3.2.99 \(\int \frac {(a+b x^2)^{7/2}}{\sqrt {c+d x^2}} \, dx\) [199]

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

[Out]

-8/105*(-2*a*d+b*c)*(11*a^2*d^2-11*a*b*c*d+6*b^2*c^2)*x*(b*x^2+a)^(1/2)/d^3/(d*x^2+c)^(1/2)+8/105*(-2*a*d+b*c)
*(11*a^2*d^2-11*a*b*c*d+6*b^2*c^2)*(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)/d^(7/2)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(d*x^2+c)^(1/2)-
1/105*(-7*a*d+3*b*c)*(15*a^2*d^2-11*a*b*c*d+8*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))*c^(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)-6/35*b*(-2*a*d+b*c)*x*(b*x^2+a)^(3/2)*(d*x^2+c)^(1/2)/d^2+1/7*b*x*(b*x^2+a)^(5/2)*(d*x^2+c
)^(1/2)/d+1/105*b*(71*a^2*d^2-71*a*b*c*d+24*b^2*c^2)*x*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/d^3

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Rubi [A]
time = 0.29, antiderivative size = 423, normalized size of antiderivative = 1.00, 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 = {427, 542, 545, 429, 506, 422} \begin {gather*} -\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 (\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 {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 (\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 {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 {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} \end {gather*}

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 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 427

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 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 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 542

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 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]

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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 3.97, size = 321, normalized size = 0.76 \begin {gather*} \frac {b \sqrt {\frac {b}{a}} d x \left (a+b x^2\right ) \left (c+d x^2\right ) \left (122 a^2 d^2+a b d \left (-89 c+66 d x^2\right )+3 b^2 \left (8 c^2-6 c d x^2+5 d^2 x^4\right )\right )-8 i b c \left (-6 b^3 c^3+23 a b^2 c^2 d-33 a^2 b c d^2+22 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 )-i \left (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\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[(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 = 0.10, size = 852, normalized size = 2.01 Too large to display

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,method=_RETURNVERBOSE)

[Out]

1/105*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)*(15*(-b/a)^(1/2)*b^4*d^4*x^9+81*(-b/a)^(1/2)*a*b^3*d^4*x^7-3*(-b/a)^(1/2
)*b^4*c*d^3*x^7+188*(-b/a)^(1/2)*a^2*b^2*d^4*x^5-26*(-b/a)^(1/2)*a*b^3*c*d^3*x^5+6*(-b/a)^(1/2)*b^4*c^2*d^2*x^
5+122*(-b/a)^(1/2)*a^3*b*d^4*x^3+99*(-b/a)^(1/2)*a^2*b^2*c*d^3*x^3-83*(-b/a)^(1/2)*a*b^3*c^2*d^2*x^3+24*(-b/a)
^(1/2)*b^4*c^3*d*x^3+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)*a^3*b*c*d^3*x-89*(-b/a)^(1/2)*a^2*b^2*c^2*d^2*x+24*(-b/a)^(1/2)*a*b^3*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((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(-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((b*x^2+a)^(7/2)/(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 {\left (a + b x^{2}\right )^{\frac {7}{2}}}{\sqrt {c + d x^{2}}}\, dx \end {gather*}

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 = 0.00, 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((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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (b\,x^2+a\right )}^{7/2}}{\sqrt {d\,x^2+c}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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