3.10.45 \(\int \frac {x^3 (a+b x^2)^{3/2}}{\sqrt {c+d x^2}} \, dx\) [945]

Optimal. Leaf size=187 \[ \frac {(b c-a d) (5 b c+a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{16 b d^3}-\frac {(5 b c+a d) \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{24 b d^2}+\frac {\left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{6 b d}-\frac {(b c-a d)^2 (5 b c+a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{16 b^{3/2} d^{7/2}} \]

[Out]

-1/16*(-a*d+b*c)^2*(a*d+5*b*c)*arctanh(d^(1/2)*(b*x^2+a)^(1/2)/b^(1/2)/(d*x^2+c)^(1/2))/b^(3/2)/d^(7/2)-1/24*(
a*d+5*b*c)*(b*x^2+a)^(3/2)*(d*x^2+c)^(1/2)/b/d^2+1/6*(b*x^2+a)^(5/2)*(d*x^2+c)^(1/2)/b/d+1/16*(-a*d+b*c)*(a*d+
5*b*c)*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/b/d^3

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Rubi [A]
time = 0.12, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {457, 81, 52, 65, 223, 212} \begin {gather*} -\frac {(b c-a d)^2 (a d+5 b c) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{16 b^{3/2} d^{7/2}}+\frac {\sqrt {a+b x^2} \sqrt {c+d x^2} (b c-a d) (a d+5 b c)}{16 b d^3}-\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2} (a d+5 b c)}{24 b d^2}+\frac {\left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{6 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)*(5*b*c + a*d)*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(16*b*d^3) - ((5*b*c + a*d)*(a + b*x^2)^(3/2)*Sqrt
[c + d*x^2])/(24*b*d^2) + ((a + b*x^2)^(5/2)*Sqrt[c + d*x^2])/(6*b*d) - ((b*c - a*d)^2*(5*b*c + a*d)*ArcTanh[(
Sqrt[d]*Sqrt[a + b*x^2])/(Sqrt[b]*Sqrt[c + d*x^2])])/(16*b^(3/2)*d^(7/2))

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 81

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 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 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 {x^3 \left (a+b x^2\right )^{3/2}}{\sqrt {c+d x^2}} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {x (a+b x)^{3/2}}{\sqrt {c+d x}} \, dx,x,x^2\right )\\ &=\frac {\left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{6 b d}-\frac {(5 b c+a d) \text {Subst}\left (\int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}} \, dx,x,x^2\right )}{12 b d}\\ &=-\frac {(5 b c+a d) \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{24 b d^2}+\frac {\left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{6 b d}+\frac {((b c-a d) (5 b c+a d)) \text {Subst}\left (\int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx,x,x^2\right )}{16 b d^2}\\ &=\frac {(b c-a d) (5 b c+a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{16 b d^3}-\frac {(5 b c+a d) \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{24 b d^2}+\frac {\left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{6 b d}-\frac {\left ((b c-a d)^2 (5 b c+a d)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^2\right )}{32 b d^3}\\ &=\frac {(b c-a d) (5 b c+a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{16 b d^3}-\frac {(5 b c+a d) \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{24 b d^2}+\frac {\left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{6 b d}-\frac {\left ((b c-a d)^2 (5 b c+a d)\right ) \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 )}{16 b^2 d^3}\\ &=\frac {(b c-a d) (5 b c+a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{16 b d^3}-\frac {(5 b c+a d) \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{24 b d^2}+\frac {\left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{6 b d}-\frac {\left ((b c-a d)^2 (5 b c+a d)\right ) \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 )}{16 b^2 d^3}\\ &=\frac {(b c-a d) (5 b c+a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{16 b d^3}-\frac {(5 b c+a d) \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{24 b d^2}+\frac {\left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{6 b d}-\frac {(b c-a d)^2 (5 b c+a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{16 b^{3/2} d^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 1.71, size = 148, normalized size = 0.79 \begin {gather*} \frac {\sqrt {a+b x^2} \sqrt {c+d x^2} \left (3 a^2 d^2+2 a b d \left (-11 c+7 d x^2\right )+b^2 \left (15 c^2-10 c d x^2+8 d^2 x^4\right )\right )}{48 b d^3}-\frac {(b c-a d)^2 (5 b c+a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {d} \sqrt {a+b x^2}}\right )}{16 b^{3/2} d^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]*(3*a^2*d^2 + 2*a*b*d*(-11*c + 7*d*x^2) + b^2*(15*c^2 - 10*c*d*x^2 + 8*d^2*x^4
)))/(48*b*d^3) - ((b*c - a*d)^2*(5*b*c + a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/(Sqrt[d]*Sqrt[a + b*x^2])])/(1
6*b^(3/2)*d^(7/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(454\) vs. \(2(155)=310\).
time = 0.12, size = 455, normalized size = 2.43

method	result	size
risch	\(\frac {\left (8 b^{2} x^{4} d^{2}+14 a b \,d^{2} x^{2}-10 b^{2} c d \,x^{2}+3 a^{2} d^{2}-22 a b c d +15 b^{2} c^{2}\right ) \sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}{48 b \,d^{3}}+\frac {\left (-\frac {\ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) a^{3}}{32 b \sqrt {b d}}-\frac {3 \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) a^{2} c}{32 d \sqrt {b d}}+\frac {9 b \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) a \,c^{2}}{32 d^{2} \sqrt {b d}}-\frac {5 b^{2} \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) c^{3}}{32 d^{3} \sqrt {b d}}\right ) \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}}{\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}\)	\(361\)
default	\(-\frac {\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\, \left (-16 b^{2} d^{2} x^{4} \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}-28 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, x^{2} a b \,d^{2} \sqrt {b d}+20 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, x^{2} c \,b^{2} d \sqrt {b d}+3 d^{3} \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{3}+9 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} c b \,d^{2}-27 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,c^{2} b^{2} d +15 b^{3} \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) c^{3}-6 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, a^{2} d^{2} \sqrt {b d}+44 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, a c b d \sqrt {b d}-30 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, c^{2} b^{2} \sqrt {b d}\right )}{96 b \,d^{3} \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}}\)	\(455\)
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \left (\frac {b \,x^{4} \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{6 d}+\frac {7 x^{2} \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\, a}{24 d}-\frac {5 b \,x^{2} \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\, c}{24 d^{2}}+\frac {\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\, a^{2}}{16 b d}-\frac {11 \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\, a c}{24 d^{2}}+\frac {5 b \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\, c^{2}}{16 d^{3}}-\frac {\ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) a^{3}}{32 b \sqrt {b d}}-\frac {3 \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) a^{2} c}{32 d \sqrt {b d}}+\frac {9 b \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) a \,c^{2}}{32 d^{2} \sqrt {b d}}-\frac {5 b^{2} \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) c^{3}}{32 d^{3} \sqrt {b d}}\right )}{\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}\)	\(473\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)*(-16*b^2*d^2*x^4*((b*x^2+a)*(d*x^2+c))^(1/2)*(b*d)^(1/2)-28*((b*x^2+a)*(
d*x^2+c))^(1/2)*x^2*a*b*d^2*(b*d)^(1/2)+20*((b*x^2+a)*(d*x^2+c))^(1/2)*x^2*c*b^2*d*(b*d)^(1/2)+3*d^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^3+9*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^2*c*b*d^2-27*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^2*b^2*d+15*b^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))*c^3-6*((b*x^2+a)*(d*x^2+c))^(1/2)*a^2*d^2*(b*d)^(1/2)+44*((b*x^2+a)*(d*x^2+c))^(1
/2)*a*c*b*d*(b*d)^(1/2)-30*((b*x^2+a)*(d*x^2+c))^(1/2)*c^2*b^2*(b*d)^(1/2))/b/d^3/((b*x^2+a)*(d*x^2+c))^(1/2)/
(b*d)^(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(x^3*(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 detail

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Fricas [A]
time = 0.71, size = 440, normalized size = 2.35 \begin {gather*} \left [\frac {3 \, {\left (5 \, b^{3} c^{3} - 9 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + a^{3} d^{3}\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 (8 \, b^{3} d^{3} x^{4} + 15 \, b^{3} c^{2} d - 22 \, a b^{2} c d^{2} + 3 \, a^{2} b d^{3} - 2 \, {\left (5 \, b^{3} c d^{2} - 7 \, a b^{2} d^{3}\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{192 \, b^{2} d^{4}}, \frac {3 \, {\left (5 \, b^{3} c^{3} - 9 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + a^{3} d^{3}\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 (8 \, b^{3} d^{3} x^{4} + 15 \, b^{3} c^{2} d - 22 \, a b^{2} c d^{2} + 3 \, a^{2} b d^{3} - 2 \, {\left (5 \, b^{3} c d^{2} - 7 \, a b^{2} d^{3}\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{96 \, b^{2} d^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/192*(3*(5*b^3*c^3 - 9*a*b^2*c^2*d + 3*a^2*b*c*d^2 + a^3*d^3)*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*(8*b^3*d^3*x^4 + 15*b^3*c^2*d - 22*a*b^2*c*d^2 + 3*a^2*b*d^3 - 2*(5*b^3*c*d^2 - 7*a*b^2*d^3)*x^2)*sqrt(
b*x^2 + a)*sqrt(d*x^2 + c))/(b^2*d^4), 1/96*(3*(5*b^3*c^3 - 9*a*b^2*c^2*d + 3*a^2*b*c*d^2 + a^3*d^3)*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*(8*b^3*d^3*x^4 + 15*b^3*c^2*d - 22*a*b^2*c*d^2 + 3*a^2*b*d^3 - 2*(5*b^3*c*d^2 - 7*a*b
^2*d^3)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(b^2*d^4)]

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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 {3}{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)**(3/2)/(d*x**2+c)**(1/2),x)

[Out]

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

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Giac [A]
time = 0.87, size = 225, normalized size = 1.20 \begin {gather*} \frac {{\left (\sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d} \sqrt {b x^{2} + a} {\left (2 \, {\left (b x^{2} + a\right )} {\left (\frac {4 \, {\left (b x^{2} + a\right )}}{b^{2} d} - \frac {5 \, b^{3} c d^{3} + a b^{2} d^{4}}{b^{4} d^{5}}\right )} + \frac {3 \, {\left (5 \, b^{4} c^{2} d^{2} - 4 \, a b^{3} c d^{3} - a^{2} b^{2} d^{4}\right )}}{b^{4} d^{5}}\right )} + \frac {3 \, {\left (5 \, b^{3} c^{3} - 9 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + a^{3} d^{3}\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 |}\right )}{\sqrt {b d} b d^{3}}\right )} b}{48 \, {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d)*sqrt(b*x^2 + a)*(2*(b*x^2 + a)*(4*(b*x^2 + a)/(b^2*d) - (5*b^3*c*d
^3 + a*b^2*d^4)/(b^4*d^5)) + 3*(5*b^4*c^2*d^2 - 4*a*b^3*c*d^3 - a^2*b^2*d^4)/(b^4*d^5)) + 3*(5*b^3*c^3 - 9*a*b
^2*c^2*d + 3*a^2*b*c*d^2 + a^3*d^3)*log(abs(-sqrt(b*x^2 + a)*sqrt(b*d) + sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d)
))/(sqrt(b*d)*b*d^3))*b/abs(b)

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

[Out]

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

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