3.10.60 \(\int \frac {x^4 (a+b x^2)^{5/2}}{\sqrt {c+d x^2}} \, dx\) [960]

Optimal. Leaf size=553 \[ \frac {\left (128 b^4 c^4-328 a b^3 c^3 d+243 a^2 b^2 c^2 d^2-25 a^3 b c d^3-10 a^4 d^4\right ) x \sqrt {a+b x^2}}{315 b^2 d^4 \sqrt {c+d x^2}}-\frac {\left (64 b^3 c^3-156 a b^2 c^2 d+105 a^2 b c d^2-5 a^3 d^3\right ) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{315 b d^4}+\frac {\left (48 b^2 c^2-115 a b c d+75 a^2 d^2\right ) x^3 \sqrt {a+b x^2} \sqrt {c+d x^2}}{315 d^3}-\frac {4 b (2 b c-3 a d) x^5 \sqrt {a+b x^2} \sqrt {c+d x^2}}{63 d^2}+\frac {b x^5 \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{9 d}-\frac {\sqrt {c} \left (128 b^4 c^4-328 a b^3 c^3 d+243 a^2 b^2 c^2 d^2-25 a^3 b c d^3-10 a^4 d^4\right ) \sqrt {a+b x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{315 b^2 d^{9/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}+\frac {c^{3/2} \left (64 b^3 c^3-156 a b^2 c^2 d+105 a^2 b c d^2-5 a^3 d^3\right ) \sqrt {a+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{315 b d^{9/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}} \]

[Out]

1/315*(-10*a^4*d^4-25*a^3*b*c*d^3+243*a^2*b^2*c^2*d^2-328*a*b^3*c^3*d+128*b^4*c^4)*x*(b*x^2+a)^(1/2)/b^2/d^4/(
d*x^2+c)^(1/2)+1/315*c^(3/2)*(-5*a^3*d^3+105*a^2*b*c*d^2-156*a*b^2*c^2*d+64*b^3*c^3)*(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^(9/2)/(c*(
b*x^2+a)/a/(d*x^2+c))^(1/2)/(d*x^2+c)^(1/2)-1/315*(-10*a^4*d^4-25*a^3*b*c*d^3+243*a^2*b^2*c^2*d^2-328*a*b^3*c^
3*d+128*b^4*c^4)*(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^(9/2)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(d*x^2+c)^(1/2)+1/9*b*x^5*(b*x
^2+a)^(3/2)*(d*x^2+c)^(1/2)/d-1/315*(-5*a^3*d^3+105*a^2*b*c*d^2-156*a*b^2*c^2*d+64*b^3*c^3)*x*(b*x^2+a)^(1/2)*
(d*x^2+c)^(1/2)/b/d^4+1/315*(75*a^2*d^2-115*a*b*c*d+48*b^2*c^2)*x^3*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/d^3-4/63*b
*(-3*a*d+2*b*c)*x^5*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/d^2

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Rubi [A]
time = 0.48, antiderivative size = 553, normalized size of antiderivative = 1.00, number of steps used = 8, 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 {x^3 \sqrt {a+b x^2} \sqrt {c+d x^2} \left (75 a^2 d^2-115 a b c d+48 b^2 c^2\right )}{315 d^3}+\frac {c^{3/2} \sqrt {a+b x^2} \left (-5 a^3 d^3+105 a^2 b c d^2-156 a b^2 c^2 d+64 b^3 c^3\right ) F\left (\text {ArcTan}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{315 b d^{9/2} \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 (-5 a^3 d^3+105 a^2 b c d^2-156 a b^2 c^2 d+64 b^3 c^3\right )}{315 b d^4}-\frac {\sqrt {c} \sqrt {a+b x^2} \left (-10 a^4 d^4-25 a^3 b c d^3+243 a^2 b^2 c^2 d^2-328 a b^3 c^3 d+128 b^4 c^4\right ) E\left (\text {ArcTan}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{315 b^2 d^{9/2} \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 (-10 a^4 d^4-25 a^3 b c d^3+243 a^2 b^2 c^2 d^2-328 a b^3 c^3 d+128 b^4 c^4\right )}{315 b^2 d^4 \sqrt {c+d x^2}}-\frac {4 b x^5 \sqrt {a+b x^2} \sqrt {c+d x^2} (2 b c-3 a d)}{63 d^2}+\frac {b x^5 \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{9 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*x^2)^(5/2))/Sqrt[c + d*x^2],x]

[Out]

((128*b^4*c^4 - 328*a*b^3*c^3*d + 243*a^2*b^2*c^2*d^2 - 25*a^3*b*c*d^3 - 10*a^4*d^4)*x*Sqrt[a + b*x^2])/(315*b
^2*d^4*Sqrt[c + d*x^2]) - ((64*b^3*c^3 - 156*a*b^2*c^2*d + 105*a^2*b*c*d^2 - 5*a^3*d^3)*x*Sqrt[a + b*x^2]*Sqrt
[c + d*x^2])/(315*b*d^4) + ((48*b^2*c^2 - 115*a*b*c*d + 75*a^2*d^2)*x^3*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(315*
d^3) - (4*b*(2*b*c - 3*a*d)*x^5*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(63*d^2) + (b*x^5*(a + b*x^2)^(3/2)*Sqrt[c +
d*x^2])/(9*d) - (Sqrt[c]*(128*b^4*c^4 - 328*a*b^3*c^3*d + 243*a^2*b^2*c^2*d^2 - 25*a^3*b*c*d^3 - 10*a^4*d^4)*S
qrt[a + b*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(315*b^2*d^(9/2)*Sqrt[(c*(a + b*x^2))/
(a*(c + d*x^2))]*Sqrt[c + d*x^2]) + (c^(3/2)*(64*b^3*c^3 - 156*a*b^2*c^2*d + 105*a^2*b*c*d^2 - 5*a^3*d^3)*Sqrt
[a + b*x^2]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(315*b*d^(9/2)*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^4 \left (a+b x^2\right )^{5/2}}{\sqrt {c+d x^2}} \, dx &=\frac {b x^5 \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{9 d}+\frac {\int \frac {x^4 \sqrt {a+b x^2} \left (-a (5 b c-9 a d)-4 b (2 b c-3 a d) x^2\right )}{\sqrt {c+d x^2}} \, dx}{9 d}\\ &=-\frac {4 b (2 b c-3 a d) x^5 \sqrt {a+b x^2} \sqrt {c+d x^2}}{63 d^2}+\frac {b x^5 \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{9 d}+\frac {\int \frac {x^4 \left (a \left (40 b^2 c^2-95 a b c d+63 a^2 d^2\right )+b \left (48 b^2 c^2-115 a b c d+75 a^2 d^2\right ) x^2\right )}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{63 d^2}\\ &=\frac {\left (48 b^2 c^2-115 a b c d+75 a^2 d^2\right ) x^3 \sqrt {a+b x^2} \sqrt {c+d x^2}}{315 d^3}-\frac {4 b (2 b c-3 a d) x^5 \sqrt {a+b x^2} \sqrt {c+d x^2}}{63 d^2}+\frac {b x^5 \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{9 d}-\frac {\int \frac {x^2 \left (3 a b c \left (48 b^2 c^2-115 a b c d+75 a^2 d^2\right )+3 b \left (64 b^3 c^3-156 a b^2 c^2 d+105 a^2 b c d^2-5 a^3 d^3\right ) x^2\right )}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{315 b d^3}\\ &=-\frac {\left (64 b^3 c^3-156 a b^2 c^2 d+105 a^2 b c d^2-5 a^3 d^3\right ) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{315 b d^4}+\frac {\left (48 b^2 c^2-115 a b c d+75 a^2 d^2\right ) x^3 \sqrt {a+b x^2} \sqrt {c+d x^2}}{315 d^3}-\frac {4 b (2 b c-3 a d) x^5 \sqrt {a+b x^2} \sqrt {c+d x^2}}{63 d^2}+\frac {b x^5 \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{9 d}+\frac {\int \frac {3 a b c \left (64 b^3 c^3-156 a b^2 c^2 d+105 a^2 b c d^2-5 a^3 d^3\right )+3 b \left (128 b^4 c^4-328 a b^3 c^3 d+243 a^2 b^2 c^2 d^2-25 a^3 b c d^3-10 a^4 d^4\right ) x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{945 b^2 d^4}\\ &=-\frac {\left (64 b^3 c^3-156 a b^2 c^2 d+105 a^2 b c d^2-5 a^3 d^3\right ) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{315 b d^4}+\frac {\left (48 b^2 c^2-115 a b c d+75 a^2 d^2\right ) x^3 \sqrt {a+b x^2} \sqrt {c+d x^2}}{315 d^3}-\frac {4 b (2 b c-3 a d) x^5 \sqrt {a+b x^2} \sqrt {c+d x^2}}{63 d^2}+\frac {b x^5 \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{9 d}+\frac {\left (a c \left (64 b^3 c^3-156 a b^2 c^2 d+105 a^2 b c d^2-5 a^3 d^3\right )\right ) \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{315 b d^4}+\frac {\left (128 b^4 c^4-328 a b^3 c^3 d+243 a^2 b^2 c^2 d^2-25 a^3 b c d^3-10 a^4 d^4\right ) \int \frac {x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{315 b d^4}\\ &=\frac {\left (128 b^4 c^4-328 a b^3 c^3 d+243 a^2 b^2 c^2 d^2-25 a^3 b c d^3-10 a^4 d^4\right ) x \sqrt {a+b x^2}}{315 b^2 d^4 \sqrt {c+d x^2}}-\frac {\left (64 b^3 c^3-156 a b^2 c^2 d+105 a^2 b c d^2-5 a^3 d^3\right ) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{315 b d^4}+\frac {\left (48 b^2 c^2-115 a b c d+75 a^2 d^2\right ) x^3 \sqrt {a+b x^2} \sqrt {c+d x^2}}{315 d^3}-\frac {4 b (2 b c-3 a d) x^5 \sqrt {a+b x^2} \sqrt {c+d x^2}}{63 d^2}+\frac {b x^5 \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{9 d}+\frac {c^{3/2} \left (64 b^3 c^3-156 a b^2 c^2 d+105 a^2 b c d^2-5 a^3 d^3\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 )}{315 b d^{9/2} \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 (128 b^4 c^4-328 a b^3 c^3 d+243 a^2 b^2 c^2 d^2-25 a^3 b c d^3-10 a^4 d^4\right )\right ) \int \frac {\sqrt {a+b x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{315 b^2 d^4}\\ &=\frac {\left (128 b^4 c^4-328 a b^3 c^3 d+243 a^2 b^2 c^2 d^2-25 a^3 b c d^3-10 a^4 d^4\right ) x \sqrt {a+b x^2}}{315 b^2 d^4 \sqrt {c+d x^2}}-\frac {\left (64 b^3 c^3-156 a b^2 c^2 d+105 a^2 b c d^2-5 a^3 d^3\right ) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{315 b d^4}+\frac {\left (48 b^2 c^2-115 a b c d+75 a^2 d^2\right ) x^3 \sqrt {a+b x^2} \sqrt {c+d x^2}}{315 d^3}-\frac {4 b (2 b c-3 a d) x^5 \sqrt {a+b x^2} \sqrt {c+d x^2}}{63 d^2}+\frac {b x^5 \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{9 d}-\frac {\sqrt {c} \left (128 b^4 c^4-328 a b^3 c^3 d+243 a^2 b^2 c^2 d^2-25 a^3 b c d^3-10 a^4 d^4\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 )}{315 b^2 d^{9/2} \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 (64 b^3 c^3-156 a b^2 c^2 d+105 a^2 b c d^2-5 a^3 d^3\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 )}{315 b d^{9/2} \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 = 4.32, size = 379, normalized size = 0.69 \begin {gather*} \frac {\sqrt {\frac {b}{a}} d x \left (a+b x^2\right ) \left (c+d x^2\right ) \left (5 a^3 d^3+15 a^2 b d^2 \left (-7 c+5 d x^2\right )+a b^2 d \left (156 c^2-115 c d x^2+95 d^2 x^4\right )+b^3 \left (-64 c^3+48 c^2 d x^2-40 c d^2 x^4+35 d^3 x^6\right )\right )+i c \left (-128 b^4 c^4+328 a b^3 c^3 d-243 a^2 b^2 c^2 d^2+25 a^3 b c d^3+10 a^4 d^4\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 (-128 b^4 c^4+392 a b^3 c^3 d-399 a^2 b^2 c^2 d^2+130 a^3 b c d^3+5 a^4 d^4\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 )}{315 b \sqrt {\frac {b}{a}} d^5 \sqrt {a+b x^2} \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(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)*(5*a^3*d^3 + 15*a^2*b*d^2*(-7*c + 5*d*x^2) + a*b^2*d*(156*c^2 - 115*c*d
*x^2 + 95*d^2*x^4) + b^3*(-64*c^3 + 48*c^2*d*x^2 - 40*c*d^2*x^4 + 35*d^3*x^6)) + I*c*(-128*b^4*c^4 + 328*a*b^3
*c^3*d - 243*a^2*b^2*c^2*d^2 + 25*a^3*b*c*d^3 + 10*a^4*d^4)*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticE[
I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)] - I*c*(-128*b^4*c^4 + 392*a*b^3*c^3*d - 399*a^2*b^2*c^2*d^2 + 130*a^3*b*c
*d^3 + 5*a^4*d^4)*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticF[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)])/(315
*b*Sqrt[b/a]*d^5*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])

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Maple [A]
time = 0.13, size = 1047, normalized size = 1.89 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(b*x^2+a)^(5/2)/(d*x^2+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/315*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)*(-25*(-b/a)^(1/2)*a*b^3*c*d^4*x^7-50*(-b/a)^(1/2)*a^2*b^2*c*d^4*x^5+49*(
-b/a)^(1/2)*a*b^3*c^2*d^3*x^5-128*((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^4*c^5+5*((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^4*c*d^4-10
*((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^4*c*d^4+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))*b^4*c^5-328*((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^3*c^4*d+130*(-b/a)^(1/2)*a*b^3*d^5*x^9-5*(-b/a)^(1/2)
*b^4*c*d^4*x^9+170*(-b/a)^(1/2)*a^2*b^2*d^5*x^7+8*(-b/a)^(1/2)*b^4*c^2*d^3*x^7+80*(-b/a)^(1/2)*a^3*b*d^5*x^5-1
05*(-b/a)^(1/2)*a^3*b*c^2*d^3*x+156*(-b/a)^(1/2)*a^2*b^2*c^3*d^2*x-25*(-b/a)^(1/2)*a^3*b*c*d^4*x^3-64*(-b/a)^(
1/2)*a^2*b^2*c^2*d^3*x^3+140*(-b/a)^(1/2)*a*b^3*c^3*d^2*x^3+35*(-b/a)^(1/2)*b^4*d^5*x^11+5*(-b/a)^(1/2)*a^4*d^
5*x^3+130*((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*b*c^2*d^3-399*
((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^2*c^3*d^2+392*((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^3*c^4*d-25*((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*b*c^2*d^3+243*((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^2*c^3*d^2-16*(-b/a)^(1/2)*b^4*c^3*d^2*x^5-64*(-b/a)^(
1/2)*b^4*c^4*d*x^3+5*(-b/a)^(1/2)*a^4*c*d^4*x-64*(-b/a)^(1/2)*a*b^3*c^4*d*x)/b/d^5/(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^4*(b*x^2+a)^(5/2)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(5/2)*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)^(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^{4} \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**4*(b*x**2+a)**(5/2)/(d*x**2+c)**(1/2),x)

[Out]

Integral(x**4*(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^4*(b*x^2+a)^(5/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(5/2)*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\,{\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^4*(a + b*x^2)^(5/2))/(c + d*x^2)^(1/2),x)

[Out]

int((x^4*(a + b*x^2)^(5/2))/(c + d*x^2)^(1/2), x)

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