Integrand size = 16, antiderivative size = 88 \[ \int \sqrt {d x} (a+b \arcsin (c x)) \, dx=\frac {4 b \sqrt {d x} \sqrt {1-c^2 x^2}}{9 c}+\frac {2 (d x)^{3/2} (a+b \arcsin (c x))}{3 d}-\frac {4 b \sqrt {d} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right ),-1\right )}{9 c^{3/2}} \]
2/3*(d*x)^(3/2)*(a+b*arcsin(c*x))/d-4/9*b*EllipticF(c^(1/2)*(d*x)^(1/2)/d^ (1/2),I)*d^(1/2)/c^(3/2)+4/9*b*(d*x)^(1/2)*(-c^2*x^2+1)^(1/2)/c
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 = 0.75 \[ \int \sqrt {d x} (a+b \arcsin (c x)) \, dx=\frac {2 \sqrt {d x} \left (3 a c x+2 b \sqrt {1-c^2 x^2}+3 b c x \arcsin (c x)-2 b \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},c^2 x^2\right )\right )}{9 c} \]
(2*Sqrt[d*x]*(3*a*c*x + 2*b*Sqrt[1 - c^2*x^2] + 3*b*c*x*ArcSin[c*x] - 2*b* Hypergeometric2F1[1/4, 1/2, 5/4, c^2*x^2]))/(9*c)
Time = 0.23 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.10, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5138, 262, 266, 762}
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.
\(\displaystyle \int \sqrt {d x} (a+b \arcsin (c x)) \, dx\) |
\(\Big \downarrow \) 5138 |
\(\displaystyle \frac {2 (d x)^{3/2} (a+b \arcsin (c x))}{3 d}-\frac {2 b c \int \frac {(d x)^{3/2}}{\sqrt {1-c^2 x^2}}dx}{3 d}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {2 (d x)^{3/2} (a+b \arcsin (c x))}{3 d}-\frac {2 b c \left (\frac {d^2 \int \frac {1}{\sqrt {d x} \sqrt {1-c^2 x^2}}dx}{3 c^2}-\frac {2 d \sqrt {1-c^2 x^2} \sqrt {d x}}{3 c^2}\right )}{3 d}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2 (d x)^{3/2} (a+b \arcsin (c x))}{3 d}-\frac {2 b c \left (\frac {2 d \int \frac {1}{\sqrt {1-c^2 x^2}}d\sqrt {d x}}{3 c^2}-\frac {2 d \sqrt {1-c^2 x^2} \sqrt {d x}}{3 c^2}\right )}{3 d}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle \frac {2 (d x)^{3/2} (a+b \arcsin (c x))}{3 d}-\frac {2 b c \left (\frac {2 d^{3/2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right ),-1\right )}{3 c^{5/2}}-\frac {2 d \sqrt {1-c^2 x^2} \sqrt {d x}}{3 c^2}\right )}{3 d}\) |
(2*(d*x)^(3/2)*(a + b*ArcSin[c*x]))/(3*d) - (2*b*c*((-2*d*Sqrt[d*x]*Sqrt[1 - c^2*x^2])/(3*c^2) + (2*d^(3/2)*EllipticF[ArcSin[(Sqrt[c]*Sqrt[d*x])/Sqr t[d]], -1])/(3*c^(5/2))))/(3*d)
3.3.5.3.1 Defintions of rubi rules used
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]
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]
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]
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]
Time = 0.41 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.35
method | result | size |
derivativedivides | \(\frac {\frac {2 \left (d x \right )^{\frac {3}{2}} a}{3}+2 b \left (\frac {\left (d x \right )^{\frac {3}{2}} \arcsin \left (c x \right )}{3}-\frac {2 c \left (-\frac {d^{2} \sqrt {d x}\, \sqrt {-c^{2} x^{2}+1}}{3 c^{2}}+\frac {d^{2} \sqrt {-c x +1}\, \sqrt {c x +1}\, \operatorname {EllipticF}\left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )}{3 c^{2} \sqrt {\frac {c}{d}}\, \sqrt {-c^{2} x^{2}+1}}\right )}{3 d}\right )}{d}\) | \(119\) |
default | \(\frac {\frac {2 \left (d x \right )^{\frac {3}{2}} a}{3}+2 b \left (\frac {\left (d x \right )^{\frac {3}{2}} \arcsin \left (c x \right )}{3}-\frac {2 c \left (-\frac {d^{2} \sqrt {d x}\, \sqrt {-c^{2} x^{2}+1}}{3 c^{2}}+\frac {d^{2} \sqrt {-c x +1}\, \sqrt {c x +1}\, \operatorname {EllipticF}\left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )}{3 c^{2} \sqrt {\frac {c}{d}}\, \sqrt {-c^{2} x^{2}+1}}\right )}{3 d}\right )}{d}\) | \(119\) |
parts | \(\frac {2 a \left (d x \right )^{\frac {3}{2}}}{3 d}+\frac {2 b \left (\frac {\left (d x \right )^{\frac {3}{2}} \arcsin \left (c x \right )}{3}-\frac {2 c \left (-\frac {d^{2} \sqrt {d x}\, \sqrt {-c^{2} x^{2}+1}}{3 c^{2}}+\frac {d^{2} \sqrt {-c x +1}\, \sqrt {c x +1}\, \operatorname {EllipticF}\left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )}{3 c^{2} \sqrt {\frac {c}{d}}\, \sqrt {-c^{2} x^{2}+1}}\right )}{3 d}\right )}{d}\) | \(121\) |
2/d*(1/3*(d*x)^(3/2)*a+b*(1/3*(d*x)^(3/2)*arcsin(c*x)-2/3*c/d*(-1/3/c^2*d^ 2*(d*x)^(1/2)*(-c^2*x^2+1)^(1/2)+1/3/c^2*d^2/(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))))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.77 \[ \int \sqrt {d x} (a+b \arcsin (c x)) \, dx=\frac {2 \, {\left (2 \, \sqrt {-c^{2} d} b {\rm weierstrassPInverse}\left (\frac {4}{c^{2}}, 0, x\right ) + {\left (3 \, b c^{3} x \arcsin \left (c x\right ) + 3 \, a c^{3} x + 2 \, \sqrt {-c^{2} x^{2} + 1} b c^{2}\right )} \sqrt {d x}\right )}}{9 \, c^{3}} \]
2/9*(2*sqrt(-c^2*d)*b*weierstrassPInverse(4/c^2, 0, x) + (3*b*c^3*x*arcsin (c*x) + 3*a*c^3*x + 2*sqrt(-c^2*x^2 + 1)*b*c^2)*sqrt(d*x))/c^3
Time = 3.26 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.97 \[ \int \sqrt {d x} (a+b \arcsin (c x)) \, dx=a \left (\begin {cases} \frac {2 \left (d x\right )^{\frac {3}{2}}}{3 d} & \text {for}\: d \neq 0 \\0 & \text {otherwise} \end {cases}\right ) - b c \left (\begin {cases} \frac {\sqrt {d} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {c^{2} x^{2} e^{2 i \pi }} \right )}}{3 \Gamma \left (\frac {9}{4}\right )} & \text {for}\: d > -\infty \wedge d < \infty \wedge d \neq 0 \\0 & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} \frac {2 \left (d x\right )^{\frac {3}{2}}}{3 d} & \text {for}\: d \neq 0 \\0 & \text {otherwise} \end {cases}\right ) \operatorname {asin}{\left (c x \right )} \]
a*Piecewise((2*(d*x)**(3/2)/(3*d), Ne(d, 0)), (0, True)) - b*c*Piecewise(( sqrt(d)*x**(5/2)*gamma(5/4)*hyper((1/2, 5/4), (9/4,), c**2*x**2*exp_polar( 2*I*pi))/(3*gamma(9/4)), (d > -oo) & (d < oo) & Ne(d, 0)), (0, True)) + b* Piecewise((2*(d*x)**(3/2)/(3*d), Ne(d, 0)), (0, True))*asin(c*x)
\[ \int \sqrt {d x} (a+b \arcsin (c x)) \, dx=\int { \sqrt {d x} {\left (b \arcsin \left (c x\right ) + a\right )} \,d x } \]
2/3*b*sqrt(d)*x^(3/2)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + 2/3*(3* b*c*integrate(1/3*sqrt(c*x + 1)*sqrt(-c*x + 1)*x^(3/2)/(c^2*x^2 - 1), x) + a*x^(3/2))*sqrt(d)
\[ \int \sqrt {d x} (a+b \arcsin (c x)) \, dx=\int { \sqrt {d x} {\left (b \arcsin \left (c x\right ) + a\right )} \,d x } \]
Timed out. \[ \int \sqrt {d x} (a+b \arcsin (c x)) \, dx=\int \left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\sqrt {d\,x} \,d x \]