\(\int (d x)^{3/2} (a+b \arcsin (c x)) \, dx\) [68]

Optimal result

Integrand size = 16, antiderivative size = 124 \[ \int (d x)^{3/2} (a+b \arcsin (c x)) \, dx=\frac {4 b (d x)^{3/2} \sqrt {1-c^2 x^2}}{25 c}+\frac {2 (d x)^{5/2} (a+b \arcsin (c x))}{5 d}-\frac {12 b d^{3/2} E\left (\left .\arcsin \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{25 c^{5/2}}+\frac {12 b d^{3/2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right ),-1\right )}{25 c^{5/2}} \] Output:

4/25*b*(d*x)^(3/2)*(-c^2*x^2+1)^(1/2)/c+2/5*(d*x)^(5/2)*(a+b*arcsin(c*x))/ 
d-12/25*b*d^(3/2)*EllipticE(c^(1/2)*(d*x)^(1/2)/d^(1/2),I)/c^(5/2)+12/25*b 
*d^(3/2)*EllipticF(c^(1/2)*(d*x)^(1/2)/d^(1/2),I)/c^(5/2)
 

Mathematica [C] (verified)

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

Time = 0.02 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.53 \[ \int (d x)^{3/2} (a+b \arcsin (c x)) \, dx=\frac {2 (d x)^{3/2} \left (5 a c x+2 b \sqrt {1-c^2 x^2}+5 b c x \arcsin (c x)-2 b \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},c^2 x^2\right )\right )}{25 c} \] Input:

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

Output:

(2*(d*x)^(3/2)*(5*a*c*x + 2*b*Sqrt[1 - c^2*x^2] + 5*b*c*x*ArcSin[c*x] - 2* 
b*Hypergeometric2F1[1/2, 3/4, 7/4, c^2*x^2]))/(25*c)
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5138, 262, 266, 836, 27, 762, 1389, 327}

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.

\(\displaystyle \int (d x)^{3/2} (a+b \arcsin (c x)) \, dx\)

\(\Big \downarrow \) 5138

\(\displaystyle \frac {2 (d x)^{5/2} (a+b \arcsin (c x))}{5 d}-\frac {2 b c \int \frac {(d x)^{5/2}}{\sqrt {1-c^2 x^2}}dx}{5 d}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {2 (d x)^{5/2} (a+b \arcsin (c x))}{5 d}-\frac {2 b c \left (\frac {3 d^2 \int \frac {\sqrt {d x}}{\sqrt {1-c^2 x^2}}dx}{5 c^2}-\frac {2 d \sqrt {1-c^2 x^2} (d x)^{3/2}}{5 c^2}\right )}{5 d}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2 (d x)^{5/2} (a+b \arcsin (c x))}{5 d}-\frac {2 b c \left (\frac {6 d \int \frac {d x}{\sqrt {1-c^2 x^2}}d\sqrt {d x}}{5 c^2}-\frac {2 d \sqrt {1-c^2 x^2} (d x)^{3/2}}{5 c^2}\right )}{5 d}\)

\(\Big \downarrow \) 836

\(\displaystyle \frac {2 (d x)^{5/2} (a+b \arcsin (c x))}{5 d}-\frac {2 b c \left (\frac {6 d \left (\frac {d \int \frac {c x d+d}{d \sqrt {1-c^2 x^2}}d\sqrt {d x}}{c}-\frac {d \int \frac {1}{\sqrt {1-c^2 x^2}}d\sqrt {d x}}{c}\right )}{5 c^2}-\frac {2 d \sqrt {1-c^2 x^2} (d x)^{3/2}}{5 c^2}\right )}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 (d x)^{5/2} (a+b \arcsin (c x))}{5 d}-\frac {2 b c \left (\frac {6 d \left (\frac {\int \frac {c x d+d}{\sqrt {1-c^2 x^2}}d\sqrt {d x}}{c}-\frac {d \int \frac {1}{\sqrt {1-c^2 x^2}}d\sqrt {d x}}{c}\right )}{5 c^2}-\frac {2 d \sqrt {1-c^2 x^2} (d x)^{3/2}}{5 c^2}\right )}{5 d}\)

\(\Big \downarrow \) 762

\(\displaystyle \frac {2 (d x)^{5/2} (a+b \arcsin (c x))}{5 d}-\frac {2 b c \left (\frac {6 d \left (\frac {\int \frac {c x d+d}{\sqrt {1-c^2 x^2}}d\sqrt {d x}}{c}-\frac {d^{3/2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right ),-1\right )}{c^{3/2}}\right )}{5 c^2}-\frac {2 d \sqrt {1-c^2 x^2} (d x)^{3/2}}{5 c^2}\right )}{5 d}\)

\(\Big \downarrow \) 1389

\(\displaystyle \frac {2 (d x)^{5/2} (a+b \arcsin (c x))}{5 d}-\frac {2 b c \left (\frac {6 d \left (\frac {d \int \frac {\sqrt {c x+1}}{\sqrt {1-c x}}d\sqrt {d x}}{c}-\frac {d^{3/2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right ),-1\right )}{c^{3/2}}\right )}{5 c^2}-\frac {2 d \sqrt {1-c^2 x^2} (d x)^{3/2}}{5 c^2}\right )}{5 d}\)

\(\Big \downarrow \) 327

\(\displaystyle \frac {2 (d x)^{5/2} (a+b \arcsin (c x))}{5 d}-\frac {2 b c \left (\frac {6 d \left (\frac {d^{3/2} E\left (\left .\arcsin \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{c^{3/2}}-\frac {d^{3/2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right ),-1\right )}{c^{3/2}}\right )}{5 c^2}-\frac {2 d \sqrt {1-c^2 x^2} (d x)^{3/2}}{5 c^2}\right )}{5 d}\)

Input:

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

Output:

(2*(d*x)^(5/2)*(a + b*ArcSin[c*x]))/(5*d) - (2*b*c*((-2*d*(d*x)^(3/2)*Sqrt 
[1 - c^2*x^2])/(5*c^2) + (6*d*((d^(3/2)*EllipticE[ArcSin[(Sqrt[c]*Sqrt[d*x 
])/Sqrt[d]], -1])/c^(3/2) - (d^(3/2)*EllipticF[ArcSin[(Sqrt[c]*Sqrt[d*x])/ 
Sqrt[d]], -1])/c^(3/2)))/(5*c^2)))/(5*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 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && 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 327

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) 
)], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0]
 

rule 762

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) 
)*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] 
 && GtQ[a, 0]
 

rule 836

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

rule 1389

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Simp[d/Sq 
rt[a]   Int[Sqrt[1 + e*(x^2/d)]/Sqrt[1 - e*(x^2/d)], x], x] /; FreeQ[{a, c, 
 d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && NegQ[c/a] && GtQ[a, 0]
 

rule 5138

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] 
:> Simp[(d*x)^(m + 1)*((a + b*ArcSin[c*x])^n/(d*(m + 1))), x] - Simp[b*c*(n 
/(d*(m + 1)))   Int[(d*x)^(m + 1)*((a + b*ArcSin[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 [A] (verified)

Time = 0.65 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.11

method	result	size
derivativedivides	\(\frac {\frac {2 \left (d x \right )^{\frac {5}{2}} a}{5}+2 b \left (\frac {\left (d x \right )^{\frac {5}{2}} \arcsin \left (c x \right )}{5}-\frac {2 c \left (-\frac {d^{2} \left (d x \right )^{\frac {3}{2}} \sqrt {-c^{2} x^{2}+1}}{5 c^{2}}-\frac {3 d^{3} \sqrt {-c x +1}\, \sqrt {c x +1}\, \left (\operatorname {EllipticF}\left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )-\operatorname {EllipticE}\left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )\right )}{5 c^{3} \sqrt {\frac {c}{d}}\, \sqrt {-c^{2} x^{2}+1}}\right )}{5 d}\right )}{d}\)	\(138\)
default	\(\frac {\frac {2 \left (d x \right )^{\frac {5}{2}} a}{5}+2 b \left (\frac {\left (d x \right )^{\frac {5}{2}} \arcsin \left (c x \right )}{5}-\frac {2 c \left (-\frac {d^{2} \left (d x \right )^{\frac {3}{2}} \sqrt {-c^{2} x^{2}+1}}{5 c^{2}}-\frac {3 d^{3} \sqrt {-c x +1}\, \sqrt {c x +1}\, \left (\operatorname {EllipticF}\left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )-\operatorname {EllipticE}\left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )\right )}{5 c^{3} \sqrt {\frac {c}{d}}\, \sqrt {-c^{2} x^{2}+1}}\right )}{5 d}\right )}{d}\)	\(138\)
parts	\(\frac {2 a \left (d x \right )^{\frac {5}{2}}}{5 d}+\frac {2 b \left (\frac {\left (d x \right )^{\frac {5}{2}} \arcsin \left (c x \right )}{5}-\frac {2 c \left (-\frac {d^{2} \left (d x \right )^{\frac {3}{2}} \sqrt {-c^{2} x^{2}+1}}{5 c^{2}}-\frac {3 d^{3} \sqrt {-c x +1}\, \sqrt {c x +1}\, \left (\operatorname {EllipticF}\left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )-\operatorname {EllipticE}\left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )\right )}{5 c^{3} \sqrt {\frac {c}{d}}\, \sqrt {-c^{2} x^{2}+1}}\right )}{5 d}\right )}{d}\)	\(140\)

Input:

int((d*x)^(3/2)*(a+b*arcsin(c*x)),x,method=_RETURNVERBOSE)
 

Output:

2/d*(1/5*(d*x)^(5/2)*a+b*(1/5*(d*x)^(5/2)*arcsin(c*x)-2/5*c/d*(-1/5/c^2*d^ 
2*(d*x)^(3/2)*(-c^2*x^2+1)^(1/2)-3/5/c^3*d^3/(c/d)^(1/2)*(-c*x+1)^(1/2)*(c 
*x+1)^(1/2)/(-c^2*x^2+1)^(1/2)*(EllipticF((d*x)^(1/2)*(c/d)^(1/2),I)-Ellip 
ticE((d*x)^(1/2)*(c/d)^(1/2),I)))))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.69 \[ \int (d x)^{3/2} (a+b \arcsin (c x)) \, dx=-\frac {2 \, {\left (6 \, \sqrt {-c^{2} d} b d {\rm weierstrassZeta}\left (\frac {4}{c^{2}}, 0, {\rm weierstrassPInverse}\left (\frac {4}{c^{2}}, 0, x\right )\right ) - {\left (5 \, b c^{3} d x^{2} \arcsin \left (c x\right ) + 5 \, a c^{3} d x^{2} + 2 \, \sqrt {-c^{2} x^{2} + 1} b c^{2} d x\right )} \sqrt {d x}\right )}}{25 \, c^{3}} \] Input:

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

Output:

-2/25*(6*sqrt(-c^2*d)*b*d*weierstrassZeta(4/c^2, 0, weierstrassPInverse(4/ 
c^2, 0, x)) - (5*b*c^3*d*x^2*arcsin(c*x) + 5*a*c^3*d*x^2 + 2*sqrt(-c^2*x^2 
 + 1)*b*c^2*d*x)*sqrt(d*x))/c^3
 

Sympy [A] (verification not implemented)

Time = 11.43 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.69 \[ \int (d x)^{3/2} (a+b \arcsin (c x)) \, dx=a \left (\begin {cases} \frac {2 \left (d x\right )^{\frac {5}{2}}}{5 d} & \text {for}\: d \neq 0 \\0 & \text {otherwise} \end {cases}\right ) - b c \left (\begin {cases} \frac {d^{\frac {3}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {c^{2} x^{2} e^{2 i \pi }} \right )}}{5 \Gamma \left (\frac {11}{4}\right )} & \text {for}\: d > -\infty \wedge d < \infty \wedge d \neq 0 \\0 & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} \frac {2 \left (d x\right )^{\frac {5}{2}}}{5 d} & \text {for}\: d \neq 0 \\0 & \text {otherwise} \end {cases}\right ) \operatorname {asin}{\left (c x \right )} \] Input:

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

Output:

a*Piecewise((2*(d*x)**(5/2)/(5*d), Ne(d, 0)), (0, True)) - b*c*Piecewise(( 
d**(3/2)*x**(7/2)*gamma(7/4)*hyper((1/2, 7/4), (11/4,), c**2*x**2*exp_pola 
r(2*I*pi))/(5*gamma(11/4)), (d > -oo) & (d < oo) & Ne(d, 0)), (0, True)) + 
 b*Piecewise((2*(d*x)**(5/2)/(5*d), Ne(d, 0)), (0, True))*asin(c*x)
 

Maxima [F]

\[ \int (d x)^{3/2} (a+b \arcsin (c x)) \, dx=\int { \left (d x\right )^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )} \,d x } \] Input:

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

Output:

2/5*b*d^(3/2)*x^(5/2)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + 2/5*(a* 
d*x^(5/2) + 5*b*c*d*integrate(1/5*sqrt(c*x + 1)*sqrt(-c*x + 1)*x^(5/2)/(c^ 
2*x^2 - 1), x))*sqrt(d)
 

Giac [F]

\[ \int (d x)^{3/2} (a+b \arcsin (c x)) \, dx=\int { \left (d x\right )^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int (d x)^{3/2} (a+b \arcsin (c x)) \, dx=\int \left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d\,x\right )}^{3/2} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(d)*d*(2*sqrt(x)*a*x**2 + 5*int(sqrt(x)*asin(c*x)*x,x)*b))/5