3.986 \(\int \frac {x^5}{(a+b x^2)^{7/2} \sqrt {c+d x^2}} \, dx\)

Optimal. Leaf size=154 \[ -\frac {\sqrt {c+d x^2} \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right )}{15 b^2 \sqrt {a+b x^2} (b c-a d)^3}-\frac {a^2 \sqrt {c+d x^2}}{5 b^2 \left (a+b x^2\right )^{5/2} (b c-a d)}+\frac {2 a \sqrt {c+d x^2} (5 b c-3 a d)}{15 b^2 \left (a+b x^2\right )^{3/2} (b c-a d)^2} \]

[Out]

-1/5*a^2*(d*x^2+c)^(1/2)/b^2/(-a*d+b*c)/(b*x^2+a)^(5/2)+2/15*a*(-3*a*d+5*b*c)*(d*x^2+c)^(1/2)/b^2/(-a*d+b*c)^2
/(b*x^2+a)^(3/2)-1/15*(3*a^2*d^2-10*a*b*c*d+15*b^2*c^2)*(d*x^2+c)^(1/2)/b^2/(-a*d+b*c)^3/(b*x^2+a)^(1/2)

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Rubi [A]  time = 0.18, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {446, 89, 78, 37} \[ -\frac {\sqrt {c+d x^2} \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right )}{15 b^2 \sqrt {a+b x^2} (b c-a d)^3}-\frac {a^2 \sqrt {c+d x^2}}{5 b^2 \left (a+b x^2\right )^{5/2} (b c-a d)}+\frac {2 a \sqrt {c+d x^2} (5 b c-3 a d)}{15 b^2 \left (a+b x^2\right )^{3/2} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Int[x^5/((a + b*x^2)^(7/2)*Sqrt[c + d*x^2]),x]

[Out]

-(a^2*Sqrt[c + d*x^2])/(5*b^2*(b*c - a*d)*(a + b*x^2)^(5/2)) + (2*a*(5*b*c - 3*a*d)*Sqrt[c + d*x^2])/(15*b^2*(
b*c - a*d)^2*(a + b*x^2)^(3/2)) - ((15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*Sqrt[c + d*x^2])/(15*b^2*(b*c - a*d)^
3*Sqrt[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 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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^5}{\left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^2}{(a+b x)^{7/2} \sqrt {c+d x}} \, dx,x,x^2\right )\\ &=-\frac {a^2 \sqrt {c+d x^2}}{5 b^2 (b c-a d) \left (a+b x^2\right )^{5/2}}+\frac {\operatorname {Subst}\left (\int \frac {-\frac {1}{2} a (5 b c-a d)+\frac {5}{2} b (b c-a d) x}{(a+b x)^{5/2} \sqrt {c+d x}} \, dx,x,x^2\right )}{5 b^2 (b c-a d)}\\ &=-\frac {a^2 \sqrt {c+d x^2}}{5 b^2 (b c-a d) \left (a+b x^2\right )^{5/2}}+\frac {2 a (5 b c-3 a d) \sqrt {c+d x^2}}{15 b^2 (b c-a d)^2 \left (a+b x^2\right )^{3/2}}+\frac {\left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{(a+b x)^{3/2} \sqrt {c+d x}} \, dx,x,x^2\right )}{30 b^2 (b c-a d)^2}\\ &=-\frac {a^2 \sqrt {c+d x^2}}{5 b^2 (b c-a d) \left (a+b x^2\right )^{5/2}}+\frac {2 a (5 b c-3 a d) \sqrt {c+d x^2}}{15 b^2 (b c-a d)^2 \left (a+b x^2\right )^{3/2}}-\frac {\left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right ) \sqrt {c+d x^2}}{15 b^2 (b c-a d)^3 \sqrt {a+b x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 91, normalized size = 0.59 \[ -\frac {\sqrt {c+d x^2} \left (a^2 \left (8 c^2-4 c d x^2+3 d^2 x^4\right )+10 a b c x^2 \left (2 c-d x^2\right )+15 b^2 c^2 x^4\right )}{15 \left (a+b x^2\right )^{5/2} (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/((a + b*x^2)^(7/2)*Sqrt[c + d*x^2]),x]

[Out]

-1/15*(Sqrt[c + d*x^2]*(15*b^2*c^2*x^4 + 10*a*b*c*x^2*(2*c - d*x^2) + a^2*(8*c^2 - 4*c*d*x^2 + 3*d^2*x^4)))/((
b*c - a*d)^3*(a + b*x^2)^(5/2))

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fricas [A]  time = 1.15, size = 262, normalized size = 1.70 \[ -\frac {{\left ({\left (15 \, b^{2} c^{2} - 10 \, a b c d + 3 \, a^{2} d^{2}\right )} x^{4} + 8 \, a^{2} c^{2} + 4 \, {\left (5 \, a b c^{2} - a^{2} c d\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{15 \, {\left (a^{3} b^{3} c^{3} - 3 \, a^{4} b^{2} c^{2} d + 3 \, a^{5} b c d^{2} - a^{6} d^{3} + {\left (b^{6} c^{3} - 3 \, a b^{5} c^{2} d + 3 \, a^{2} b^{4} c d^{2} - a^{3} b^{3} d^{3}\right )} x^{6} + 3 \, {\left (a b^{5} c^{3} - 3 \, a^{2} b^{4} c^{2} d + 3 \, a^{3} b^{3} c d^{2} - a^{4} b^{2} d^{3}\right )} x^{4} + 3 \, {\left (a^{2} b^{4} c^{3} - 3 \, a^{3} b^{3} c^{2} d + 3 \, a^{4} b^{2} c d^{2} - a^{5} b d^{3}\right )} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^2+a)^(7/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

-1/15*((15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*x^4 + 8*a^2*c^2 + 4*(5*a*b*c^2 - a^2*c*d)*x^2)*sqrt(b*x^2 + a)*sq
rt(d*x^2 + c)/(a^3*b^3*c^3 - 3*a^4*b^2*c^2*d + 3*a^5*b*c*d^2 - a^6*d^3 + (b^6*c^3 - 3*a*b^5*c^2*d + 3*a^2*b^4*
c*d^2 - a^3*b^3*d^3)*x^6 + 3*(a*b^5*c^3 - 3*a^2*b^4*c^2*d + 3*a^3*b^3*c*d^2 - a^4*b^2*d^3)*x^4 + 3*(a^2*b^4*c^
3 - 3*a^3*b^3*c^2*d + 3*a^4*b^2*c*d^2 - a^5*b*d^3)*x^2)

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giac [B]  time = 0.87, size = 597, normalized size = 3.88 \[ -\frac {2 \, {\left (15 \, \sqrt {b d} b^{8} c^{4} - 40 \, \sqrt {b d} a b^{7} c^{3} d + 38 \, \sqrt {b d} a^{2} b^{6} c^{2} d^{2} - 16 \, \sqrt {b d} a^{3} b^{5} c d^{3} + 3 \, \sqrt {b d} a^{4} b^{4} d^{4} - 60 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} b^{6} c^{3} + 80 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a b^{5} c^{2} d - 20 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a^{2} b^{4} c d^{2} + 90 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{4} b^{4} c^{2} - 40 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{4} a b^{3} c d + 30 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{4} a^{2} b^{2} d^{2} - 60 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{6} b^{2} c + 15 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{8}\right )}}{15 \, {\left (b^{2} c - a b d - {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2}\right )}^{5} b {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^2+a)^(7/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

-2/15*(15*sqrt(b*d)*b^8*c^4 - 40*sqrt(b*d)*a*b^7*c^3*d + 38*sqrt(b*d)*a^2*b^6*c^2*d^2 - 16*sqrt(b*d)*a^3*b^5*c
*d^3 + 3*sqrt(b*d)*a^4*b^4*d^4 - 60*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^6*c^3 + 80*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*a*b^5*c^2*d
- 20*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*a^2*b^4*c*d^2 + 90*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^4*c^2 - 40*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*a*b^3*c*d + 30*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*a^2*b^2*d^2 - 60*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c
+ (b*x^2 + a)*b*d - a*b*d))^6*b^2*c + 15*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d -
a*b*d))^8)/((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)^5*b*abs(b
))

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maple [A]  time = 0.01, size = 119, normalized size = 0.77 \[ \frac {\sqrt {d \,x^{2}+c}\, \left (3 a^{2} d^{2} x^{4}-10 a b c d \,x^{4}+15 b^{2} c^{2} x^{4}-4 a^{2} c d \,x^{2}+20 a b \,c^{2} x^{2}+8 a^{2} c^{2}\right )}{15 \left (b \,x^{2}+a \right )^{\frac {5}{2}} \left (a^{3} d^{3}-3 a^{2} c \,d^{2} b +3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(b*x^2+a)^(7/2)/(d*x^2+c)^(1/2),x)

[Out]

1/15*(d*x^2+c)^(1/2)*(3*a^2*d^2*x^4-10*a*b*c*d*x^4+15*b^2*c^2*x^4-4*a^2*c*d*x^2+20*a*b*c^2*x^2+8*a^2*c^2)/(b*x
^2+a)^(5/2)/(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^2+a)^(7/2)/(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 = 1.60, size = 220, normalized size = 1.43 \[ \frac {\sqrt {b\,x^2+a}\,\left (\frac {8\,a^2\,c^3}{15\,b^3\,{\left (a\,d-b\,c\right )}^3}+\frac {x^4\,\left (-a^2\,c\,d^2+10\,a\,b\,c^2\,d+15\,b^2\,c^3\right )}{15\,b^3\,{\left (a\,d-b\,c\right )}^3}+\frac {x^6\,\left (3\,a^2\,d^3-10\,a\,b\,c\,d^2+15\,b^2\,c^2\,d\right )}{15\,b^3\,{\left (a\,d-b\,c\right )}^3}+\frac {4\,a\,c^2\,x^2\,\left (a\,d+5\,b\,c\right )}{15\,b^3\,{\left (a\,d-b\,c\right )}^3}\right )}{x^6\,\sqrt {d\,x^2+c}+\frac {a^3\,\sqrt {d\,x^2+c}}{b^3}+\frac {3\,a\,x^4\,\sqrt {d\,x^2+c}}{b}+\frac {3\,a^2\,x^2\,\sqrt {d\,x^2+c}}{b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/((a + b*x^2)^(7/2)*(c + d*x^2)^(1/2)),x)

[Out]

((a + b*x^2)^(1/2)*((8*a^2*c^3)/(15*b^3*(a*d - b*c)^3) + (x^4*(15*b^2*c^3 - a^2*c*d^2 + 10*a*b*c^2*d))/(15*b^3
*(a*d - b*c)^3) + (x^6*(3*a^2*d^3 + 15*b^2*c^2*d - 10*a*b*c*d^2))/(15*b^3*(a*d - b*c)^3) + (4*a*c^2*x^2*(a*d +
5*b*c))/(15*b^3*(a*d - b*c)^3)))/(x^6*(c + d*x^2)^(1/2) + (a^3*(c + d*x^2)^(1/2))/b^3 + (3*a*x^4*(c + d*x^2)^
(1/2))/b + (3*a^2*x^2*(c + d*x^2)^(1/2))/b^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5}}{\left (a + b x^{2}\right )^{\frac {7}{2}} \sqrt {c + d x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(b*x**2+a)**(7/2)/(d*x**2+c)**(1/2),x)

[Out]

Integral(x**5/((a + b*x**2)**(7/2)*sqrt(c + d*x**2)), x)

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