3.207 \(\int \frac{(a+b x^2)^{7/2}}{(c+d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=445 \[ \frac{b \sqrt{c} \sqrt{a+b x^2} \left (45 a^2 d^2-61 a b c d+24 b^2 c^2\right ) \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right ),1-\frac{b c}{a d}\right )}{15 d^{7/2} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac{b x \sqrt{a+b x^2} \sqrt{c+d x^2} \left (15 a^2 d^2-43 a b c d+24 b^2 c^2\right )}{15 c d^3}+\frac{x \sqrt{a+b x^2} \left (103 a^2 b c d^2-15 a^3 d^3-128 a b^2 c^2 d+48 b^3 c^3\right )}{15 c d^3 \sqrt{c+d x^2}}-\frac{\sqrt{a+b x^2} \left (103 a^2 b c d^2-15 a^3 d^3-128 a b^2 c^2 d+48 b^3 c^3\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{15 \sqrt{c} d^{7/2} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac{b x \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2} (6 b c-5 a d)}{5 c d^2}-\frac{x \left (a+b x^2\right )^{5/2} (b c-a d)}{c d \sqrt{c+d x^2}} \]

[Out]

((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])/(15*c*d^3*Sqrt[c + d*x^2]) -
((b*c - a*d)*x*(a + b*x^2)^(5/2))/(c*d*Sqrt[c + d*x^2]) - (b*(24*b^2*c^2 - 43*a*b*c*d + 15*a^2*d^2)*x*Sqrt[a
+ b*x^2]*Sqrt[c + d*x^2])/(15*c*d^3) + (b*(6*b*c - 5*a*d)*x*(a + b*x^2)^(3/2)*Sqrt[c + d*x^2])/(5*c*d^2) - ((4
8*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)/Sqrt[
c]], 1 - (b*c)/(a*d)])/(15*Sqrt[c]*d^(7/2)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2]) + (b*Sqrt[c]
*(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)
])/(15*d^(7/2)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.405257, antiderivative size = 445, 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 = {413, 528, 531, 418, 492, 411} \[ -\frac{b x \sqrt{a+b x^2} \sqrt{c+d x^2} \left (15 a^2 d^2-43 a b c d+24 b^2 c^2\right )}{15 c d^3}+\frac{x \sqrt{a+b x^2} \left (103 a^2 b c d^2-15 a^3 d^3-128 a b^2 c^2 d+48 b^3 c^3\right )}{15 c d^3 \sqrt{c+d x^2}}+\frac{b \sqrt{c} \sqrt{a+b x^2} \left (45 a^2 d^2-61 a b c d+24 b^2 c^2\right ) F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{15 d^{7/2} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac{\sqrt{a+b x^2} \left (103 a^2 b c d^2-15 a^3 d^3-128 a b^2 c^2 d+48 b^3 c^3\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{15 \sqrt{c} d^{7/2} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac{b x \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2} (6 b c-5 a d)}{5 c d^2}-\frac{x \left (a+b x^2\right )^{5/2} (b c-a d)}{c d \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(7/2)/(c + d*x^2)^(3/2),x]

[Out]

((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])/(15*c*d^3*Sqrt[c + d*x^2]) -
((b*c - a*d)*x*(a + b*x^2)^(5/2))/(c*d*Sqrt[c + d*x^2]) - (b*(24*b^2*c^2 - 43*a*b*c*d + 15*a^2*d^2)*x*Sqrt[a
+ b*x^2]*Sqrt[c + d*x^2])/(15*c*d^3) + (b*(6*b*c - 5*a*d)*x*(a + b*x^2)^(3/2)*Sqrt[c + d*x^2])/(5*c*d^2) - ((4
8*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)/Sqrt[
c]], 1 - (b*c)/(a*d)])/(15*Sqrt[c]*d^(7/2)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2]) + (b*Sqrt[c]
*(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)
])/(15*d^(7/2)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2])

Rule 413

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((a*d - c*b)*x*(a + b*x^n)^
(p + 1)*(c + d*x^n)^(q - 1))/(a*b*n*(p + 1)), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[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}}{\left (c+d x^2\right )^{3/2}} \, dx &=-\frac{(b c-a d) x \left (a+b x^2\right )^{5/2}}{c d \sqrt{c+d x^2}}+\frac{\int \frac{\left (a+b x^2\right )^{3/2} \left (a b c+b (6 b c-5 a d) x^2\right )}{\sqrt{c+d x^2}} \, dx}{c d}\\ &=-\frac{(b c-a d) x \left (a+b x^2\right )^{5/2}}{c d \sqrt{c+d x^2}}+\frac{b (6 b c-5 a d) x \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{5 c d^2}+\frac{\int \frac{\sqrt{a+b x^2} \left (-2 a b c (3 b c-5 a d)-b \left (24 b^2 c^2-43 a b c d+15 a^2 d^2\right ) x^2\right )}{\sqrt{c+d x^2}} \, dx}{5 c d^2}\\ &=-\frac{(b c-a d) x \left (a+b x^2\right )^{5/2}}{c d \sqrt{c+d x^2}}-\frac{b \left (24 b^2 c^2-43 a b c d+15 a^2 d^2\right ) x \sqrt{a+b x^2} \sqrt{c+d x^2}}{15 c d^3}+\frac{b (6 b c-5 a d) x \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{5 c d^2}+\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}{15 c d^3}\\ &=-\frac{(b c-a d) x \left (a+b x^2\right )^{5/2}}{c d \sqrt{c+d x^2}}-\frac{b \left (24 b^2 c^2-43 a b c d+15 a^2 d^2\right ) x \sqrt{a+b x^2} \sqrt{c+d x^2}}{15 c d^3}+\frac{b (6 b c-5 a d) x \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{5 c d^2}+\frac{\left (a b \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}{15 d^3}+\frac{\left (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 )\right ) \int \frac{x^2}{\sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx}{15 c 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}}{15 c d^3 \sqrt{c+d x^2}}-\frac{(b c-a d) x \left (a+b x^2\right )^{5/2}}{c d \sqrt{c+d x^2}}-\frac{b \left (24 b^2 c^2-43 a b c d+15 a^2 d^2\right ) x \sqrt{a+b x^2} \sqrt{c+d x^2}}{15 c d^3}+\frac{b (6 b c-5 a d) x \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{5 c d^2}+\frac{b \sqrt{c} \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 )}{15 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 (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{\sqrt{a+b x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{15 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}}{15 c d^3 \sqrt{c+d x^2}}-\frac{(b c-a d) x \left (a+b x^2\right )^{5/2}}{c d \sqrt{c+d x^2}}-\frac{b \left (24 b^2 c^2-43 a b c d+15 a^2 d^2\right ) x \sqrt{a+b x^2} \sqrt{c+d x^2}}{15 c d^3}+\frac{b (6 b c-5 a d) x \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{5 c d^2}-\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 ) \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 \sqrt{c} 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{b \sqrt{c} \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 )}{15 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 = 1.1849, size = 318, normalized size = 0.71 \[ \frac{4 i b c \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} \left (41 a^2 b c d^2-15 a^3 d^3-38 a b^2 c^2 d+12 b^3 c^3\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{\frac{b}{a}}\right ),\frac{a d}{b c}\right )+d x \sqrt{\frac{b}{a}} \left (a+b x^2\right ) \left (-45 a^2 b c d^2+15 a^3 d^3+a b^2 c d \left (61 c+16 d x^2\right )-3 b^3 c \left (8 c^2+2 c d x^2-d^2 x^4\right )\right )+i b c \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} \left (-103 a^2 b c d^2+15 a^3 d^3+128 a b^2 c^2 d-48 b^3 c^3\right ) E\left (i \sinh ^{-1}\left (\sqrt{\frac{b}{a}} x\right )|\frac{a d}{b c}\right )}{15 c d^4 \sqrt{\frac{b}{a}} \sqrt{a+b x^2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(7/2)/(c + d*x^2)^(3/2),x]

[Out]

(Sqrt[b/a]*d*x*(a + b*x^2)*(-45*a^2*b*c*d^2 + 15*a^3*d^3 + a*b^2*c*d*(61*c + 16*d*x^2) - 3*b^3*c*(8*c^2 + 2*c*
d*x^2 - d^2*x^4)) + I*b*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]*S
qrt[1 + (d*x^2)/c]*EllipticE[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)] + (4*I)*b*c*(12*b^3*c^3 - 38*a*b^2*c^2*d + 4
1*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)])/(15*Sqrt[b/a]*c*d^4*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])

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Maple [A]  time = 0.043, size = 755, normalized size = 1.7 \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)^(3/2),x)

[Out]

1/15*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)*(3*(-b/a)^(1/2)*x^7*b^4*c*d^3+19*(-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+15*(-b/a)^(1/2)*x^3*a^3*b*d^4-29*(-b/a)^(1/2)*x^3*a^2*b^2*c*d^3+55*(-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+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*b*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^2*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^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^4-15*((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*b*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^2*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^3*c^3*d+48*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*El
lipticE(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*b^4*c^4+15*x*a^4*d^4*(-b/a)^(1/2)-45*(-b/a)^(1/2)*x*a^3*b*c*d^3+61*(-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)/c

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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}}}{{\left (d x^{2} + c\right )}^{\frac{3}{2}}}\,{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)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(7/2)/(d*x^2 + c)^(3/2), 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}}{d^{2} x^{4} + 2 \, c d x^{2} + c^{2}}, 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)^(3/2),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)/(d^2*x^4 + 2*c*d*x^2 + c^
2), 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}}}{\left (c + d x^{2}\right )^{\frac{3}{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)**(3/2),x)

[Out]

Integral((a + b*x**2)**(7/2)/(c + d*x**2)**(3/2), 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}}}{{\left (d x^{2} + c\right )}^{\frac{3}{2}}}\,{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)^(3/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(7/2)/(d*x^2 + c)^(3/2), x)