Integrand size = 16, antiderivative size = 125 \[ \int \frac {a+b \arcsin (c x)}{(d x)^{5/2}} \, dx=-\frac {4 b c \sqrt {1-c^2 x^2}}{3 d^2 \sqrt {d x}}-\frac {2 (a+b \arcsin (c x))}{3 d (d x)^{3/2}}-\frac {4 b c^{3/2} E\left (\left .\arcsin \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{3 d^{5/2}}+\frac {4 b c^{3/2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right ),-1\right )}{3 d^{5/2}} \]
-2/3*(a+b*arcsin(c*x))/d/(d*x)^(3/2)-4/3*b*c^(3/2)*EllipticE(c^(1/2)*(d*x) ^(1/2)/d^(1/2),I)/d^(5/2)+4/3*b*c^(3/2)*EllipticF(c^(1/2)*(d*x)^(1/2)/d^(1 /2),I)/d^(5/2)-4/3*b*c*(-c^2*x^2+1)^(1/2)/d^2/(d*x)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.34 \[ \int \frac {a+b \arcsin (c x)}{(d x)^{5/2}} \, dx=-\frac {2 x \left (a+b \arcsin (c x)+2 b c x \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},c^2 x^2\right )\right )}{3 (d x)^{5/2}} \]
(-2*x*(a + b*ArcSin[c*x] + 2*b*c*x*Hypergeometric2F1[-1/4, 1/2, 3/4, c^2*x ^2]))/(3*(d*x)^(5/2))
Time = 0.31 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5138, 264, 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.
\(\displaystyle \int \frac {a+b \arcsin (c x)}{(d x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 5138 |
\(\displaystyle \frac {2 b c \int \frac {1}{(d x)^{3/2} \sqrt {1-c^2 x^2}}dx}{3 d}-\frac {2 (a+b \arcsin (c x))}{3 d (d x)^{3/2}}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {2 b c \left (-\frac {c^2 \int \frac {\sqrt {d x}}{\sqrt {1-c^2 x^2}}dx}{d^2}-\frac {2 \sqrt {1-c^2 x^2}}{d \sqrt {d x}}\right )}{3 d}-\frac {2 (a+b \arcsin (c x))}{3 d (d x)^{3/2}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2 b c \left (-\frac {2 c^2 \int \frac {d x}{\sqrt {1-c^2 x^2}}d\sqrt {d x}}{d^3}-\frac {2 \sqrt {1-c^2 x^2}}{d \sqrt {d x}}\right )}{3 d}-\frac {2 (a+b \arcsin (c x))}{3 d (d x)^{3/2}}\) |
\(\Big \downarrow \) 836 |
\(\displaystyle \frac {2 b c \left (-\frac {2 c^2 \left (\frac {d \int \frac {c x d+d}{d \sqrt {1-c^2 x^2}}d\sqrt {d x}}{c}-\frac {d \int \frac {1}{\sqrt {1-c^2 x^2}}d\sqrt {d x}}{c}\right )}{d^3}-\frac {2 \sqrt {1-c^2 x^2}}{d \sqrt {d x}}\right )}{3 d}-\frac {2 (a+b \arcsin (c x))}{3 d (d x)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 b c \left (-\frac {2 c^2 \left (\frac {\int \frac {c x d+d}{\sqrt {1-c^2 x^2}}d\sqrt {d x}}{c}-\frac {d \int \frac {1}{\sqrt {1-c^2 x^2}}d\sqrt {d x}}{c}\right )}{d^3}-\frac {2 \sqrt {1-c^2 x^2}}{d \sqrt {d x}}\right )}{3 d}-\frac {2 (a+b \arcsin (c x))}{3 d (d x)^{3/2}}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle \frac {2 b c \left (-\frac {2 c^2 \left (\frac {\int \frac {c x d+d}{\sqrt {1-c^2 x^2}}d\sqrt {d x}}{c}-\frac {d^{3/2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right ),-1\right )}{c^{3/2}}\right )}{d^3}-\frac {2 \sqrt {1-c^2 x^2}}{d \sqrt {d x}}\right )}{3 d}-\frac {2 (a+b \arcsin (c x))}{3 d (d x)^{3/2}}\) |
\(\Big \downarrow \) 1389 |
\(\displaystyle \frac {2 b c \left (-\frac {2 c^2 \left (\frac {d \int \frac {\sqrt {c x+1}}{\sqrt {1-c x}}d\sqrt {d x}}{c}-\frac {d^{3/2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right ),-1\right )}{c^{3/2}}\right )}{d^3}-\frac {2 \sqrt {1-c^2 x^2}}{d \sqrt {d x}}\right )}{3 d}-\frac {2 (a+b \arcsin (c x))}{3 d (d x)^{3/2}}\) |
\(\Big \downarrow \) 327 |
\(\displaystyle \frac {2 b c \left (-\frac {2 c^2 \left (\frac {d^{3/2} E\left (\left .\arcsin \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{c^{3/2}}-\frac {d^{3/2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right ),-1\right )}{c^{3/2}}\right )}{d^3}-\frac {2 \sqrt {1-c^2 x^2}}{d \sqrt {d x}}\right )}{3 d}-\frac {2 (a+b \arcsin (c x))}{3 d (d x)^{3/2}}\) |
(-2*(a + b*ArcSin[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^(3/2)*EllipticE[ArcSin[(Sqrt[c]*Sqrt[d*x])/Sq rt[d]], -1])/c^(3/2) - (d^(3/2)*EllipticF[ArcSin[(Sqrt[c]*Sqrt[d*x])/Sqrt[ d]], -1])/c^(3/2)))/d^3))/(3*d)
3.3.8.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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[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]
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[(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]
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]
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 = 0.30 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.03
method | result | size |
derivativedivides | \(\frac {-\frac {2 a}{3 \left (d x \right )^{\frac {3}{2}}}+2 b \left (-\frac {\arcsin \left (c x \right )}{3 \left (d x \right )^{\frac {3}{2}}}+\frac {2 c \left (-\frac {\sqrt {-c^{2} x^{2}+1}}{\sqrt {d x}}+\frac {c \sqrt {-c x +1}\, \sqrt {c x +1}\, \left (\operatorname {EllipticF}\left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )-\operatorname {EllipticE}\left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )\right )}{d \sqrt {\frac {c}{d}}\, \sqrt {-c^{2} x^{2}+1}}\right )}{3 d}\right )}{d}\) | \(129\) |
default | \(\frac {-\frac {2 a}{3 \left (d x \right )^{\frac {3}{2}}}+2 b \left (-\frac {\arcsin \left (c x \right )}{3 \left (d x \right )^{\frac {3}{2}}}+\frac {2 c \left (-\frac {\sqrt {-c^{2} x^{2}+1}}{\sqrt {d x}}+\frac {c \sqrt {-c x +1}\, \sqrt {c x +1}\, \left (\operatorname {EllipticF}\left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )-\operatorname {EllipticE}\left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )\right )}{d \sqrt {\frac {c}{d}}\, \sqrt {-c^{2} x^{2}+1}}\right )}{3 d}\right )}{d}\) | \(129\) |
parts | \(-\frac {2 a}{3 \left (d x \right )^{\frac {3}{2}} d}+\frac {2 b \left (-\frac {\arcsin \left (c x \right )}{3 \left (d x \right )^{\frac {3}{2}}}+\frac {2 c \left (-\frac {\sqrt {-c^{2} x^{2}+1}}{\sqrt {d x}}+\frac {c \sqrt {-c x +1}\, \sqrt {c x +1}\, \left (\operatorname {EllipticF}\left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )-\operatorname {EllipticE}\left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )\right )}{d \sqrt {\frac {c}{d}}\, \sqrt {-c^{2} x^{2}+1}}\right )}{3 d}\right )}{d}\) | \(131\) |
2/d*(-1/3*a/(d*x)^(3/2)+b*(-1/3/(d*x)^(3/2)*arcsin(c*x)+2/3*c/d*(-(-c^2*x^ 2+1)^(1/2)/(d*x)^(1/2)+c/d/(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)-EllipticE((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 = 70, normalized size of antiderivative = 0.56 \[ \int \frac {a+b \arcsin (c x)}{(d x)^{5/2}} \, dx=-\frac {2 \, {\left (2 \, \sqrt {-c^{2} d} b c x^{2} {\rm weierstrassZeta}\left (\frac {4}{c^{2}}, 0, {\rm weierstrassPInverse}\left (\frac {4}{c^{2}}, 0, x\right )\right ) + {\left (2 \, \sqrt {-c^{2} x^{2} + 1} b c x + b \arcsin \left (c x\right ) + a\right )} \sqrt {d x}\right )}}{3 \, d^{3} x^{2}} \]
-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)*b*c*x + b*arcsin(c*x) + a)*sqrt(d*x ))/(d^3*x^2)
Exception generated. \[ \int \frac {a+b \arcsin (c x)}{(d x)^{5/2}} \, dx=\text {Exception raised: TypeError} \]
\[ \int \frac {a+b \arcsin (c x)}{(d x)^{5/2}} \, dx=\int { \frac {b \arcsin \left (c x\right ) + a}{\left (d x\right )^{\frac {5}{2}}} \,d x } \]
-2/3*(b*sqrt(d)*x^(3/2)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + (3*b* c*d^3*x^(5/2)*integrate(1/3*sqrt(c*x + 1)*sqrt(-c*x + 1)*sqrt(x)/(c^2*d^3* x^4 - d^3*x^2), x) + a*x)*sqrt(d)*sqrt(x))/(d^3*x^3)
\[ \int \frac {a+b \arcsin (c x)}{(d x)^{5/2}} \, dx=\int { \frac {b \arcsin \left (c x\right ) + a}{\left (d x\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {a+b \arcsin (c x)}{(d x)^{5/2}} \, dx=\int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{{\left (d\,x\right )}^{5/2}} \,d x \]