3.10.40 \(\int \frac {x^4 \sqrt {a+b x^2}}{\sqrt {c+d x^2}} \, dx\) [940]

Optimal. Leaf size=343 \[ \frac {\left (8 b^2 c^2-3 a b c d-2 a^2 d^2\right ) x \sqrt {a+b x^2}}{15 b^2 d^2 \sqrt {c+d x^2}}-\frac {(4 b c-a d) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{15 b d^2}+\frac {x^3 \sqrt {a+b x^2} \sqrt {c+d x^2}}{5 d}-\frac {\sqrt {c} \left (8 b^2 c^2-3 a b c d-2 a^2 d^2\right ) \sqrt {a+b x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 b^2 d^{5/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}+\frac {c^{3/2} (4 b c-a d) \sqrt {a+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 b d^{5/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}} \]

[Out]

1/15*(-2*a^2*d^2-3*a*b*c*d+8*b^2*c^2)*x*(b*x^2+a)^(1/2)/b^2/d^2/(d*x^2+c)^(1/2)+1/15*c^(3/2)*(-a*d+4*b*c)*(1/(
1+d*x^2/c))^(1/2)*(1+d*x^2/c)^(1/2)*EllipticF(x*d^(1/2)/c^(1/2)/(1+d*x^2/c)^(1/2),(1-b*c/a/d)^(1/2))*(b*x^2+a)
^(1/2)/b/d^(5/2)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(d*x^2+c)^(1/2)-1/15*(-2*a^2*d^2-3*a*b*c*d+8*b^2*c^2)*(1/(1+d
*x^2/c))^(1/2)*(1+d*x^2/c)^(1/2)*EllipticE(x*d^(1/2)/c^(1/2)/(1+d*x^2/c)^(1/2),(1-b*c/a/d)^(1/2))*c^(1/2)*(b*x
^2+a)^(1/2)/b^2/d^(5/2)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(d*x^2+c)^(1/2)-1/15*(-a*d+4*b*c)*x*(b*x^2+a)^(1/2)*(d
*x^2+c)^(1/2)/b/d^2+1/5*x^3*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/d

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Rubi [A]
time = 0.21, antiderivative size = 343, 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 = {489, 596, 545, 429, 506, 422} \begin {gather*} -\frac {\sqrt {c} \sqrt {a+b x^2} \left (-2 a^2 d^2-3 a b c d+8 b^2 c^2\right ) E\left (\text {ArcTan}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 b^2 d^{5/2} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {x \sqrt {a+b x^2} \left (-2 a^2 d^2-3 a b c d+8 b^2 c^2\right )}{15 b^2 d^2 \sqrt {c+d x^2}}+\frac {c^{3/2} \sqrt {a+b x^2} (4 b c-a d) F\left (\text {ArcTan}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 b d^{5/2} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {x \sqrt {a+b x^2} \sqrt {c+d x^2} (4 b c-a d)}{15 b d^2}+\frac {x^3 \sqrt {a+b x^2} \sqrt {c+d x^2}}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*Sqrt[a + b*x^2])/Sqrt[c + d*x^2],x]

[Out]

((8*b^2*c^2 - 3*a*b*c*d - 2*a^2*d^2)*x*Sqrt[a + b*x^2])/(15*b^2*d^2*Sqrt[c + d*x^2]) - ((4*b*c - a*d)*x*Sqrt[a
+ b*x^2]*Sqrt[c + d*x^2])/(15*b*d^2) + (x^3*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(5*d) - (Sqrt[c]*(8*b^2*c^2 - 3*
a*b*c*d - 2*a^2*d^2)*Sqrt[a + b*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(15*b^2*d^(5/2)*
Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2]) + (c^(3/2)*(4*b*c - a*d)*Sqrt[a + b*x^2]*EllipticF[ArcT
an[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(15*b*d^(5/2)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2]
)

Rule 422

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]/(c*Rt[d/c, 2]*Sq
rt[c + d*x^2]*Sqrt[c*((a + b*x^2)/(a*(c + d*x^2)))]))*EllipticE[ArcTan[Rt[d/c, 2]*x], 1 - b*(c/(a*d))], x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rule 429

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]/(a*Rt[d/c, 2]*
Sqrt[c + d*x^2]*Sqrt[c*((a + b*x^2)/(a*(c + d*x^2)))]))*EllipticF[ArcTan[Rt[d/c, 2]*x], 1 - b*(c/(a*d))], x] /
; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

Rule 489

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

Rule 506

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[x*(Sqrt[a + b*x^2]/(b*Sqrt
[c + d*x^2])), x] - Dist[c/b, Int[Sqrt[a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]

Rule 545

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

Rule 596

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

Rubi steps

\begin {align*} \int \frac {x^4 \sqrt {a+b x^2}}{\sqrt {c+d x^2}} \, dx &=\frac {x^3 \sqrt {a+b x^2} \sqrt {c+d x^2}}{5 d}-\frac {\int \frac {x^2 \left (3 a c+(4 b c-a d) x^2\right )}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{5 d}\\ &=-\frac {(4 b c-a d) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{15 b d^2}+\frac {x^3 \sqrt {a+b x^2} \sqrt {c+d x^2}}{5 d}+\frac {\int \frac {a c (4 b c-a d)+\left (8 b^2 c^2-3 a b c d-2 a^2 d^2\right ) x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{15 b d^2}\\ &=-\frac {(4 b c-a d) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{15 b d^2}+\frac {x^3 \sqrt {a+b x^2} \sqrt {c+d x^2}}{5 d}+\frac {(a c (4 b c-a d)) \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{15 b d^2}+\frac {\left (8 b^2 c^2-3 a b c d-2 a^2 d^2\right ) \int \frac {x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{15 b d^2}\\ &=\frac {\left (8 b^2 c^2-3 a b c d-2 a^2 d^2\right ) x \sqrt {a+b x^2}}{15 b^2 d^2 \sqrt {c+d x^2}}-\frac {(4 b c-a d) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{15 b d^2}+\frac {x^3 \sqrt {a+b x^2} \sqrt {c+d x^2}}{5 d}+\frac {c^{3/2} (4 b c-a d) \sqrt {a+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 b d^{5/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}-\frac {\left (c \left (8 b^2 c^2-3 a b c d-2 a^2 d^2\right )\right ) \int \frac {\sqrt {a+b x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{15 b^2 d^2}\\ &=\frac {\left (8 b^2 c^2-3 a b c d-2 a^2 d^2\right ) x \sqrt {a+b x^2}}{15 b^2 d^2 \sqrt {c+d x^2}}-\frac {(4 b c-a d) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{15 b d^2}+\frac {x^3 \sqrt {a+b x^2} \sqrt {c+d x^2}}{5 d}-\frac {\sqrt {c} \left (8 b^2 c^2-3 a b c d-2 a^2 d^2\right ) \sqrt {a+b x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 b^2 d^{5/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}+\frac {c^{3/2} (4 b c-a d) \sqrt {a+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 b d^{5/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 1.60, size = 246, normalized size = 0.72 \begin {gather*} \frac {\sqrt {\frac {b}{a}} d x \left (a+b x^2\right ) \left (c+d x^2\right ) \left (-4 b c+a d+3 b d x^2\right )+i c \left (-8 b^2 c^2+3 a b c d+2 a^2 d^2\right ) \sqrt {1+\frac {b x^2}{a}} \sqrt {1+\frac {d x^2}{c}} E\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )-i c \left (-8 b^2 c^2+7 a b c d+a^2 d^2\right ) \sqrt {1+\frac {b x^2}{a}} \sqrt {1+\frac {d x^2}{c}} F\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )}{15 b \sqrt {\frac {b}{a}} d^3 \sqrt {a+b x^2} \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*Sqrt[a + b*x^2])/Sqrt[c + d*x^2],x]

[Out]

(Sqrt[b/a]*d*x*(a + b*x^2)*(c + d*x^2)*(-4*b*c + a*d + 3*b*d*x^2) + I*c*(-8*b^2*c^2 + 3*a*b*c*d + 2*a^2*d^2)*S
qrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticE[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)] - I*c*(-8*b^2*c^2 + 7*a*
b*c*d + a^2*d^2)*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticF[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)])/(15*b
*Sqrt[b/a]*d^3*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])

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Maple [A]
time = 0.14, size = 526, normalized size = 1.53

method	result	size
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \left (\frac {x^{3} \sqrt {b d \,x^{4}+a d \,x^{2}+c \,x^{2} b +a c}}{5 d}+\frac {\left (a -\frac {4 a d +4 b c}{5 d}\right ) x \sqrt {b d \,x^{4}+a d \,x^{2}+c \,x^{2} b +a c}}{3 b d}-\frac {\left (a -\frac {4 a d +4 b c}{5 d}\right ) a c \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \EllipticF \left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )}{3 b d \sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+c \,x^{2} b +a c}}-\frac {\left (-\frac {3 a c}{5 d}-\frac {\left (a -\frac {4 a d +4 b c}{5 d}\right ) \left (2 a d +2 b c \right )}{3 b d}\right ) c \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \left (\EllipticF \left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )-\EllipticE \left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )\right )}{\sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+c \,x^{2} b +a c}\, d}\right )}{\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}\)	\(394\)
risch	\(\frac {x \left (3 b d \,x^{2}+a d -4 b c \right ) \sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}{15 b \,d^{2}}-\frac {\left (-\frac {\left (2 a^{2} d^{2}+3 a b c d -8 b^{2} c^{2}\right ) c \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \left (\EllipticF \left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )-\EllipticE \left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )\right )}{\sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+c \,x^{2} b +a c}\, d}+\frac {a^{2} c d \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \EllipticF \left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )}{\sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+c \,x^{2} b +a c}}-\frac {4 a b \,c^{2} \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \EllipticF \left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )}{\sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+c \,x^{2} b +a c}}\right ) \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}}{15 b \,d^{2} \sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}\)	\(417\)
default	\(\frac {\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\, \left (3 \sqrt {-\frac {b}{a}}\, b^{2} d^{3} x^{7}+4 \sqrt {-\frac {b}{a}}\, a b \,d^{3} x^{5}-\sqrt {-\frac {b}{a}}\, b^{2} c \,d^{2} x^{5}+\sqrt {-\frac {b}{a}}\, a^{2} d^{3} x^{3}-4 \sqrt {-\frac {b}{a}}\, b^{2} c^{2} d \,x^{3}+\sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, \EllipticF \left (x \sqrt {-\frac {b}{a}}, \sqrt {\frac {a d}{b c}}\right ) a^{2} c \,d^{2}+7 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, \EllipticF \left (x \sqrt {-\frac {b}{a}}, \sqrt {\frac {a d}{b c}}\right ) a b \,c^{2} d -8 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, \EllipticF \left (x \sqrt {-\frac {b}{a}}, \sqrt {\frac {a d}{b c}}\right ) b^{2} c^{3}-2 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, \EllipticE \left (x \sqrt {-\frac {b}{a}}, \sqrt {\frac {a d}{b c}}\right ) a^{2} c \,d^{2}-3 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, \EllipticE \left (x \sqrt {-\frac {b}{a}}, \sqrt {\frac {a d}{b c}}\right ) a b \,c^{2} d +8 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, \EllipticE \left (x \sqrt {-\frac {b}{a}}, \sqrt {\frac {a d}{b c}}\right ) b^{2} c^{3}+\sqrt {-\frac {b}{a}}\, a^{2} c \,d^{2} x -4 \sqrt {-\frac {b}{a}}\, a b \,c^{2} d x \right )}{15 \left (b d \,x^{4}+a d \,x^{2}+c \,x^{2} b +a c \right ) d^{3} b \sqrt {-\frac {b}{a}}}\)	\(526\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/15*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)*(3*(-b/a)^(1/2)*b^2*d^3*x^7+4*(-b/a)^(1/2)*a*b*d^3*x^5-(-b/a)^(1/2)*b^2*c
*d^2*x^5+(-b/a)^(1/2)*a^2*d^3*x^3-4*(-b/a)^(1/2)*b^2*c^2*d*x^3+((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*Ellipti
cF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*a^2*c*d^2+7*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticF(x*(-b/a)^(1/2
),(a*d/b/c)^(1/2))*a*b*c^2*d-8*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2
))*b^2*c^3-2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*a^2*c*d^2-3*((b
*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*a*b*c^2*d+8*((b*x^2+a)/a)^(1/2)
*((d*x^2+c)/c)^(1/2)*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*b^2*c^3+(-b/a)^(1/2)*a^2*c*d^2*x-4*(-b/a)^(1/2)
*a*b*c^2*d*x)/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)/d^3/b/(-b/a)^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*x^2 + a)*x^4/sqrt(d*x^2 + c), x)

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Fricas [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(x^4*(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >> Symbolic function elliptic_ec takes exactly 1 arguments (2 given)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4} \sqrt {a + b x^{2}}}{\sqrt {c + d x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(b*x**2+a)**(1/2)/(d*x**2+c)**(1/2),x)

[Out]

Integral(x**4*sqrt(a + b*x**2)/sqrt(c + d*x**2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*x^2 + a)*x^4/sqrt(d*x^2 + c), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^4\,\sqrt {b\,x^2+a}}{\sqrt {d\,x^2+c}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(a + b*x^2)^(1/2))/(c + d*x^2)^(1/2),x)

[Out]

int((x^4*(a + b*x^2)^(1/2))/(c + d*x^2)^(1/2), x)

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