3.10.61 \(\int \frac {x^2 (a+b x^2)^{5/2}}{\sqrt {c+d x^2}} \, dx\) [961]

Optimal. Leaf size=436 \[ -\frac {\left (48 b^3 c^3-128 a b^2 c^2 d+103 a^2 b c d^2-15 a^3 d^3\right ) x \sqrt {a+b x^2}}{105 b d^3 \sqrt {c+d x^2}}+\frac {\left (24 b^2 c^2-61 a b c d+45 a^2 d^2\right ) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{105 d^3}-\frac {2 b (3 b c-5 a d) x^3 \sqrt {a+b x^2} \sqrt {c+d x^2}}{35 d^2}+\frac {b x^3 \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{7 d}+\frac {\sqrt {c} \left (48 b^3 c^3-128 a b^2 c^2 d+103 a^2 b c d^2-15 a^3 d^3\right ) \sqrt {a+b x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{105 b d^{7/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}-\frac {c^{3/2} \left (24 b^2 c^2-61 a b c d+45 a^2 d^2\right ) \sqrt {a+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{105 d^{7/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}} \]

[Out]

-1/105*(-15*a^3*d^3+103*a^2*b*c*d^2-128*a*b^2*c^2*d+48*b^3*c^3)*x*(b*x^2+a)^(1/2)/b/d^3/(d*x^2+c)^(1/2)-1/105*
c^(3/2)*(45*a^2*d^2-61*a*b*c*d+24*b^2*c^2)*(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)/d^(7/2)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(d*x^2+c)^(1/2)+
1/105*(-15*a^3*d^3+103*a^2*b*c*d^2-128*a*b^2*c^2*d+48*b^3*c^3)*(1/(1+d*x^2/c))^(1/2)*(1+d*x^2/c)^(1/2)*Ellipti
cE(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/d^(7/2)/(c*(b*x^2+a)/a/(d*
x^2+c))^(1/2)/(d*x^2+c)^(1/2)+1/7*b*x^3*(b*x^2+a)^(3/2)*(d*x^2+c)^(1/2)/d+1/105*(45*a^2*d^2-61*a*b*c*d+24*b^2*
c^2)*x*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/d^3-2/35*b*(-5*a*d+3*b*c)*x^3*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/d^2

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Rubi [A]
time = 0.32, antiderivative size = 436, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {488, 595, 596, 545, 429, 506, 422} \begin {gather*} -\frac {c^{3/2} \sqrt {a+b x^2} \left (45 a^2 d^2-61 a b c d+24 b^2 c^2\right ) F\left (\text {ArcTan}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{105 d^{7/2} \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} \left (45 a^2 d^2-61 a b c d+24 b^2 c^2\right )}{105 d^3}+\frac {\sqrt {c} \sqrt {a+b x^2} \left (-15 a^3 d^3+103 a^2 b c d^2-128 a b^2 c^2 d+48 b^3 c^3\right ) E\left (\text {ArcTan}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{105 b d^{7/2} \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 (-15 a^3 d^3+103 a^2 b c d^2-128 a b^2 c^2 d+48 b^3 c^3\right )}{105 b d^3 \sqrt {c+d x^2}}-\frac {2 b x^3 \sqrt {a+b x^2} \sqrt {c+d x^2} (3 b c-5 a d)}{35 d^2}+\frac {b x^3 \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{7 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/105*((48*b^3*c^3 - 128*a*b^2*c^2*d + 103*a^2*b*c*d^2 - 15*a^3*d^3)*x*Sqrt[a + b*x^2])/(b*d^3*Sqrt[c + d*x^2
]) + ((24*b^2*c^2 - 61*a*b*c*d + 45*a^2*d^2)*x*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(105*d^3) - (2*b*(3*b*c - 5*a*
d)*x^3*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(35*d^2) + (b*x^3*(a + b*x^2)^(3/2)*Sqrt[c + d*x^2])/(7*d) + (Sqrt[c]*
(48*b^3*c^3 - 128*a*b^2*c^2*d + 103*a^2*b*c*d^2 - 15*a^3*d^3)*Sqrt[a + b*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqr
t[c]], 1 - (b*c)/(a*d)])/(105*b*d^(7/2)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2]) - (c^(3/2)*(24*
b^2*c^2 - 61*a*b*c*d + 45*a^2*d^2)*Sqrt[a + b*x^2]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(1
05*d^(7/2)*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 488

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[d*(e*x)^
(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q - 1)/(b*e*(m + n*(p + q) + 1))), x] + Dist[1/(b*(m + n*(p + q) + 1
)), Int[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^(q - 2)*Simp[c*((c*b - a*d)*(m + 1) + c*b*n*(p + q)) + (d*(c*b - a*d
)*(m + 1) + d*n*(q - 1)*(b*c - a*d) + c*b*d*n*(p + q))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && N
eQ[b*c - a*d, 0] && IGtQ[n, 0] && GtQ[q, 1] && 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 595

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

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^2 \left (a+b x^2\right )^{5/2}}{\sqrt {c+d x^2}} \, dx &=\frac {b x^3 \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{7 d}+\frac {\int \frac {x^2 \sqrt {a+b x^2} \left (-a (3 b c-7 a d)-2 b (3 b c-5 a d) x^2\right )}{\sqrt {c+d x^2}} \, dx}{7 d}\\ &=-\frac {2 b (3 b c-5 a d) x^3 \sqrt {a+b x^2} \sqrt {c+d x^2}}{35 d^2}+\frac {b x^3 \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{7 d}+\frac {\int \frac {x^2 \left (a \left (18 b^2 c^2-45 a b c d+35 a^2 d^2\right )+b \left (24 b^2 c^2-61 a b c d+45 a^2 d^2\right ) x^2\right )}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{35 d^2}\\ &=\frac {\left (24 b^2 c^2-61 a b c d+45 a^2 d^2\right ) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{105 d^3}-\frac {2 b (3 b c-5 a d) x^3 \sqrt {a+b x^2} \sqrt {c+d x^2}}{35 d^2}+\frac {b x^3 \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{7 d}-\frac {\int \frac {a b c \left (24 b^2 c^2-61 a b c d+45 a^2 d^2\right )+b \left (48 b^3 c^3-128 a b^2 c^2 d+103 a^2 b c d^2-15 a^3 d^3\right ) x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{105 b d^3}\\ &=\frac {\left (24 b^2 c^2-61 a b c d+45 a^2 d^2\right ) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{105 d^3}-\frac {2 b (3 b c-5 a d) x^3 \sqrt {a+b x^2} \sqrt {c+d x^2}}{35 d^2}+\frac {b x^3 \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{7 d}-\frac {\left (a c \left (24 b^2 c^2-61 a b c d+45 a^2 d^2\right )\right ) \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{105 d^3}-\frac {\left (48 b^3 c^3-128 a b^2 c^2 d+103 a^2 b c d^2-15 a^3 d^3\right ) \int \frac {x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{105 d^3}\\ &=-\frac {\left (48 b^3 c^3-128 a b^2 c^2 d+103 a^2 b c d^2-15 a^3 d^3\right ) x \sqrt {a+b x^2}}{105 b d^3 \sqrt {c+d x^2}}+\frac {\left (24 b^2 c^2-61 a b c d+45 a^2 d^2\right ) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{105 d^3}-\frac {2 b (3 b c-5 a d) x^3 \sqrt {a+b x^2} \sqrt {c+d x^2}}{35 d^2}+\frac {b x^3 \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{7 d}-\frac {c^{3/2} \left (24 b^2 c^2-61 a b c d+45 a^2 d^2\right ) \sqrt {a+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{105 d^{7/2} \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 (48 b^3 c^3-128 a b^2 c^2 d+103 a^2 b c d^2-15 a^3 d^3\right )\right ) \int \frac {\sqrt {a+b x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{105 b d^3}\\ &=-\frac {\left (48 b^3 c^3-128 a b^2 c^2 d+103 a^2 b c d^2-15 a^3 d^3\right ) x \sqrt {a+b x^2}}{105 b d^3 \sqrt {c+d x^2}}+\frac {\left (24 b^2 c^2-61 a b c d+45 a^2 d^2\right ) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{105 d^3}-\frac {2 b (3 b c-5 a d) x^3 \sqrt {a+b x^2} \sqrt {c+d x^2}}{35 d^2}+\frac {b x^3 \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{7 d}+\frac {\sqrt {c} \left (48 b^3 c^3-128 a b^2 c^2 d+103 a^2 b c d^2-15 a^3 d^3\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 )}{105 b d^{7/2} \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} \left (24 b^2 c^2-61 a b c d+45 a^2 d^2\right ) \sqrt {a+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{105 d^{7/2} \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 = 3.06, size = 306, normalized size = 0.70 \begin {gather*} \frac {\sqrt {\frac {b}{a}} d x \left (a+b x^2\right ) \left (c+d x^2\right ) \left (45 a^2 d^2+a b d \left (-61 c+45 d x^2\right )+3 b^2 \left (8 c^2-6 c d x^2+5 d^2 x^4\right )\right )-i c \left (-48 b^3 c^3+128 a b^2 c^2 d-103 a^2 b c d^2+15 a^3 d^3\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 )+4 i c \left (-12 b^3 c^3+38 a b^2 c^2 d-41 a^2 b c d^2+15 a^3 d^3\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 )}{105 \sqrt {\frac {b}{a}} d^4 \sqrt {a+b x^2} \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b/a]*d*x*(a + b*x^2)*(c + d*x^2)*(45*a^2*d^2 + a*b*d*(-61*c + 45*d*x^2) + 3*b^2*(8*c^2 - 6*c*d*x^2 + 5*d
^2*x^4)) - I*c*(-48*b^3*c^3 + 128*a*b^2*c^2*d - 103*a^2*b*c*d^2 + 15*a^3*d^3)*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*
x^2)/c]*EllipticE[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)] + (4*I)*c*(-12*b^3*c^3 + 38*a*b^2*c^2*d - 41*a^2*b*c*d^
2 + 15*a^3*d^3)*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticF[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)])/(105*S
qrt[b/a]*d^4*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])

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Maple [A]
time = 0.13, size = 782, normalized size = 1.79 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/105*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)*(-15*(-b/a)^(1/2)*b^3*d^4*x^9-60*(-b/a)^(1/2)*a*b^2*d^4*x^7+3*(-b/a)^(1
/2)*b^3*c*d^3*x^7-90*(-b/a)^(1/2)*a^2*b*d^4*x^5+19*(-b/a)^(1/2)*a*b^2*c*d^3*x^5-6*(-b/a)^(1/2)*b^3*c^2*d^2*x^5
-45*(-b/a)^(1/2)*a^3*d^4*x^3-29*(-b/a)^(1/2)*a^2*b*c*d^3*x^3+55*(-b/a)^(1/2)*a*b^2*c^2*d^2*x^3-24*(-b/a)^(1/2)
*b^3*c^3*d*x^3+60*((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^3*c*d^3-
164*((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^2*b*c^2*d^2+152*((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^2*c^3*d-48*((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^3*c^4-15*((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^3*c*d^3+103*((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*b*c^2*d^2-128*((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^2*c^3*d+48*((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^3*c^4-45*(-b/a)^(1/2)*a^3*c*d^3*x+61*(-b/a)^(1/2)*a^2*b*c^2*d^2*x-24*(-b/a)^(1/2)*a*b^2*c^3*d*x)/d^
4/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)/(-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^2*(b*x^2+a)^(5/2)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(5/2)*x^2/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^2*(b*x^2+a)^(5/2)/(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^{2} \left (a + b x^{2}\right )^{\frac {5}{2}}}{\sqrt {c + d x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate((b*x^2 + a)^(5/2)*x^2/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^2\,{\left (b\,x^2+a\right )}^{5/2}}{\sqrt {d\,x^2+c}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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