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

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

[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])

________________________________________________________________________________________

Rubi [A]  time = 0.718035, antiderivative size = 553, normalized size of antiderivative = 1., 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 = {477, 581, 582, 531, 418, 492, 411} \[ \frac{x^3 \sqrt{a+b x^2} \sqrt{c+d x^2} \left (75 a^2 d^2-115 a b c d+48 b^2 c^2\right )}{315 d^3}-\frac{x \sqrt{a+b x^2} \sqrt{c+d x^2} \left (105 a^2 b c d^2-5 a^3 d^3-156 a b^2 c^2 d+64 b^3 c^3\right )}{315 b d^4}+\frac{x \sqrt{a+b x^2} \left (243 a^2 b^2 c^2 d^2-25 a^3 b c d^3-10 a^4 d^4-328 a b^3 c^3 d+128 b^4 c^4\right )}{315 b^2 d^4 \sqrt{c+d x^2}}+\frac{c^{3/2} \sqrt{a+b x^2} \left (105 a^2 b c d^2-5 a^3 d^3-156 a b^2 c^2 d+64 b^3 c^3\right ) F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{315 b d^{9/2} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac{\sqrt{c} \sqrt{a+b x^2} \left (243 a^2 b^2 c^2 d^2-25 a^3 b c d^3-10 a^4 d^4-328 a b^3 c^3 d+128 b^4 c^4\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{315 b^2 d^{9/2} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac{4 b x^5 \sqrt{a+b x^2} \sqrt{c+d x^2} (2 b c-3 a d)}{63 d^2}+\frac{b x^5 \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{9 d} \]

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 477

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 581

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 582

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]

Rule 531

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 418

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

Rule 492

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(x*Sqrt[a + b*x^2])/(b*Sqr
t[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 411

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

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*}

Mathematica [C]  time = 1.66892, size = 379, normalized size = 0.69 \[ \frac{-i c \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} \left (-399 a^2 b^2 c^2 d^2+130 a^3 b c d^3+5 a^4 d^4+392 a b^3 c^3 d-128 b^4 c^4\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{\frac{b}{a}}\right ),\frac{a d}{b c}\right )+d x \sqrt{\frac{b}{a}} \left (a+b x^2\right ) \left (c+d x^2\right ) \left (15 a^2 b d^2 \left (5 d x^2-7 c\right )+5 a^3 d^3+a b^2 d \left (156 c^2-115 c d x^2+95 d^2 x^4\right )+b^3 \left (48 c^2 d x^2-64 c^3-40 c d^2 x^4+35 d^3 x^6\right )\right )+i c \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} \left (-243 a^2 b^2 c^2 d^2+25 a^3 b c d^3+10 a^4 d^4+328 a b^3 c^3 d-128 b^4 c^4\right ) E\left (i \sinh ^{-1}\left (\sqrt{\frac{b}{a}} x\right )|\frac{a d}{b c}\right )}{315 b d^5 \sqrt{\frac{b}{a}} \sqrt{a+b x^2} \sqrt{c+d x^2}} \]

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 = 0.028, size = 1047, normalized size = 1.9 \begin{align*} \text{result too large to display} \end{align*}

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)

[Out]

1/315*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)*(-25*(-b/a)^(1/2)*x^7*a*b^3*c*d^4-328*((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+392*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*Ellipt
icF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*a*b^3*c^4*d+130*(-b/a)^(1/2)*x^9*a*b^3*d^5-64*(-b/a)^(1/2)*x*a*b^3*c^4*d-2
5*((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+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+35*(-b/a)^(1/2)*x^11*b^4*d^5+5*(-b/a)^
(1/2)*x^3*a^4*d^5-50*(-b/a)^(1/2)*x^5*a^2*b^2*c*d^4+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+49*(-b/a)^(1/2)*x^5*a*b^3*c^2*d^3-25*(-b/a)^(1/2)*x^3*a^3*b*c*d^4-64*(-b/a)^(1/2)*x^3*a^2
*b^2*c^2*d^3+140*(-b/a)^(1/2)*x^3*a*b^3*c^3*d^2-105*(-b/a)^(1/2)*x*a^3*b*c^2*d^3+156*(-b/a)^(1/2)*x*a^2*b^2*c^
3*d^2-128*((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+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+80*(-b/a)^(1/2)*x^5*a^3*b*
d^5-16*(-b/a)^(1/2)*x^5*b^4*c^3*d^2+5*(-b/a)^(1/2)*x*a^4*c*d^4-64*(-b/a)^(1/2)*x^3*b^4*c^4*d-5*(-b/a)^(1/2)*x^
9*b^4*c*d^4+170*(-b/a)^(1/2)*x^7*a^2*b^2*d^5+8*(-b/a)^(1/2)*x^7*b^4*c^2*d^3)/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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{\frac{5}{2}} x^{4}}{\sqrt{d x^{2} + c}}\,{d x} \end{align*}

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]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b^{2} x^{8} + 2 \, a b x^{6} + a^{2} x^{4}\right )} \sqrt{b x^{2} + a}}{\sqrt{d x^{2} + c}}, x\right ) \end{align*}

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]

integral((b^2*x^8 + 2*a*b*x^6 + a^2*x^4)*sqrt(b*x^2 + a)/sqrt(d*x^2 + c), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4} \left (a + b x^{2}\right )^{\frac{5}{2}}}{\sqrt{c + d x^{2}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{\frac{5}{2}} x^{4}}{\sqrt{d x^{2} + c}}\,{d x} \end{align*}

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)