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

Optimal. Leaf size=129 \[ -\frac {a^2 \sqrt {c+d x^2}}{b^2 \sqrt {a+b x^2} (b c-a d)}-\frac {(3 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{2 b^{5/2} d^{3/2}}+\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{2 b^2 d} \]

________________________________________________________________________________________

Rubi [A]  time = 0.16, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {446, 89, 80, 63, 217, 206} \begin {gather*} -\frac {a^2 \sqrt {c+d x^2}}{b^2 \sqrt {a+b x^2} (b c-a d)}-\frac {(3 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{2 b^{5/2} d^{3/2}}+\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{2 b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 80

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

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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 )^{3/2} \sqrt {c+d x^2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^2}{(a+b x)^{3/2} \sqrt {c+d x}} \, dx,x,x^2\right )\\ &=-\frac {a^2 \sqrt {c+d x^2}}{b^2 (b c-a d) \sqrt {a+b x^2}}+\frac {\operatorname {Subst}\left (\int \frac {-\frac {1}{2} a (b c-a d)+\frac {1}{2} b (b c-a d) x}{\sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^2\right )}{b^2 (b c-a d)}\\ &=-\frac {a^2 \sqrt {c+d x^2}}{b^2 (b c-a d) \sqrt {a+b x^2}}+\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{2 b^2 d}-\frac {(b c+3 a d) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^2\right )}{4 b^2 d}\\ &=-\frac {a^2 \sqrt {c+d x^2}}{b^2 (b c-a d) \sqrt {a+b x^2}}+\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{2 b^2 d}-\frac {(b c+3 a d) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b x^2}\right )}{2 b^3 d}\\ &=-\frac {a^2 \sqrt {c+d x^2}}{b^2 (b c-a d) \sqrt {a+b x^2}}+\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{2 b^2 d}-\frac {(b c+3 a d) \operatorname {Subst}\left (\int \frac {1}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b x^2}}{\sqrt {c+d x^2}}\right )}{2 b^3 d}\\ &=-\frac {a^2 \sqrt {c+d x^2}}{b^2 (b c-a d) \sqrt {a+b x^2}}+\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{2 b^2 d}-\frac {(b c+3 a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{2 b^{5/2} d^{3/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.35, size = 185, normalized size = 1.43 \begin {gather*} \frac {\sqrt {a+b x^2} \sqrt {b c-a d} \left (-3 a^2 d^2+2 a b c d+b^2 c^2\right ) \sqrt {\frac {b \left (c+d x^2\right )}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b c-a d}}\right )-b \sqrt {d} \left (c+d x^2\right ) \left (-3 a^2 d+a b \left (c-d x^2\right )+b^2 c x^2\right )}{2 b^3 d^{3/2} \sqrt {a+b x^2} \sqrt {c+d x^2} (a d-b c)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 2.33, size = 169, normalized size = 1.31 \begin {gather*} \frac {\sqrt {c+d x^2} \left (-\frac {2 a^2 b d \left (c+d x^2\right )}{a+b x^2}+3 a^2 d^2-2 a b c d+b^2 c^2\right )}{2 b^2 d \sqrt {a+b x^2} (b c-a d) \left (\frac {b \left (c+d x^2\right )}{a+b x^2}-d\right )}+\frac {(-3 a d-b c) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {d} \sqrt {a+b x^2}}\right )}{2 b^{5/2} d^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x^2]*(b^2*c^2 - 2*a*b*c*d + 3*a^2*d^2 - (2*a^2*b*d*(c + d*x^2))/(a + b*x^2)))/(2*b^2*d*(b*c - a*d)
*Sqrt[a + b*x^2]*(-d + (b*(c + d*x^2))/(a + b*x^2))) + ((-(b*c) - 3*a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/(Sq
rt[d]*Sqrt[a + b*x^2])])/(2*b^(5/2)*d^(3/2))

________________________________________________________________________________________

fricas [B]  time = 1.45, size = 498, normalized size = 3.86 \begin {gather*} \left [\frac {{\left (a b^{2} c^{2} + 2 \, a^{2} b c d - 3 \, a^{3} d^{2} + {\left (b^{3} c^{2} + 2 \, a b^{2} c d - 3 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{2} - 4 \, {\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {b d}\right ) + 4 \, {\left (a b^{2} c d - 3 \, a^{2} b d^{2} + {\left (b^{3} c d - a b^{2} d^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{8 \, {\left (a b^{4} c d^{2} - a^{2} b^{3} d^{3} + {\left (b^{5} c d^{2} - a b^{4} d^{3}\right )} x^{2}\right )}}, \frac {{\left (a b^{2} c^{2} + 2 \, a^{2} b c d - 3 \, a^{3} d^{2} + {\left (b^{3} c^{2} + 2 \, a b^{2} c d - 3 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {-b d}}{2 \, {\left (b^{2} d^{2} x^{4} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x^{2}\right )}}\right ) + 2 \, {\left (a b^{2} c d - 3 \, a^{2} b d^{2} + {\left (b^{3} c d - a b^{2} d^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{4 \, {\left (a b^{4} c d^{2} - a^{2} b^{3} d^{3} + {\left (b^{5} c d^{2} - a b^{4} d^{3}\right )} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*((a*b^2*c^2 + 2*a^2*b*c*d - 3*a^3*d^2 + (b^3*c^2 + 2*a*b^2*c*d - 3*a^2*b*d^2)*x^2)*sqrt(b*d)*log(8*b^2*d^
2*x^4 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 8*(b^2*c*d + a*b*d^2)*x^2 - 4*(2*b*d*x^2 + b*c + a*d)*sqrt(b*x^2 + a)*
sqrt(d*x^2 + c)*sqrt(b*d)) + 4*(a*b^2*c*d - 3*a^2*b*d^2 + (b^3*c*d - a*b^2*d^2)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^
2 + c))/(a*b^4*c*d^2 - a^2*b^3*d^3 + (b^5*c*d^2 - a*b^4*d^3)*x^2), 1/4*((a*b^2*c^2 + 2*a^2*b*c*d - 3*a^3*d^2 +
(b^3*c^2 + 2*a*b^2*c*d - 3*a^2*b*d^2)*x^2)*sqrt(-b*d)*arctan(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt(b*x^2 + a)*sqrt
(d*x^2 + c)*sqrt(-b*d)/(b^2*d^2*x^4 + a*b*c*d + (b^2*c*d + a*b*d^2)*x^2)) + 2*(a*b^2*c*d - 3*a^2*b*d^2 + (b^3*
c*d - a*b^2*d^2)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(a*b^4*c*d^2 - a^2*b^3*d^3 + (b^5*c*d^2 - a*b^4*d^3)*x^
2)]

________________________________________________________________________________________

giac [A]  time = 0.54, size = 192, normalized size = 1.49 \begin {gather*} -\frac {2 \, \sqrt {b d} a^{2}}{{\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 )} b {\left | b \right |}} + \frac {\sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d} \sqrt {b x^{2} + a} {\left | b \right |}}{2 \, b^{4} d} + \frac {{\left (\sqrt {b d} b c + 3 \, \sqrt {b d} a d\right )} \log \left ({\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 )}{4 \, b^{2} d^{2} {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.05, size = 553, normalized size = 4.29 \begin {gather*} -\frac {\left (3 a^{2} b \,d^{2} x^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-2 a \,b^{2} c d \,x^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-b^{3} c^{2} x^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+3 a^{3} d^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-2 a^{2} b c d \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-a \,b^{2} c^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}\, a b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}\, b^{2} c \,x^{2}-6 \sqrt {b d}\, \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, a^{2} d +2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}\, a b c \right ) \sqrt {d \,x^{2}+c}}{4 \sqrt {b \,x^{2}+a}\, \sqrt {b d}\, \left (a d -b c \right ) \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, b^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(3*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x^2*a^2*b*d^
2-2*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x^2*a*b^2*c*d-ln
(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x^2*b^3*c^2-2*((b*x^2+
a)*(d*x^2+c))^(1/2)*(b*d)^(1/2)*x^2*a*b*d+2*((b*x^2+a)*(d*x^2+c))^(1/2)*(b*d)^(1/2)*x^2*b^2*c+3*ln(1/2*(2*b*d*
x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*a^3*d^2-2*ln(1/2*(2*b*d*x^2+a*d+b*
c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*a^2*b*c*d-ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x
^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*a*b^2*c^2-6*(b*d)^(1/2)*((b*x^2+a)*(d*x^2+c))^(1/2)*a^
2*d+2*((b*x^2+a)*(d*x^2+c))^(1/2)*(b*d)^(1/2)*a*b*c)*(d*x^2+c)^(1/2)/b^2/(b*x^2+a)^(1/2)/d/(b*d)^(1/2)/(a*d-b*
c)/((b*x^2+a)*(d*x^2+c))^(1/2)

________________________________________________________________________________________

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^5/(b*x^2+a)^(3/2)/(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?

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^5}{{\left (b\,x^2+a\right )}^{3/2}\,\sqrt {d\,x^2+c}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________