3.10.58 \(\int \frac {(a+b x^2)^{5/2}}{x^3 \sqrt {c+d x^2}} \, dx\) [958]

Optimal. Leaf size=187 \[ \frac {b (b c+a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{2 c d}-\frac {a \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{2 c x^2}-\frac {a^{3/2} (5 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 c^{3/2}}-\frac {b^{3/2} (b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{2 d^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.16, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {457, 100, 159, 163, 65, 223, 212, 95, 214} \begin {gather*} -\frac {a^{3/2} (5 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 c^{3/2}}-\frac {b^{3/2} (b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{2 d^{3/2}}-\frac {a \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{2 c x^2}+\frac {b \sqrt {a+b x^2} \sqrt {c+d x^2} (a d+b c)}{2 c d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

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 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 100

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

Rule 159

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2
*n, 2*p]

Rule 163

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 212

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

Rule 214

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

Rule 223

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 457

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

________________________________________________________________________________________

Mathematica [A]
time = 2.23, size = 166, normalized size = 0.89 \begin {gather*} \frac {a^{3/2} (-5 b c+a d) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c+d x^2}}{\sqrt {c} \sqrt {a+b x^2}}\right )+\frac {\sqrt {c} \left (\frac {\sqrt {d} \sqrt {a+b x^2} \left (-a^2 d+b^2 c x^2\right ) \sqrt {c+d x^2}}{x^2}-b^{3/2} c (b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {d} \sqrt {a+b x^2}}\right )\right )}{d^{3/2}}}{2 c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(373\) vs. \(2(147)=294\).
time = 0.17, size = 374, normalized size = 2.00

method	result	size
risch	\(-\frac {a^{2} \sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}{2 c \,x^{2}}+\frac {\left (\frac {b^{2} \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{2 d}+\frac {5 b^{2} \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) a}{4 \sqrt {b d}}-\frac {c \,b^{3} \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right )}{4 d \sqrt {b d}}+\frac {a^{3} \ln \left (\frac {2 a c +\left (a d +b c \right ) x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{x^{2}}\right ) d}{4 c \sqrt {a c}}-\frac {5 a^{2} \ln \left (\frac {2 a c +\left (a d +b c \right ) x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{x^{2}}\right ) b}{4 \sqrt {a c}}\right ) \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}}{\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}\)	\(343\)
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \left (\frac {b^{2} \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{2 d}+\frac {5 b^{2} \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) a}{4 \sqrt {b d}}-\frac {c \,b^{3} \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right )}{4 d \sqrt {b d}}-\frac {5 a^{2} \ln \left (\frac {2 a c +\left (a d +b c \right ) x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{x^{2}}\right ) b}{4 \sqrt {a c}}-\frac {a^{2} \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{2 c \,x^{2}}+\frac {a^{3} \ln \left (\frac {2 a c +\left (a d +b c \right ) x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{x^{2}}\right ) d}{4 c \sqrt {a c}}\right )}{\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}\)	\(347\)
default	\(\frac {\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\, \left (5 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{2} c d \,x^{2} \sqrt {a c}-\ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{3} c^{2} x^{2} \sqrt {a c}+\ln \left (\frac {a d \,x^{2}+c \,x^{2} b +2 \sqrt {a c}\, \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}+2 a c}{x^{2}}\right ) a^{3} d^{2} x^{2} \sqrt {b d}-5 \ln \left (\frac {a d \,x^{2}+c \,x^{2} b +2 \sqrt {a c}\, \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}+2 a c}{x^{2}}\right ) a^{2} b c d \,x^{2} \sqrt {b d}+2 b^{2} c \,x^{2} \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {a c}\, \sqrt {b d}-2 a^{2} d \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {a c}\, \sqrt {b d}\right )}{4 c \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, x^{2} \sqrt {b d}\, \sqrt {a c}\, d}\)	\(374\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/c*(5*ln(1/2*(2*b*d*x^2+2*((b*x^2+a)*(d*x^2+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/
(b*d)^(1/2))*a*b^2*c*d*x^2*(a*c)^(1/2)-ln(1/2*(2*b*d*x^2+2*((b*x^2+a)*(d*x^2+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b
*d)^(1/2))*b^3*c^2*x^2*(a*c)^(1/2)+ln((a*d*x^2+c*x^2*b+2*(a*c)^(1/2)*((b*x^2+a)*(d*x^2+c))^(1/2)+2*a*c)/x^2)*a
^3*d^2*x^2*(b*d)^(1/2)-5*ln((a*d*x^2+c*x^2*b+2*(a*c)^(1/2)*((b*x^2+a)*(d*x^2+c))^(1/2)+2*a*c)/x^2)*a^2*b*c*d*x
^2*(b*d)^(1/2)+2*b^2*c*x^2*((b*x^2+a)*(d*x^2+c))^(1/2)*(a*c)^(1/2)*(b*d)^(1/2)-2*a^2*d*((b*x^2+a)*(d*x^2+c))^(
1/2)*(a*c)^(1/2)*(b*d)^(1/2))/((b*x^2+a)*(d*x^2+c))^(1/2)/x^2/(b*d)^(1/2)/(a*c)^(1/2)/d

________________________________________________________________________________________

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((b*x^2+a)^(5/2)/x^3/(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 detail

________________________________________________________________________________________

Fricas [A]
time = 1.95, size = 1097, normalized size = 5.87 \begin {gather*} \left [-\frac {{\left (b^{2} c^{2} - 5 \, a b c d\right )} x^{2} \sqrt {\frac {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^{2} x^{2} + b c d + a d^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {\frac {b}{d}}\right ) + {\left (5 \, a b c d - a^{2} d^{2}\right )} x^{2} \sqrt {\frac {a}{c}} \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x^{2} + 4 \, {\left (2 \, a c^{2} + {\left (b c^{2} + a c d\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {\frac {a}{c}}}{x^{4}}\right ) - 4 \, {\left (b^{2} c x^{2} - a^{2} d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{8 \, c d x^{2}}, \frac {2 \, {\left (b^{2} c^{2} - 5 \, a b c d\right )} x^{2} \sqrt {-\frac {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 {-\frac {b}{d}}}{2 \, {\left (b^{2} d x^{4} + a b c + {\left (b^{2} c + a b d\right )} x^{2}\right )}}\right ) - {\left (5 \, a b c d - a^{2} d^{2}\right )} x^{2} \sqrt {\frac {a}{c}} \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x^{2} + 4 \, {\left (2 \, a c^{2} + {\left (b c^{2} + a c d\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {\frac {a}{c}}}{x^{4}}\right ) + 4 \, {\left (b^{2} c x^{2} - a^{2} d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{8 \, c d x^{2}}, \frac {2 \, {\left (5 \, a b c d - a^{2} d^{2}\right )} x^{2} \sqrt {-\frac {a}{c}} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {-\frac {a}{c}}}{2 \, {\left (a b d x^{4} + a^{2} c + {\left (a b c + a^{2} d\right )} x^{2}\right )}}\right ) - {\left (b^{2} c^{2} - 5 \, a b c d\right )} x^{2} \sqrt {\frac {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^{2} x^{2} + b c d + a d^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {\frac {b}{d}}\right ) + 4 \, {\left (b^{2} c x^{2} - a^{2} d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{8 \, c d x^{2}}, \frac {{\left (5 \, a b c d - a^{2} d^{2}\right )} x^{2} \sqrt {-\frac {a}{c}} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {-\frac {a}{c}}}{2 \, {\left (a b d x^{4} + a^{2} c + {\left (a b c + a^{2} d\right )} x^{2}\right )}}\right ) + {\left (b^{2} c^{2} - 5 \, a b c d\right )} x^{2} \sqrt {-\frac {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 {-\frac {b}{d}}}{2 \, {\left (b^{2} d x^{4} + a b c + {\left (b^{2} c + a b d\right )} x^{2}\right )}}\right ) + 2 \, {\left (b^{2} c x^{2} - a^{2} d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{4 \, c d x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*((b^2*c^2 - 5*a*b*c*d)*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^2*x^2 + b*c*d + a*d^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(b/d)) + (5*a*b*c*d - a^2*d^2
)*x^2*sqrt(a/c)*log(((b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^4 + 8*a^2*c^2 + 8*(a*b*c^2 + a^2*c*d)*x^2 + 4*(2*a*c^2
+ (b*c^2 + a*c*d)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(a/c))/x^4) - 4*(b^2*c*x^2 - a^2*d)*sqrt(b*x^2 + a)
*sqrt(d*x^2 + c))/(c*d*x^2), 1/8*(2*(b^2*c^2 - 5*a*b*c*d)*x^2*sqrt(-b/d)*arctan(1/2*(2*b*d*x^2 + b*c + a*d)*sq
rt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-b/d)/(b^2*d*x^4 + a*b*c + (b^2*c + a*b*d)*x^2)) - (5*a*b*c*d - a^2*d^2)*x^
2*sqrt(a/c)*log(((b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^4 + 8*a^2*c^2 + 8*(a*b*c^2 + a^2*c*d)*x^2 + 4*(2*a*c^2 + (b
*c^2 + a*c*d)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(a/c))/x^4) + 4*(b^2*c*x^2 - a^2*d)*sqrt(b*x^2 + a)*sqr
t(d*x^2 + c))/(c*d*x^2), 1/8*(2*(5*a*b*c*d - a^2*d^2)*x^2*sqrt(-a/c)*arctan(1/2*((b*c + a*d)*x^2 + 2*a*c)*sqrt
(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-a/c)/(a*b*d*x^4 + a^2*c + (a*b*c + a^2*d)*x^2)) - (b^2*c^2 - 5*a*b*c*d)*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^2*x^2 + b*c
*d + a*d^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(b/d)) + 4*(b^2*c*x^2 - a^2*d)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)
)/(c*d*x^2), 1/4*((5*a*b*c*d - a^2*d^2)*x^2*sqrt(-a/c)*arctan(1/2*((b*c + a*d)*x^2 + 2*a*c)*sqrt(b*x^2 + a)*sq
rt(d*x^2 + c)*sqrt(-a/c)/(a*b*d*x^4 + a^2*c + (a*b*c + a^2*d)*x^2)) + (b^2*c^2 - 5*a*b*c*d)*x^2*sqrt(-b/d)*arc
tan(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*x^4 + a*b*c + (b^2*c + a*b*d
)*x^2)) + 2*(b^2*c*x^2 - a^2*d)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(c*d*x^2)]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 558 vs. \(2 (147) = 294\).
time = 1.45, size = 558, normalized size = 2.98 \begin {gather*} \frac {b {\left (\frac {2 \, \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d} \sqrt {b x^{2} + a} b}{d} + \frac {{\left (\sqrt {b d} b^{2} c - 5 \, \sqrt {b d} a b 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 )}{d^{2}} - \frac {2 \, {\left (5 \, \sqrt {b d} a^{2} b^{2} c - \sqrt {b d} a^{3} b d\right )} \arctan \left (-\frac {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}}{2 \, \sqrt {-a b c d} b}\right )}{\sqrt {-a b c d} b c} - \frac {4 \, {\left (\sqrt {b d} a^{2} b^{4} c^{2} - 2 \, \sqrt {b d} a^{3} b^{3} c d + \sqrt {b d} a^{4} b^{2} d^{2} - \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^{2} c - \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^{3} b d\right )}}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 2 \, {\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^{2} c - 2 \, {\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 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}\right )} c}\right )}}{4 \, {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*(2*sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d)*sqrt(b*x^2 + a)*b/d + (sqrt(b*d)*b^2*c - 5*sqrt(b*d)*a*b*d)*log
((sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2)/d^2 - 2*(5*sqrt(b*d)*a^2*b^2*c - sqrt(
b*d)*a^3*b*d)*arctan(-1/2*(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)/(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)*b*c) - 4*(sqrt(b*d)*a^2*b^4*c^2 - 2*sqrt(b*d)*a^3*b^3*c*d + sqrt(b*d)*
a^4*b^2*d^2 - 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^2*c - 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^3*b*d)/((b^4*c^2 - 2*a*b^3*c*d +
a^2*b^2*d^2 - 2*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*b^2*c - 2*(sqrt(b*x^2 +
a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*a*b*d + (sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*
x^2 + a)*b*d - a*b*d))^4)*c))/abs(b)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________