3.10.47 \(\int \frac {(a+b x^2)^{3/2}}{x \sqrt {c+d x^2}} \, dx\) [947]

Optimal. Leaf size=133 \[ \frac {b \sqrt {a+b x^2} \sqrt {c+d x^2}}{2 d}-\frac {a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{\sqrt {c}}-\frac {\sqrt {b} (b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{2 d^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {457, 104, 163, 65, 223, 212, 95, 214} \begin {gather*} -\frac {a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{\sqrt {c}}-\frac {\sqrt {b} (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 d^{3/2}}+\frac {b \sqrt {a+b x^2} \sqrt {c+d x^2}}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(2*d) - (a^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x^2])/(Sqrt[a]*Sqrt[c + d*x^2
])])/Sqrt[c] - (Sqrt[b]*(b*c - 3*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 104

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(m + n + p + 1))), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 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 )^{3/2}}{x \sqrt {c+d x^2}} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {(a+b x)^{3/2}}{x \sqrt {c+d x}} \, dx,x,x^2\right )\\ &=\frac {b \sqrt {a+b x^2} \sqrt {c+d x^2}}{2 d}+\frac {\text {Subst}\left (\int \frac {a^2 d-\frac {1}{2} b (b c-3 a d) x}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^2\right )}{2 d}\\ &=\frac {b \sqrt {a+b x^2} \sqrt {c+d x^2}}{2 d}+\frac {1}{2} a^2 \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^2\right )-\frac {(b (b c-3 a d)) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^2\right )}{4 d}\\ &=\frac {b \sqrt {a+b x^2} \sqrt {c+d x^2}}{2 d}+a^2 \text {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x^2}}{\sqrt {c+d x^2}}\right )-\frac {(b c-3 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 {b \sqrt {a+b x^2} \sqrt {c+d x^2}}{2 d}-\frac {a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{\sqrt {c}}-\frac {(b c-3 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 \sqrt {a+b x^2} \sqrt {c+d x^2}}{2 d}-\frac {a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{\sqrt {c}}-\frac {\sqrt {b} (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 d^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 1.23, size = 132, normalized size = 0.99 \begin {gather*} \frac {1}{2} \left (\frac {b \sqrt {a+b x^2} \sqrt {c+d x^2}}{d}-\frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c+d x^2}}{\sqrt {c} \sqrt {a+b x^2}}\right )}{\sqrt {c}}-\frac {\sqrt {b} (b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {d} \sqrt {a+b x^2}}\right )}{d^{3/2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(251\) vs. \(2(101)=202\).
time = 0.12, size = 252, normalized size = 1.89

method	result	size
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \left (\frac {b \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{2 d}+\frac {3 a b \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right )}{4 \sqrt {b d}}-\frac {b^{2} \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) c}{4 d \sqrt {b d}}-\frac {a^{2} \ln \left (\frac {2 a c +\left (a d +b c \right ) x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{x^{2}}\right )}{2 \sqrt {a c}}\right )}{\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}\)	\(243\)
default	\(\frac {\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\, \left (3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, a b d -\ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, b^{2} c -2 \sqrt {b d}\, \ln \left (\frac {a d \,x^{2}+c \,x^{2} b +2 \sqrt {a c}\, \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}+2 a c}{x^{2}}\right ) a^{2} d +2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}\, \sqrt {a c}\, b \right )}{4 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, d \sqrt {b d}\, \sqrt {a c}}\)	\(252\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.25, size = 918, normalized size = 6.90 \begin {gather*} \left [\frac {2 \, a d \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 ) - {\left (b c - 3 \, a d\right )} \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 \, \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} b}{8 \, d}, \frac {a d \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 ) + {\left (b c - 3 \, a d\right )} \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 \, \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} b}{4 \, d}, \frac {4 \, a d \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 c - 3 \, a d\right )} \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 \, \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} b}{8 \, d}, \frac {2 \, a d \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 c - 3 \, a d\right )} \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 \, \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} b}{4 \, d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(2*a*d*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) - (b*c - 3*a*d)*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*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*b)/d, 1/4*(a*d*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) + (b*c - 3*a*d)*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*x^4 + a*b*c + (b^2*c + a*b*d)*x^2)) + 2*sqrt(b*x^2 + a)*sqrt(d*x
^2 + c)*b)/d, 1/8*(4*a*d*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*c - 3*a*d)*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)*s
qrt(b/d)) + 4*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*b)/d, 1/4*(2*a*d*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*c - 3*a*d)*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*x^4 + a*b*c + (b^2*c
+ a*b*d)*x^2)) + 2*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*b)/d]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:index.cc index_m i_lex_is_greater Error: Bad Argument Value

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

[Out]

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

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