3.1.95 \(\int x (d-c^2 d x^2)^{5/2} (a+b \text {ArcSin}(c x)) \, dx\) [95]

Optimal. Leaf size=202 \[ \frac {b d^2 x \sqrt {d-c^2 d x^2}}{7 c \sqrt {1-c^2 x^2}}-\frac {b c d^2 x^3 \sqrt {d-c^2 d x^2}}{7 \sqrt {1-c^2 x^2}}+\frac {3 b c^3 d^2 x^5 \sqrt {d-c^2 d x^2}}{35 \sqrt {1-c^2 x^2}}-\frac {b c^5 d^2 x^7 \sqrt {d-c^2 d x^2}}{49 \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{7/2} (a+b \text {ArcSin}(c x))}{7 c^2 d} \]

[Out]

-1/7*(-c^2*d*x^2+d)^(7/2)*(a+b*arcsin(c*x))/c^2/d+1/7*b*d^2*x*(-c^2*d*x^2+d)^(1/2)/c/(-c^2*x^2+1)^(1/2)-1/7*b*
c*d^2*x^3*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)+3/35*b*c^3*d^2*x^5*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)-1
/49*b*c^5*d^2*x^7*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {4767, 200} \begin {gather*} -\frac {\left (d-c^2 d x^2\right )^{7/2} (a+b \text {ArcSin}(c x))}{7 c^2 d}+\frac {b d^2 x \sqrt {d-c^2 d x^2}}{7 c \sqrt {1-c^2 x^2}}-\frac {b c d^2 x^3 \sqrt {d-c^2 d x^2}}{7 \sqrt {1-c^2 x^2}}-\frac {b c^5 d^2 x^7 \sqrt {d-c^2 d x^2}}{49 \sqrt {1-c^2 x^2}}+\frac {3 b c^3 d^2 x^5 \sqrt {d-c^2 d x^2}}{35 \sqrt {1-c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d^2*x*Sqrt[d - c^2*d*x^2])/(7*c*Sqrt[1 - c^2*x^2]) - (b*c*d^2*x^3*Sqrt[d - c^2*d*x^2])/(7*Sqrt[1 - c^2*x^2]
) + (3*b*c^3*d^2*x^5*Sqrt[d - c^2*d*x^2])/(35*Sqrt[1 - c^2*x^2]) - (b*c^5*d^2*x^7*Sqrt[d - c^2*d*x^2])/(49*Sqr
t[1 - c^2*x^2]) - ((d - c^2*d*x^2)^(7/2)*(a + b*ArcSin[c*x]))/(7*c^2*d)

Rule 200

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rule 4767

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

Rubi steps

\begin {align*} \int x \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx &=-\frac {\left (d-c^2 d x^2\right )^{7/2} \left (a+b \sin ^{-1}(c x)\right )}{7 c^2 d}+\frac {\left (b d^2 \sqrt {d-c^2 d x^2}\right ) \int \left (1-c^2 x^2\right )^3 \, dx}{7 c \sqrt {1-c^2 x^2}}\\ &=-\frac {\left (d-c^2 d x^2\right )^{7/2} \left (a+b \sin ^{-1}(c x)\right )}{7 c^2 d}+\frac {\left (b d^2 \sqrt {d-c^2 d x^2}\right ) \int \left (1-3 c^2 x^2+3 c^4 x^4-c^6 x^6\right ) \, dx}{7 c \sqrt {1-c^2 x^2}}\\ &=\frac {b d^2 x \sqrt {d-c^2 d x^2}}{7 c \sqrt {1-c^2 x^2}}-\frac {b c d^2 x^3 \sqrt {d-c^2 d x^2}}{7 \sqrt {1-c^2 x^2}}+\frac {3 b c^3 d^2 x^5 \sqrt {d-c^2 d x^2}}{35 \sqrt {1-c^2 x^2}}-\frac {b c^5 d^2 x^7 \sqrt {d-c^2 d x^2}}{49 \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{7/2} \left (a+b \sin ^{-1}(c x)\right )}{7 c^2 d}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 113, normalized size = 0.56 \begin {gather*} \frac {d^2 \sqrt {d-c^2 d x^2} \left (35 a \left (-1+c^2 x^2\right )^4+b c x \sqrt {1-c^2 x^2} \left (-35+35 c^2 x^2-21 c^4 x^4+5 c^6 x^6\right )+35 b \left (-1+c^2 x^2\right )^4 \text {ArcSin}(c x)\right )}{245 c^2 \left (-1+c^2 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*Sqrt[d - c^2*d*x^2]*(35*a*(-1 + c^2*x^2)^4 + b*c*x*Sqrt[1 - c^2*x^2]*(-35 + 35*c^2*x^2 - 21*c^4*x^4 + 5*c
^6*x^6) + 35*b*(-1 + c^2*x^2)^4*ArcSin[c*x]))/(245*c^2*(-1 + c^2*x^2))

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Maple [C] Result contains complex when optimal does not.
time = 0.15, size = 717, normalized size = 3.55

method	result	size
default	\(-\frac {a \left (-c^{2} d \,x^{2}+d \right )^{\frac {7}{2}}}{7 c^{2} d}+b \left (\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (64 c^{8} x^{8}-144 c^{6} x^{6}-64 i \sqrt {-c^{2} x^{2}+1}\, x^{7} c^{7}+104 c^{4} x^{4}+112 i \sqrt {-c^{2} x^{2}+1}\, x^{5} c^{5}-25 c^{2} x^{2}-56 i \sqrt {-c^{2} x^{2}+1}\, x^{3} c^{3}+7 i \sqrt {-c^{2} x^{2}+1}\, x c +1\right ) \left (i+7 \arcsin \left (c x \right )\right ) d^{2}}{6272 c^{2} \left (c^{2} x^{2}-1\right )}-\frac {5 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (c^{2} x^{2}-i \sqrt {-c^{2} x^{2}+1}\, x c -1\right ) \left (\arcsin \left (c x \right )+i\right ) d^{2}}{128 c^{2} \left (c^{2} x^{2}-1\right )}-\frac {5 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i \sqrt {-c^{2} x^{2}+1}\, x c +c^{2} x^{2}-1\right ) \left (\arcsin \left (c x \right )-i\right ) d^{2}}{128 c^{2} \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (4 i \sqrt {-c^{2} x^{2}+1}\, x^{3} c^{3}+4 c^{4} x^{4}-3 i \sqrt {-c^{2} x^{2}+1}\, x c -5 c^{2} x^{2}+1\right ) \left (-i+3 \arcsin \left (c x \right )\right ) d^{2}}{128 c^{2} \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i \sqrt {-c^{2} x^{2}+1}\, x c +c^{2} x^{2}-1\right ) \left (11 i+70 \arcsin \left (c x \right )\right ) \cos \left (6 \arcsin \left (c x \right )\right ) d^{2}}{7840 c^{2} \left (c^{2} x^{2}-1\right )}-\frac {3 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i x^{2} c^{2}-c x \sqrt {-c^{2} x^{2}+1}-i\right ) \left (9 i+35 \arcsin \left (c x \right )\right ) \sin \left (6 \arcsin \left (c x \right )\right ) d^{2}}{15680 c^{2} \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i \sqrt {-c^{2} x^{2}+1}\, x c +c^{2} x^{2}-1\right ) \left (i+5 \arcsin \left (c x \right )\right ) \cos \left (4 \arcsin \left (c x \right )\right ) d^{2}}{160 c^{2} \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i x^{2} c^{2}-c x \sqrt {-c^{2} x^{2}+1}-i\right ) \left (3 i+5 \arcsin \left (c x \right )\right ) \sin \left (4 \arcsin \left (c x \right )\right ) d^{2}}{320 c^{2} \left (c^{2} x^{2}-1\right )}\right )\)	\(717\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7*a/c^2/d*(-c^2*d*x^2+d)^(7/2)+b*(1/6272*(-d*(c^2*x^2-1))^(1/2)*(64*c^8*x^8-144*c^6*x^6-64*I*(-c^2*x^2+1)^(
1/2)*x^7*c^7+104*c^4*x^4+112*I*(-c^2*x^2+1)^(1/2)*x^5*c^5-25*c^2*x^2-56*I*(-c^2*x^2+1)^(1/2)*x^3*c^3+7*I*(-c^2
*x^2+1)^(1/2)*x*c+1)*(I+7*arcsin(c*x))*d^2/c^2/(c^2*x^2-1)-5/128*(-d*(c^2*x^2-1))^(1/2)*(c^2*x^2-I*(-c^2*x^2+1
)^(1/2)*x*c-1)*(arcsin(c*x)+I)*d^2/c^2/(c^2*x^2-1)-5/128*(-d*(c^2*x^2-1))^(1/2)*(I*(-c^2*x^2+1)^(1/2)*x*c+c^2*
x^2-1)*(arcsin(c*x)-I)*d^2/c^2/(c^2*x^2-1)+1/128*(-d*(c^2*x^2-1))^(1/2)*(4*I*(-c^2*x^2+1)^(1/2)*x^3*c^3+4*c^4*
x^4-3*I*(-c^2*x^2+1)^(1/2)*x*c-5*c^2*x^2+1)*(-I+3*arcsin(c*x))*d^2/c^2/(c^2*x^2-1)-1/7840*(-d*(c^2*x^2-1))^(1/
2)*(I*(-c^2*x^2+1)^(1/2)*x*c+c^2*x^2-1)*(11*I+70*arcsin(c*x))*cos(6*arcsin(c*x))*d^2/c^2/(c^2*x^2-1)-3/15680*(
-d*(c^2*x^2-1))^(1/2)*(I*x^2*c^2-c*x*(-c^2*x^2+1)^(1/2)-I)*(9*I+35*arcsin(c*x))*sin(6*arcsin(c*x))*d^2/c^2/(c^
2*x^2-1)-1/160*(-d*(c^2*x^2-1))^(1/2)*(I*(-c^2*x^2+1)^(1/2)*x*c+c^2*x^2-1)*(I+5*arcsin(c*x))*cos(4*arcsin(c*x)
)*d^2/c^2/(c^2*x^2-1)-1/320*(-d*(c^2*x^2-1))^(1/2)*(I*x^2*c^2-c*x*(-c^2*x^2+1)^(1/2)-I)*(3*I+5*arcsin(c*x))*si
n(4*arcsin(c*x))*d^2/c^2/(c^2*x^2-1))

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Maxima [A]
time = 0.49, size = 98, normalized size = 0.49 \begin {gather*} -\frac {{\left (-c^{2} d x^{2} + d\right )}^{\frac {7}{2}} b \arcsin \left (c x\right )}{7 \, c^{2} d} - \frac {{\left (-c^{2} d x^{2} + d\right )}^{\frac {7}{2}} a}{7 \, c^{2} d} - \frac {{\left (5 \, c^{6} d^{\frac {7}{2}} x^{7} - 21 \, c^{4} d^{\frac {7}{2}} x^{5} + 35 \, c^{2} d^{\frac {7}{2}} x^{3} - 35 \, d^{\frac {7}{2}} x\right )} b}{245 \, c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7*(-c^2*d*x^2 + d)^(7/2)*b*arcsin(c*x)/(c^2*d) - 1/7*(-c^2*d*x^2 + d)^(7/2)*a/(c^2*d) - 1/245*(5*c^6*d^(7/2
)*x^7 - 21*c^4*d^(7/2)*x^5 + 35*c^2*d^(7/2)*x^3 - 35*d^(7/2)*x)*b/(c*d)

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Fricas [A]
time = 1.90, size = 215, normalized size = 1.06 \begin {gather*} \frac {{\left (5 \, b c^{7} d^{2} x^{7} - 21 \, b c^{5} d^{2} x^{5} + 35 \, b c^{3} d^{2} x^{3} - 35 \, b c d^{2} x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} + 35 \, {\left (a c^{8} d^{2} x^{8} - 4 \, a c^{6} d^{2} x^{6} + 6 \, a c^{4} d^{2} x^{4} - 4 \, a c^{2} d^{2} x^{2} + a d^{2} + {\left (b c^{8} d^{2} x^{8} - 4 \, b c^{6} d^{2} x^{6} + 6 \, b c^{4} d^{2} x^{4} - 4 \, b c^{2} d^{2} x^{2} + b d^{2}\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} d x^{2} + d}}{245 \, {\left (c^{4} x^{2} - c^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/245*((5*b*c^7*d^2*x^7 - 21*b*c^5*d^2*x^5 + 35*b*c^3*d^2*x^3 - 35*b*c*d^2*x)*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x
^2 + 1) + 35*(a*c^8*d^2*x^8 - 4*a*c^6*d^2*x^6 + 6*a*c^4*d^2*x^4 - 4*a*c^2*d^2*x^2 + a*d^2 + (b*c^8*d^2*x^8 - 4
*b*c^6*d^2*x^6 + 6*b*c^4*d^2*x^4 - 4*b*c^2*d^2*x^2 + b*d^2)*arcsin(c*x))*sqrt(-c^2*d*x^2 + d))/(c^4*x^2 - c^2)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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