3.5.8 \(\int \frac {x^5}{\sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))^2} \, dx\) [408]

Optimal. Leaf size=204 \[ -\frac {x^5}{b c (a+b \text {ArcSin}(c x))}+\frac {5 \cos \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a+b \text {ArcSin}(c x)}{b}\right )}{8 b^2 c^6}-\frac {15 \cos \left (\frac {3 a}{b}\right ) \text {CosIntegral}\left (\frac {3 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b^2 c^6}+\frac {5 \cos \left (\frac {5 a}{b}\right ) \text {CosIntegral}\left (\frac {5 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b^2 c^6}+\frac {5 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \text {ArcSin}(c x)}{b}\right )}{8 b^2 c^6}-\frac {15 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b^2 c^6}+\frac {5 \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b^2 c^6} \]

[Out]

-x^5/b/c/(a+b*arcsin(c*x))+5/8*Ci((a+b*arcsin(c*x))/b)*cos(a/b)/b^2/c^6-15/16*Ci(3*(a+b*arcsin(c*x))/b)*cos(3*
a/b)/b^2/c^6+5/16*Ci(5*(a+b*arcsin(c*x))/b)*cos(5*a/b)/b^2/c^6+5/8*Si((a+b*arcsin(c*x))/b)*sin(a/b)/b^2/c^6-15
/16*Si(3*(a+b*arcsin(c*x))/b)*sin(3*a/b)/b^2/c^6+5/16*Si(5*(a+b*arcsin(c*x))/b)*sin(5*a/b)/b^2/c^6

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Rubi [A]
time = 0.29, antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {4807, 4731, 4491, 3384, 3380, 3383} \begin {gather*} \frac {5 \cos \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a+b \text {ArcSin}(c x)}{b}\right )}{8 b^2 c^6}-\frac {15 \cos \left (\frac {3 a}{b}\right ) \text {CosIntegral}\left (\frac {3 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b^2 c^6}+\frac {5 \cos \left (\frac {5 a}{b}\right ) \text {CosIntegral}\left (\frac {5 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b^2 c^6}+\frac {5 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \text {ArcSin}(c x)}{b}\right )}{8 b^2 c^6}-\frac {15 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b^2 c^6}+\frac {5 \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b^2 c^6}-\frac {x^5}{b c (a+b \text {ArcSin}(c x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5/(Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2),x]

[Out]

-(x^5/(b*c*(a + b*ArcSin[c*x]))) + (5*Cos[a/b]*CosIntegral[(a + b*ArcSin[c*x])/b])/(8*b^2*c^6) - (15*Cos[(3*a)
/b]*CosIntegral[(3*(a + b*ArcSin[c*x]))/b])/(16*b^2*c^6) + (5*Cos[(5*a)/b]*CosIntegral[(5*(a + b*ArcSin[c*x]))
/b])/(16*b^2*c^6) + (5*Sin[a/b]*SinIntegral[(a + b*ArcSin[c*x])/b])/(8*b^2*c^6) - (15*Sin[(3*a)/b]*SinIntegral
[(3*(a + b*ArcSin[c*x]))/b])/(16*b^2*c^6) + (5*Sin[(5*a)/b]*SinIntegral[(5*(a + b*ArcSin[c*x]))/b])/(16*b^2*c^
6)

Rule 3380

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3383

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3384

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[c*(f/d) + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 4491

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 4731

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/(b*c^(m + 1)), Subst[Int[x^n*Sin[-
a/b + x/b]^m*Cos[-a/b + x/b], x], x, a + b*ArcSin[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 4807

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[
((f*x)^m/(b*c*(n + 1)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSin[c*x])^(n + 1), x] - Dist[f*(m/(b
*c*(n + 1)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]], Int[(f*x)^(m - 1)*(a + b*ArcSin[c*x])^(n + 1), x], x] /;
FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && LtQ[n, -1]

Rubi steps

\begin {align*} \int \frac {x^5}{\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2} \, dx &=-\frac {x^5}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {5 \int \frac {x^4}{a+b \sin ^{-1}(c x)} \, dx}{b c}\\ &=-\frac {x^5}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {5 \text {Subst}\left (\int \frac {\cos (x) \sin ^4(x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b c^6}\\ &=-\frac {x^5}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {5 \text {Subst}\left (\int \left (\frac {\cos (x)}{8 (a+b x)}-\frac {3 \cos (3 x)}{16 (a+b x)}+\frac {\cos (5 x)}{16 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{b c^6}\\ &=-\frac {x^5}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {5 \text {Subst}\left (\int \frac {\cos (5 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c^6}+\frac {5 \text {Subst}\left (\int \frac {\cos (x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 b c^6}-\frac {15 \text {Subst}\left (\int \frac {\cos (3 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c^6}\\ &=-\frac {x^5}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {\left (5 \cos \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 b c^6}-\frac {\left (15 \cos \left (\frac {3 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c^6}+\frac {\left (5 \cos \left (\frac {5 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c^6}+\frac {\left (5 \sin \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 b c^6}-\frac {\left (15 \sin \left (\frac {3 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c^6}+\frac {\left (5 \sin \left (\frac {5 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c^6}\\ &=-\frac {x^5}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {5 \cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{8 b^2 c^6}-\frac {15 \cos \left (\frac {3 a}{b}\right ) \text {Ci}\left (\frac {3 a}{b}+3 \sin ^{-1}(c x)\right )}{16 b^2 c^6}+\frac {5 \cos \left (\frac {5 a}{b}\right ) \text {Ci}\left (\frac {5 a}{b}+5 \sin ^{-1}(c x)\right )}{16 b^2 c^6}+\frac {5 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{8 b^2 c^6}-\frac {15 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \sin ^{-1}(c x)\right )}{16 b^2 c^6}+\frac {5 \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 a}{b}+5 \sin ^{-1}(c x)\right )}{16 b^2 c^6}\\ \end {align*}

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Mathematica [A]
time = 0.25, size = 157, normalized size = 0.77 \begin {gather*} -\frac {x^5}{b c (a+b \text {ArcSin}(c x))}+\frac {5 \left (2 \cos \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a}{b}+\text {ArcSin}(c x)\right )-3 \cos \left (\frac {3 a}{b}\right ) \text {CosIntegral}\left (3 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )+\cos \left (\frac {5 a}{b}\right ) \text {CosIntegral}\left (5 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )+2 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\text {ArcSin}(c x)\right )-3 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )+\sin \left (\frac {5 a}{b}\right ) \text {Si}\left (5 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )\right )}{16 b^2 c^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5/(Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2),x]

[Out]

-(x^5/(b*c*(a + b*ArcSin[c*x]))) + (5*(2*Cos[a/b]*CosIntegral[a/b + ArcSin[c*x]] - 3*Cos[(3*a)/b]*CosIntegral[
3*(a/b + ArcSin[c*x])] + Cos[(5*a)/b]*CosIntegral[5*(a/b + ArcSin[c*x])] + 2*Sin[a/b]*SinIntegral[a/b + ArcSin
[c*x]] - 3*Sin[(3*a)/b]*SinIntegral[3*(a/b + ArcSin[c*x])] + Sin[(5*a)/b]*SinIntegral[5*(a/b + ArcSin[c*x])]))
/(16*b^2*c^6)

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Maple [A]
time = 0.12, size = 341, normalized size = 1.67

method	result	size
default	\(\frac {5 \arcsin \left (c x \right ) \sinIntegral \left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right ) b +5 \arcsin \left (c x \right ) \cosineIntegral \left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right ) b -15 \arcsin \left (c x \right ) \sinIntegral \left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right ) b -15 \arcsin \left (c x \right ) \cosineIntegral \left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right ) b +10 \arcsin \left (c x \right ) \sinIntegral \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) b +10 \arcsin \left (c x \right ) \cosineIntegral \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) b +5 \sinIntegral \left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right ) a +5 \cosineIntegral \left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right ) a -15 \sinIntegral \left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right ) a -15 \cosineIntegral \left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right ) a +10 \sinIntegral \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) a +10 \cosineIntegral \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) a -10 x b c -\sin \left (5 \arcsin \left (c x \right )\right ) b +5 \sin \left (3 \arcsin \left (c x \right )\right ) b}{16 c^{6} \left (a +b \arcsin \left (c x \right )\right ) b^{2}}\)	\(341\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a+b*arcsin(c*x))^2/(-c^2*x^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/16/c^6*(5*arcsin(c*x)*Si(5*arcsin(c*x)+5*a/b)*sin(5*a/b)*b+5*arcsin(c*x)*Ci(5*arcsin(c*x)+5*a/b)*cos(5*a/b)*
b-15*arcsin(c*x)*Si(3*arcsin(c*x)+3*a/b)*sin(3*a/b)*b-15*arcsin(c*x)*Ci(3*arcsin(c*x)+3*a/b)*cos(3*a/b)*b+10*a
rcsin(c*x)*Si(arcsin(c*x)+a/b)*sin(a/b)*b+10*arcsin(c*x)*Ci(arcsin(c*x)+a/b)*cos(a/b)*b+5*Si(5*arcsin(c*x)+5*a
/b)*sin(5*a/b)*a+5*Ci(5*arcsin(c*x)+5*a/b)*cos(5*a/b)*a-15*Si(3*arcsin(c*x)+3*a/b)*sin(3*a/b)*a-15*Ci(3*arcsin
(c*x)+3*a/b)*cos(3*a/b)*a+10*Si(arcsin(c*x)+a/b)*sin(a/b)*a+10*Ci(arcsin(c*x)+a/b)*cos(a/b)*a-10*x*b*c-sin(5*a
rcsin(c*x))*b+5*sin(3*arcsin(c*x))*b)/(a+b*arcsin(c*x))/b^2

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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(x^5/(a+b*arcsin(c*x))^2/(-c^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

-(x^5 - 5*(b^2*c*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + a*b*c)*integrate(x^4/(b^2*c*arctan2(c*x, sqrt(c*
x + 1)*sqrt(-c*x + 1)) + a*b*c), x))/(b^2*c*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + a*b*c)

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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(x^5/(a+b*arcsin(c*x))^2/(-c^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-c^2*x^2 + 1)*x^5/(a^2*c^2*x^2 + (b^2*c^2*x^2 - b^2)*arcsin(c*x)^2 - a^2 + 2*(a*b*c^2*x^2 - a*b
)*arcsin(c*x)), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{5}}{\sqrt {- \left (c x - 1\right ) \left (c x + 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(a+b*asin(c*x))**2/(-c**2*x**2+1)**(1/2),x)

[Out]

Integral(x**5/(sqrt(-(c*x - 1)*(c*x + 1))*(a + b*asin(c*x))**2), x)

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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(x^5/(a+b*arcsin(c*x))^2/(-c^2*x^2+1)^(1/2),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 \frac {x^5}{{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,\sqrt {1-c^2\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/((a + b*asin(c*x))^2*(1 - c^2*x^2)^(1/2)),x)

[Out]

int(x^5/((a + b*asin(c*x))^2*(1 - c^2*x^2)^(1/2)), x)

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