3.199 \(\int \frac{(a+b x^2)^{7/2}}{\sqrt{c+d x^2}} \, dx\)

Optimal. Leaf size=423 \[ -\frac{\sqrt{c} \sqrt{a+b x^2} (3 b c-7 a d) \left (15 a^2 d^2-11 a b c d+8 b^2 c^2\right ) \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right ),1-\frac{b c}{a d}\right )}{105 d^{7/2} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac{b x \sqrt{a+b x^2} \sqrt{c+d x^2} \left (71 a^2 d^2-71 a b c d+24 b^2 c^2\right )}{105 d^3}-\frac{8 x \sqrt{a+b x^2} (b c-2 a d) \left (11 a^2 d^2-11 a b c d+6 b^2 c^2\right )}{105 d^3 \sqrt{c+d x^2}}+\frac{8 \sqrt{c} \sqrt{a+b x^2} (b c-2 a d) \left (11 a^2 d^2-11 a b c d+6 b^2 c^2\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{105 d^{7/2} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac{6 b x \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2} (b c-2 a d)}{35 d^2}+\frac{b x \left (a+b x^2\right )^{5/2} \sqrt{c+d x^2}}{7 d} \]

[Out]

(-8*(b*c - 2*a*d)*(6*b^2*c^2 - 11*a*b*c*d + 11*a^2*d^2)*x*Sqrt[a + b*x^2])/(105*d^3*Sqrt[c + d*x^2]) + (b*(24*
b^2*c^2 - 71*a*b*c*d + 71*a^2*d^2)*x*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(105*d^3) - (6*b*(b*c - 2*a*d)*x*(a + b*
x^2)^(3/2)*Sqrt[c + d*x^2])/(35*d^2) + (b*x*(a + b*x^2)^(5/2)*Sqrt[c + d*x^2])/(7*d) + (8*Sqrt[c]*(b*c - 2*a*d
)*(6*b^2*c^2 - 11*a*b*c*d + 11*a^2*d^2)*Sqrt[a + b*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)
])/(105*d^(7/2)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2]) - (Sqrt[c]*(3*b*c - 7*a*d)*(8*b^2*c^2 -
11*a*b*c*d + 15*a^2*d^2)*Sqrt[a + b*x^2]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(105*d^(7/2
)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.431295, antiderivative size = 423, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {416, 528, 531, 418, 492, 411} \[ \frac{b x \sqrt{a+b x^2} \sqrt{c+d x^2} \left (71 a^2 d^2-71 a b c d+24 b^2 c^2\right )}{105 d^3}-\frac{8 x \sqrt{a+b x^2} (b c-2 a d) \left (11 a^2 d^2-11 a b c d+6 b^2 c^2\right )}{105 d^3 \sqrt{c+d x^2}}-\frac{\sqrt{c} \sqrt{a+b x^2} (3 b c-7 a d) \left (15 a^2 d^2-11 a b c d+8 b^2 c^2\right ) F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{105 d^{7/2} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac{8 \sqrt{c} \sqrt{a+b x^2} (b c-2 a d) \left (11 a^2 d^2-11 a b c d+6 b^2 c^2\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{105 d^{7/2} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac{6 b x \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2} (b c-2 a d)}{35 d^2}+\frac{b x \left (a+b x^2\right )^{5/2} \sqrt{c+d x^2}}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(7/2)/Sqrt[c + d*x^2],x]

[Out]

(-8*(b*c - 2*a*d)*(6*b^2*c^2 - 11*a*b*c*d + 11*a^2*d^2)*x*Sqrt[a + b*x^2])/(105*d^3*Sqrt[c + d*x^2]) + (b*(24*
b^2*c^2 - 71*a*b*c*d + 71*a^2*d^2)*x*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(105*d^3) - (6*b*(b*c - 2*a*d)*x*(a + b*
x^2)^(3/2)*Sqrt[c + d*x^2])/(35*d^2) + (b*x*(a + b*x^2)^(5/2)*Sqrt[c + d*x^2])/(7*d) + (8*Sqrt[c]*(b*c - 2*a*d
)*(6*b^2*c^2 - 11*a*b*c*d + 11*a^2*d^2)*Sqrt[a + b*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)
])/(105*d^(7/2)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2]) - (Sqrt[c]*(3*b*c - 7*a*d)*(8*b^2*c^2 -
11*a*b*c*d + 15*a^2*d^2)*Sqrt[a + b*x^2]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(105*d^(7/2
)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2])

Rule 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rule 528

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

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{\left (a+b x^2\right )^{7/2}}{\sqrt{c+d x^2}} \, dx &=\frac{b x \left (a+b x^2\right )^{5/2} \sqrt{c+d x^2}}{7 d}+\frac{\int \frac{\left (a+b x^2\right )^{3/2} \left (-a (b c-7 a d)-6 b (b c-2 a d) x^2\right )}{\sqrt{c+d x^2}} \, dx}{7 d}\\ &=-\frac{6 b (b c-2 a d) x \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{35 d^2}+\frac{b x \left (a+b x^2\right )^{5/2} \sqrt{c+d x^2}}{7 d}+\frac{\int \frac{\sqrt{a+b x^2} \left (a \left (6 b^2 c^2-17 a b c d+35 a^2 d^2\right )+b \left (24 b^2 c^2-71 a b c d+71 a^2 d^2\right ) x^2\right )}{\sqrt{c+d x^2}} \, dx}{35 d^2}\\ &=\frac{b \left (24 b^2 c^2-71 a b c d+71 a^2 d^2\right ) x \sqrt{a+b x^2} \sqrt{c+d x^2}}{105 d^3}-\frac{6 b (b c-2 a d) x \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{35 d^2}+\frac{b x \left (a+b x^2\right )^{5/2} \sqrt{c+d x^2}}{7 d}+\frac{\int \frac{-a (3 b c-7 a d) \left (8 b^2 c^2-11 a b c d+15 a^2 d^2\right )-8 b (b c-2 a d) \left (6 b^2 c^2-11 a b c d+11 a^2 d^2\right ) x^2}{\sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx}{105 d^3}\\ &=\frac{b \left (24 b^2 c^2-71 a b c d+71 a^2 d^2\right ) x \sqrt{a+b x^2} \sqrt{c+d x^2}}{105 d^3}-\frac{6 b (b c-2 a d) x \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{35 d^2}+\frac{b x \left (a+b x^2\right )^{5/2} \sqrt{c+d x^2}}{7 d}-\frac{\left (8 b (b c-2 a d) \left (6 b^2 c^2-11 a b c d+11 a^2 d^2\right )\right ) \int \frac{x^2}{\sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx}{105 d^3}-\frac{\left (a (3 b c-7 a d) \left (8 b^2 c^2-11 a b c d+15 a^2 d^2\right )\right ) \int \frac{1}{\sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx}{105 d^3}\\ &=-\frac{8 (b c-2 a d) \left (6 b^2 c^2-11 a b c d+11 a^2 d^2\right ) x \sqrt{a+b x^2}}{105 d^3 \sqrt{c+d x^2}}+\frac{b \left (24 b^2 c^2-71 a b c d+71 a^2 d^2\right ) x \sqrt{a+b x^2} \sqrt{c+d x^2}}{105 d^3}-\frac{6 b (b c-2 a d) x \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{35 d^2}+\frac{b x \left (a+b x^2\right )^{5/2} \sqrt{c+d x^2}}{7 d}-\frac{\sqrt{c} (3 b c-7 a d) \left (8 b^2 c^2-11 a b c d+15 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 (8 c (b c-2 a d) \left (6 b^2 c^2-11 a b c d+11 a^2 d^2\right )\right ) \int \frac{\sqrt{a+b x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{105 d^3}\\ &=-\frac{8 (b c-2 a d) \left (6 b^2 c^2-11 a b c d+11 a^2 d^2\right ) x \sqrt{a+b x^2}}{105 d^3 \sqrt{c+d x^2}}+\frac{b \left (24 b^2 c^2-71 a b c d+71 a^2 d^2\right ) x \sqrt{a+b x^2} \sqrt{c+d x^2}}{105 d^3}-\frac{6 b (b c-2 a d) x \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{35 d^2}+\frac{b x \left (a+b x^2\right )^{5/2} \sqrt{c+d x^2}}{7 d}+\frac{8 \sqrt{c} (b c-2 a d) \left (6 b^2 c^2-11 a b c d+11 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 )}{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{\sqrt{c} (3 b c-7 a d) \left (8 b^2 c^2-11 a b c d+15 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*}

Mathematica [C]  time = 1.48904, size = 321, normalized size = 0.76 \[ \frac{-i \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} \left (353 a^2 b^2 c^2 d^2-298 a^3 b c d^3+105 a^4 d^4-208 a b^3 c^3 d+48 b^4 c^4\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{\frac{b}{a}}\right ),\frac{a d}{b c}\right )+b d x \sqrt{\frac{b}{a}} \left (a+b x^2\right ) \left (c+d x^2\right ) \left (122 a^2 d^2+a b d \left (66 d x^2-89 c\right )+3 b^2 \left (8 c^2-6 c d x^2+5 d^2 x^4\right )\right )-8 i b c \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} \left (-33 a^2 b c d^2+22 a^3 d^3+23 a b^2 c^2 d-6 b^3 c^3\right ) E\left (i \sinh ^{-1}\left (\sqrt{\frac{b}{a}} x\right )|\frac{a d}{b c}\right )}{105 d^4 \sqrt{\frac{b}{a}} \sqrt{a+b x^2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(7/2)/Sqrt[c + d*x^2],x]

[Out]

(b*Sqrt[b/a]*d*x*(a + b*x^2)*(c + d*x^2)*(122*a^2*d^2 + a*b*d*(-89*c + 66*d*x^2) + 3*b^2*(8*c^2 - 6*c*d*x^2 +
5*d^2*x^4)) - (8*I)*b*c*(-6*b^3*c^3 + 23*a*b^2*c^2*d - 33*a^2*b*c*d^2 + 22*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)] - I*(48*b^4*c^4 - 208*a*b^3*c^3*d + 353*a^2*b^2*c
^2*d^2 - 298*a^3*b*c*d^3 + 105*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)])/(105*Sqrt[b/a]*d^4*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])

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Maple [A]  time = 0.023, size = 852, normalized size = 2. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(7/2)/(d*x^2+c)^(1/2),x)

[Out]

1/105*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)*(15*(-b/a)^(1/2)*x^9*b^4*d^4+81*(-b/a)^(1/2)*x^7*a*b^3*d^4-3*(-b/a)^(1/2
)*x^7*b^4*c*d^3+188*(-b/a)^(1/2)*x^5*a^2*b^2*d^4-26*(-b/a)^(1/2)*x^5*a*b^3*c*d^3+6*(-b/a)^(1/2)*x^5*b^4*c^2*d^
2+122*(-b/a)^(1/2)*x^3*a^3*b*d^4+99*(-b/a)^(1/2)*x^3*a^2*b^2*c*d^3-83*(-b/a)^(1/2)*x^3*a*b^3*c^2*d^2+24*(-b/a)
^(1/2)*x^3*b^4*c^3*d+105*((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
*d^4-298*((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*d^3+353*((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^2*d^2-208*((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^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^4*c^4+176*((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*d^3-264*((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^2*d^2+184*((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^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^4*c^4+122*(-b/a)^(1/2)*x*a^3*b*c*d^3-89*(-b/a)^(1/2)*x*a^2*b^2*c^2*d^2+24*(-b/a)^(1/2)*x*a*b^3*c^3*
d)/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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{\frac{7}{2}}}{\sqrt{d x^{2} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(7/2)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(7/2)/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^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\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((b*x^2+a)^(7/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

integral((b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a^3)*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{\left (a + b x^{2}\right )^{\frac{7}{2}}}{\sqrt{c + d x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(7/2)/(d*x**2+c)**(1/2),x)

[Out]

Integral((a + b*x**2)**(7/2)/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{7}{2}}}{\sqrt{d x^{2} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(7/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(7/2)/sqrt(d*x^2 + c), x)