3.8.97 \(\int \frac {x^3}{(a+b x^2)^{5/2} \sqrt {c+d x^2}} \, dx\)

Optimal. Leaf size=89 \[ \frac {a \sqrt {c+d x^2}}{3 b \left (a+b x^2\right )^{3/2} (b c-a d)}-\frac {\sqrt {c+d x^2} (3 b c-a d)}{3 b \sqrt {a+b x^2} (b c-a d)^2} \]

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Rubi [A]  time = 0.07, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {446, 78, 37} \begin {gather*} \frac {a \sqrt {c+d x^2}}{3 b \left (a+b x^2\right )^{3/2} (b c-a d)}-\frac {\sqrt {c+d x^2} (3 b c-a d)}{3 b \sqrt {a+b x^2} (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/((a + b*x^2)^(5/2)*Sqrt[c + d*x^2]),x]

[Out]

(a*Sqrt[c + d*x^2])/(3*b*(b*c - a*d)*(a + b*x^2)^(3/2)) - ((3*b*c - a*d)*Sqrt[c + d*x^2])/(3*b*(b*c - a*d)^2*S
qrt[a + b*x^2])

Rule 37

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

Rule 78

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

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 54, normalized size = 0.61 \begin {gather*} \frac {\sqrt {c+d x^2} \left (-2 a c+a d x^2-3 b c x^2\right )}{3 \left (a+b x^2\right )^{3/2} (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/((a + b*x^2)^(5/2)*Sqrt[c + d*x^2]),x]

[Out]

(Sqrt[c + d*x^2]*(-2*a*c - 3*b*c*x^2 + a*d*x^2))/(3*(b*c - a*d)^2*(a + b*x^2)^(3/2))

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IntegrateAlgebraic [A]  time = 2.28, size = 64, normalized size = 0.72 \begin {gather*} \frac {\frac {a \left (c+d x^2\right )^{3/2}}{\left (a+b x^2\right )^{3/2}}-\frac {3 c \sqrt {c+d x^2}}{\sqrt {a+b x^2}}}{3 (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^3/((a + b*x^2)^(5/2)*Sqrt[c + d*x^2]),x]

[Out]

((-3*c*Sqrt[c + d*x^2])/Sqrt[a + b*x^2] + (a*(c + d*x^2)^(3/2))/(a + b*x^2)^(3/2))/(3*(b*c - a*d)^2)

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fricas [A]  time = 1.55, size = 128, normalized size = 1.44 \begin {gather*} -\frac {{\left ({\left (3 \, b c - a d\right )} x^{2} + 2 \, a c\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{3 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2} + {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} x^{4} + 2 \, {\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^2+a)^(5/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

-1/3*((3*b*c - a*d)*x^2 + 2*a*c)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)/(a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2 + (b^4*c
^2 - 2*a*b^3*c*d + a^2*b^2*d^2)*x^4 + 2*(a*b^3*c^2 - 2*a^2*b^2*c*d + a^3*b*d^2)*x^2)

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giac [B]  time = 0.63, size = 214, normalized size = 2.40 \begin {gather*} -\frac {2 \, {\left (3 \, \sqrt {b d} b^{5} c^{2} - 4 \, \sqrt {b d} a b^{4} c d + \sqrt {b d} a^{2} b^{3} d^{2} - 6 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} b^{3} c + 3 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{4} b\right )}}{3 \, {\left (b^{2} c - a b d - {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2}\right )}^{3} b {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^2+a)^(5/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

-2/3*(3*sqrt(b*d)*b^5*c^2 - 4*sqrt(b*d)*a*b^4*c*d + sqrt(b*d)*a^2*b^3*d^2 - 6*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(
b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*b^3*c + 3*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c +
(b*x^2 + a)*b*d - a*b*d))^4*b)/((b^2*c - a*b*d - (sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a
*b*d))^2)^3*b*abs(b))

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maple [A]  time = 0.01, size = 63, normalized size = 0.71 \begin {gather*} -\frac {\sqrt {d \,x^{2}+c}\, \left (-a d \,x^{2}+3 b c \,x^{2}+2 a c \right )}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(b*x^2+a)^(5/2)/(d*x^2+c)^(1/2),x)

[Out]

-1/3*(d*x^2+c)^(1/2)*(-a*d*x^2+3*b*c*x^2+2*a*c)/(b*x^2+a)^(3/2)/(a^2*d^2-2*a*b*c*d+b^2*c^2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^2+a)^(5/2)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c zero or nonzero?

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mupad [B]  time = 1.41, size = 139, normalized size = 1.56 \begin {gather*} -\frac {\sqrt {b\,x^2+a}\,\left (\frac {2\,a\,c^2}{3\,b^2\,{\left (a\,d-b\,c\right )}^2}+\frac {x^2\,\left (3\,b\,c^2+a\,d\,c\right )}{3\,b^2\,{\left (a\,d-b\,c\right )}^2}-\frac {x^4\,\left (a\,d^2-3\,b\,c\,d\right )}{3\,b^2\,{\left (a\,d-b\,c\right )}^2}\right )}{x^4\,\sqrt {d\,x^2+c}+\frac {a^2\,\sqrt {d\,x^2+c}}{b^2}+\frac {2\,a\,x^2\,\sqrt {d\,x^2+c}}{b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/((a + b*x^2)^(5/2)*(c + d*x^2)^(1/2)),x)

[Out]

-((a + b*x^2)^(1/2)*((2*a*c^2)/(3*b^2*(a*d - b*c)^2) + (x^2*(3*b*c^2 + a*c*d))/(3*b^2*(a*d - b*c)^2) - (x^4*(a
*d^2 - 3*b*c*d))/(3*b^2*(a*d - b*c)^2)))/(x^4*(c + d*x^2)^(1/2) + (a^2*(c + d*x^2)^(1/2))/b^2 + (2*a*x^2*(c +
d*x^2)^(1/2))/b)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\left (a + b x^{2}\right )^{\frac {5}{2}} \sqrt {c + d x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3/((a + b*x**2)**(5/2)*sqrt(c + d*x**2)), x)

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