2)) - sqrt(2)*log(c*x + sqrt(2)*sqrt(c)*sqr 
t(x) + 1)/(sqrt(c)*d^(5/2)) + sqrt(2)*log(c*x - sqrt(2)*sqrt(c)*sqrt(x) + 
1)/(sqrt(c)*d^(5/2))) - 12*c*integrate(1/3/((c^3*d^(5/2)*x^3 + c*d^(5/2)*x 
 + (c^2*d^(5/2)*x^2 + d^(5/2))*sqrt(c^2*x^2 + 1))*x^(3/2)), x) + 4*log(c*x 
 + sqrt(c^2*x^2 + 1))/(d^(5/2)*x^(3/2)))*b - 2/3*a/((d*x)^(3/2)*d)
 

Giac [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{(d x)^{5/2}} \, dx=\int { \frac {b \operatorname {arsinh}\left (c x\right ) + a}{\left (d x\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((a+b*arcsinh(c*x))/(d*x)^(5/2),x, algorithm="giac")
 

Output:

integrate((b*arcsinh(c*x) + a)/(d*x)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \text {arcsinh}(c x)}{(d x)^{5/2}} \, dx=\int \frac {a+b\,\mathrm {asinh}\left (c\,x\right )}{{\left (d\,x\right )}^{5/2}} \,d x \] Input:

int((a + b*asinh(c*x))/(d*x)^(5/2),x)
 

Output:

int((a + b*asinh(c*x))/(d*x)^(5/2), x)
 

Reduce [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{(d x)^{5/2}} \, dx=\frac {3 \sqrt {x}\, \left (\int \frac {\mathit {asinh} \left (c x \right )}{\sqrt {x}\, x^{2}}d x \right ) b x -2 a}{3 \sqrt {x}\, \sqrt {d}\, d^{2} x} \] Input:

int((a+b*asinh(c*x))/(d*x)^(5/2),x)
 

Output:

(3*sqrt(x)*int(asinh(c*x)/(sqrt(x)*x**2),x)*b*x - 2*a)/(3*sqrt(x)*sqrt(d)* 
d**2*x)