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

Optimal. Leaf size=137 \[ \frac {(b c-a d) (a d+3 b c) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{8 b^{3/2} d^{5/2}}-\frac {\sqrt {a+b x^2} \sqrt {c+d x^2} (a d+3 b c)}{8 b d^2}+\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 b d} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 137, 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, 80, 50, 63, 217, 206} \[ \frac {(b c-a d) (a d+3 b c) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{8 b^{3/2} d^{5/2}}-\frac {\sqrt {a+b x^2} \sqrt {c+d x^2} (a d+3 b c)}{8 b d^2}+\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 50

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 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 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^3 \sqrt {a+b x^2}}{\sqrt {c+d x^2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x \sqrt {a+b x}}{\sqrt {c+d x}} \, dx,x,x^2\right )\\ &=\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 b d}-\frac {(3 b c+a d) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx,x,x^2\right )}{8 b d}\\ &=-\frac {(3 b c+a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 b d^2}+\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 b d}+\frac {((b c-a d) (3 b c+a d)) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^2\right )}{16 b d^2}\\ &=-\frac {(3 b c+a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 b d^2}+\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 b d}+\frac {((b c-a d) (3 b c+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 )}{8 b^2 d^2}\\ &=-\frac {(3 b c+a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 b d^2}+\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 b d}+\frac {((b c-a d) (3 b c+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 )}{8 b^2 d^2}\\ &=-\frac {(3 b c+a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 b d^2}+\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 b d}+\frac {(b c-a d) (3 b c+a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{8 b^{3/2} d^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.31, size = 138, normalized size = 1.01 \[ \frac {b \sqrt {d} \sqrt {a+b x^2} \left (c+d x^2\right ) \left (a d-3 b c+2 b d x^2\right )+(a d+3 b c) (b c-a d)^{3/2} \sqrt {\frac {b \left (c+d x^2\right )}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b c-a d}}\right )}{8 b^2 d^{5/2} \sqrt {c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[d]*Sqrt[a + b*x^2]*(c + d*x^2)*(-3*b*c + a*d + 2*b*d*x^2) + (b*c - a*d)^(3/2)*(3*b*c + a*d)*Sqrt[(b*(c
+ d*x^2))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x^2])/Sqrt[b*c - a*d]])/(8*b^2*d^(5/2)*Sqrt[c + d*x^2])

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fricas [A]  time = 1.66, size = 334, normalized size = 2.44 \[ \left [-\frac {{\left (3 \, b^{2} c^{2} - 2 \, a b c d - a^{2} d^{2}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{2} - 4 \, {\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {b d}\right ) - 4 \, {\left (2 \, b^{2} d^{2} x^{2} - 3 \, b^{2} c d + a b d^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{32 \, b^{2} d^{3}}, -\frac {{\left (3 \, b^{2} c^{2} - 2 \, a b c d - a^{2} d^{2}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {-b d}}{2 \, {\left (b^{2} d^{2} x^{4} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x^{2}\right )}}\right ) - 2 \, {\left (2 \, b^{2} d^{2} x^{2} - 3 \, b^{2} c d + a b d^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{16 \, b^{2} d^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/32*((3*b^2*c^2 - 2*a*b*c*d - a^2*d^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*(2*b^2*d^2*x^2
- 3*b^2*c*d + a*b*d^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(b^2*d^3), -1/16*((3*b^2*c^2 - 2*a*b*c*d - a^2*d^2)*sq
rt(-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*(2*b^2*d^2*x^2 - 3*b^2*c*d + a*b*d^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(b^2*d^
3)]

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giac [A]  time = 0.56, size = 159, normalized size = 1.16 \[ \frac {{\left (\sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d} \sqrt {b x^{2} + a} {\left (\frac {2 \, {\left (b x^{2} + a\right )}}{b^{2} d} - \frac {3 \, b^{3} c d + a b^{2} d^{2}}{b^{4} d^{3}}\right )} - \frac {{\left (3 \, b^{2} c^{2} - 2 \, a b c d - a^{2} d^{2}\right )} \log \left ({\left | -\sqrt {b x^{2} + a} \sqrt {b d} + \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} b d^{2}}\right )} b}{8 \, {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d)*sqrt(b*x^2 + a)*(2*(b*x^2 + a)/(b^2*d) - (3*b^3*c*d + a*b^2*d^2)/(b
^4*d^3)) - (3*b^2*c^2 - 2*a*b*c*d - a^2*d^2)*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^2))*b/abs(b)

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maple [B]  time = 0.02, size = 339, normalized size = 2.47 \[ -\frac {\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\, \left (a^{2} d^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {x^{4} b d +x^{2} a d +b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+2 a b c d \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {x^{4} b d +x^{2} a d +b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-3 b^{2} c^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {x^{4} b d +x^{2} a d +b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-4 \sqrt {b d}\, \sqrt {x^{4} b d +x^{2} a d +b c \,x^{2}+a c}\, b d \,x^{2}-2 \sqrt {b d}\, \sqrt {x^{4} b d +x^{2} a d +b c \,x^{2}+a c}\, a d +6 \sqrt {b d}\, \sqrt {x^{4} b d +x^{2} a d +b c \,x^{2}+a c}\, b c \right )}{16 \sqrt {x^{4} b d +x^{2} a d +b c \,x^{2}+a c}\, \sqrt {b d}\, b \,d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)*(-4*(b*d)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*x^2*b*d+d^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+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*c*b*d-3*b^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))*c^2-2*(b*d)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^
(1/2)*a*d+6*(b*d)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*b*c)/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)/d^2/b/(b*
d)^(1/2)

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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^3*(b*x^2+a)^(1/2)/(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 = 25.18, size = 639, normalized size = 4.66 \[ \frac {\frac {\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )\,\left (\frac {a^2\,b^2\,d^2}{4}+\frac {a\,b^3\,c\,d}{2}-\frac {3\,b^4\,c^2}{4}\right )}{d^6\,\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}+\frac {{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^3\,\left (\frac {7\,a^2\,b\,d^2}{4}+\frac {23\,a\,b^2\,c\,d}{2}+\frac {11\,b^3\,c^2}{4}\right )}{d^5\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^3}+\frac {{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^5\,\left (\frac {7\,a^2\,d^2}{4}+\frac {23\,a\,b\,c\,d}{2}+\frac {11\,b^2\,c^2}{4}\right )}{d^4\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^5}+\frac {{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^7\,\left (\frac {a^2\,d^2}{4}+\frac {a\,b\,c\,d}{2}-\frac {3\,b^2\,c^2}{4}\right )}{b\,d^3\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^7}-\frac {4\,a^{3/2}\,\sqrt {c}\,{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^6}{d^2\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^6}-\frac {\sqrt {a}\,\sqrt {c}\,{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^4\,\left (16\,c\,b^2+8\,a\,d\,b\right )}{d^4\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^4}-\frac {4\,a^{3/2}\,b^2\,\sqrt {c}\,{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^2}{d^4\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^2}}{\frac {{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^8}{{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^8}+\frac {b^4}{d^4}-\frac {4\,b^3\,{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^2}{d^3\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^2}+\frac {6\,b^2\,{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^4}{d^2\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^4}-\frac {4\,b\,{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^6}{d\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^6}}-\frac {\mathrm {atanh}\left (\frac {\sqrt {d}\,\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}{\sqrt {b}\,\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}\right )\,\left (a\,d-b\,c\right )\,\left (a\,d+3\,b\,c\right )}{4\,b^{3/2}\,d^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((((a + b*x^2)^(1/2) - a^(1/2))*((a^2*b^2*d^2)/4 - (3*b^4*c^2)/4 + (a*b^3*c*d)/2))/(d^6*((c + d*x^2)^(1/2) - c
^(1/2))) + (((a + b*x^2)^(1/2) - a^(1/2))^3*((11*b^3*c^2)/4 + (7*a^2*b*d^2)/4 + (23*a*b^2*c*d)/2))/(d^5*((c +
d*x^2)^(1/2) - c^(1/2))^3) + (((a + b*x^2)^(1/2) - a^(1/2))^5*((7*a^2*d^2)/4 + (11*b^2*c^2)/4 + (23*a*b*c*d)/2
))/(d^4*((c + d*x^2)^(1/2) - c^(1/2))^5) + (((a + b*x^2)^(1/2) - a^(1/2))^7*((a^2*d^2)/4 - (3*b^2*c^2)/4 + (a*
b*c*d)/2))/(b*d^3*((c + d*x^2)^(1/2) - c^(1/2))^7) - (4*a^(3/2)*c^(1/2)*((a + b*x^2)^(1/2) - a^(1/2))^6)/(d^2*
((c + d*x^2)^(1/2) - c^(1/2))^6) - (a^(1/2)*c^(1/2)*((a + b*x^2)^(1/2) - a^(1/2))^4*(16*b^2*c + 8*a*b*d))/(d^4
*((c + d*x^2)^(1/2) - c^(1/2))^4) - (4*a^(3/2)*b^2*c^(1/2)*((a + b*x^2)^(1/2) - a^(1/2))^2)/(d^4*((c + d*x^2)^
(1/2) - c^(1/2))^2))/(((a + b*x^2)^(1/2) - a^(1/2))^8/((c + d*x^2)^(1/2) - c^(1/2))^8 + b^4/d^4 - (4*b^3*((a +
b*x^2)^(1/2) - a^(1/2))^2)/(d^3*((c + d*x^2)^(1/2) - c^(1/2))^2) + (6*b^2*((a + b*x^2)^(1/2) - a^(1/2))^4)/(d
^2*((c + d*x^2)^(1/2) - c^(1/2))^4) - (4*b*((a + b*x^2)^(1/2) - a^(1/2))^6)/(d*((c + d*x^2)^(1/2) - c^(1/2))^6
)) - (atanh((d^(1/2)*((a + b*x^2)^(1/2) - a^(1/2)))/(b^(1/2)*((c + d*x^2)^(1/2) - c^(1/2))))*(a*d - b*c)*(a*d
+ 3*b*c))/(4*b^(3/2)*d^(5/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \sqrt {a + b x^{2}}}{\sqrt {c + d x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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