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

Optimal. Leaf size=217 \[ -\frac {a^2 \left (c+d x^2\right )^{7/2}}{c x}-\frac {5 c^2 \left (b^2 c^2-16 a d (3 a d+b c)\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{128 d^{3/2}}-\frac {x \left (c+d x^2\right )^{5/2} \left (b^2 c^2-16 a d (3 a d+b c)\right )}{48 c d}-\frac {5 x \left (c+d x^2\right )^{3/2} \left (b^2 c^2-16 a d (3 a d+b c)\right )}{192 d}-\frac {5 c x \sqrt {c+d x^2} \left (b^2 c^2-16 a d (3 a d+b c)\right )}{128 d}+\frac {b^2 x \left (c+d x^2\right )^{7/2}}{8 d} \]

[Out]

-5/192*(b^2*c^2-16*a*d*(3*a*d+b*c))*x*(d*x^2+c)^(3/2)/d-1/48*(b^2*c^2-16*a*d*(3*a*d+b*c))*x*(d*x^2+c)^(5/2)/c/
d-a^2*(d*x^2+c)^(7/2)/c/x+1/8*b^2*x*(d*x^2+c)^(7/2)/d-5/128*c^2*(b^2*c^2-16*a*d*(3*a*d+b*c))*arctanh(x*d^(1/2)
/(d*x^2+c)^(1/2))/d^(3/2)-5/128*c*(b^2*c^2-16*a*d*(3*a*d+b*c))*x*(d*x^2+c)^(1/2)/d

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Rubi [A]  time = 0.14, antiderivative size = 214, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {462, 388, 195, 217, 206} \[ -\frac {a^2 \left (c+d x^2\right )^{7/2}}{c x}-\frac {5 c^2 \left (b^2 c^2-16 a d (3 a d+b c)\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{128 d^{3/2}}-\frac {5 x \left (c+d x^2\right )^{3/2} \left (b^2 c^2-16 a d (3 a d+b c)\right )}{192 d}-\frac {5 c x \sqrt {c+d x^2} \left (b^2 c^2-16 a d (3 a d+b c)\right )}{128 d}-\frac {1}{48} x \left (c+d x^2\right )^{5/2} \left (\frac {b^2 c}{d}-\frac {16 a (3 a d+b c)}{c}\right )+\frac {b^2 x \left (c+d x^2\right )^{7/2}}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*c*(b^2*c^2 - 16*a*d*(b*c + 3*a*d))*x*Sqrt[c + d*x^2])/(128*d) - (5*(b^2*c^2 - 16*a*d*(b*c + 3*a*d))*x*(c +
d*x^2)^(3/2))/(192*d) - (((b^2*c)/d - (16*a*(b*c + 3*a*d))/c)*x*(c + d*x^2)^(5/2))/48 - (a^2*(c + d*x^2)^(7/2
))/(c*x) + (b^2*x*(c + d*x^2)^(7/2))/(8*d) - (5*c^2*(b^2*c^2 - 16*a*d*(b*c + 3*a*d))*ArcTanh[(Sqrt[d]*x)/Sqrt[
c + d*x^2]])/(128*d^(3/2))

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 388

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

Rule 462

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(c^2*(e*x)^(
m + 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^2} \, dx &=-\frac {a^2 \left (c+d x^2\right )^{7/2}}{c x}+\frac {\int \left (2 a (b c+3 a d)+b^2 c x^2\right ) \left (c+d x^2\right )^{5/2} \, dx}{c}\\ &=-\frac {a^2 \left (c+d x^2\right )^{7/2}}{c x}+\frac {b^2 x \left (c+d x^2\right )^{7/2}}{8 d}-\frac {\left (b^2 c^2-16 a d (b c+3 a d)\right ) \int \left (c+d x^2\right )^{5/2} \, dx}{8 c d}\\ &=-\frac {1}{48} \left (\frac {b^2 c}{d}-\frac {16 a (b c+3 a d)}{c}\right ) x \left (c+d x^2\right )^{5/2}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{c x}+\frac {b^2 x \left (c+d x^2\right )^{7/2}}{8 d}-\frac {\left (5 \left (b^2 c^2-16 a d (b c+3 a d)\right )\right ) \int \left (c+d x^2\right )^{3/2} \, dx}{48 d}\\ &=-\frac {5 \left (b^2 c^2-16 a d (b c+3 a d)\right ) x \left (c+d x^2\right )^{3/2}}{192 d}-\frac {1}{48} \left (\frac {b^2 c}{d}-\frac {16 a (b c+3 a d)}{c}\right ) x \left (c+d x^2\right )^{5/2}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{c x}+\frac {b^2 x \left (c+d x^2\right )^{7/2}}{8 d}-\frac {\left (5 c \left (b^2 c^2-16 a d (b c+3 a d)\right )\right ) \int \sqrt {c+d x^2} \, dx}{64 d}\\ &=-\frac {5 c \left (b^2 c^2-16 a d (b c+3 a d)\right ) x \sqrt {c+d x^2}}{128 d}-\frac {5 \left (b^2 c^2-16 a d (b c+3 a d)\right ) x \left (c+d x^2\right )^{3/2}}{192 d}-\frac {1}{48} \left (\frac {b^2 c}{d}-\frac {16 a (b c+3 a d)}{c}\right ) x \left (c+d x^2\right )^{5/2}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{c x}+\frac {b^2 x \left (c+d x^2\right )^{7/2}}{8 d}-\frac {\left (5 c^2 \left (b^2 c^2-16 a d (b c+3 a d)\right )\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{128 d}\\ &=-\frac {5 c \left (b^2 c^2-16 a d (b c+3 a d)\right ) x \sqrt {c+d x^2}}{128 d}-\frac {5 \left (b^2 c^2-16 a d (b c+3 a d)\right ) x \left (c+d x^2\right )^{3/2}}{192 d}-\frac {1}{48} \left (\frac {b^2 c}{d}-\frac {16 a (b c+3 a d)}{c}\right ) x \left (c+d x^2\right )^{5/2}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{c x}+\frac {b^2 x \left (c+d x^2\right )^{7/2}}{8 d}-\frac {\left (5 c^2 \left (b^2 c^2-16 a d (b c+3 a d)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{128 d}\\ &=-\frac {5 c \left (b^2 c^2-16 a d (b c+3 a d)\right ) x \sqrt {c+d x^2}}{128 d}-\frac {5 \left (b^2 c^2-16 a d (b c+3 a d)\right ) x \left (c+d x^2\right )^{3/2}}{192 d}-\frac {1}{48} \left (\frac {b^2 c}{d}-\frac {16 a (b c+3 a d)}{c}\right ) x \left (c+d x^2\right )^{5/2}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{c x}+\frac {b^2 x \left (c+d x^2\right )^{7/2}}{8 d}-\frac {5 c^2 \left (b^2 c^2-16 a d (b c+3 a d)\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{128 d^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 174, normalized size = 0.80 \[ \sqrt {c+d x^2} \left (\frac {1}{192} x^3 \left (48 a^2 d^2+208 a b c d+59 b^2 c^2\right )+\frac {c x \left (144 a^2 d^2+176 a b c d+5 b^2 c^2\right )}{128 d}-\frac {a^2 c^2}{x}+\frac {1}{48} b d x^5 (16 a d+17 b c)+\frac {1}{8} b^2 d^2 x^7\right )-\frac {5 c^2 \left (-48 a^2 d^2-16 a b c d+b^2 c^2\right ) \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )}{128 d^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[c + d*x^2]*(-((a^2*c^2)/x) + (c*(5*b^2*c^2 + 176*a*b*c*d + 144*a^2*d^2)*x)/(128*d) + ((59*b^2*c^2 + 208*a
*b*c*d + 48*a^2*d^2)*x^3)/192 + (b*d*(17*b*c + 16*a*d)*x^5)/48 + (b^2*d^2*x^7)/8) - (5*c^2*(b^2*c^2 - 16*a*b*c
*d - 48*a^2*d^2)*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]])/(128*d^(3/2))

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fricas [A]  time = 0.60, size = 375, normalized size = 1.73 \[ \left [-\frac {15 \, {\left (b^{2} c^{4} - 16 \, a b c^{3} d - 48 \, a^{2} c^{2} d^{2}\right )} \sqrt {d} x \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) - 2 \, {\left (48 \, b^{2} d^{4} x^{8} + 8 \, {\left (17 \, b^{2} c d^{3} + 16 \, a b d^{4}\right )} x^{6} - 384 \, a^{2} c^{2} d^{2} + 2 \, {\left (59 \, b^{2} c^{2} d^{2} + 208 \, a b c d^{3} + 48 \, a^{2} d^{4}\right )} x^{4} + 3 \, {\left (5 \, b^{2} c^{3} d + 176 \, a b c^{2} d^{2} + 144 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{768 \, d^{2} x}, \frac {15 \, {\left (b^{2} c^{4} - 16 \, a b c^{3} d - 48 \, a^{2} c^{2} d^{2}\right )} \sqrt {-d} x \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (48 \, b^{2} d^{4} x^{8} + 8 \, {\left (17 \, b^{2} c d^{3} + 16 \, a b d^{4}\right )} x^{6} - 384 \, a^{2} c^{2} d^{2} + 2 \, {\left (59 \, b^{2} c^{2} d^{2} + 208 \, a b c d^{3} + 48 \, a^{2} d^{4}\right )} x^{4} + 3 \, {\left (5 \, b^{2} c^{3} d + 176 \, a b c^{2} d^{2} + 144 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{384 \, d^{2} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(5/2)/x^2,x, algorithm="fricas")

[Out]

[-1/768*(15*(b^2*c^4 - 16*a*b*c^3*d - 48*a^2*c^2*d^2)*sqrt(d)*x*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c
) - 2*(48*b^2*d^4*x^8 + 8*(17*b^2*c*d^3 + 16*a*b*d^4)*x^6 - 384*a^2*c^2*d^2 + 2*(59*b^2*c^2*d^2 + 208*a*b*c*d^
3 + 48*a^2*d^4)*x^4 + 3*(5*b^2*c^3*d + 176*a*b*c^2*d^2 + 144*a^2*c*d^3)*x^2)*sqrt(d*x^2 + c))/(d^2*x), 1/384*(
15*(b^2*c^4 - 16*a*b*c^3*d - 48*a^2*c^2*d^2)*sqrt(-d)*x*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + (48*b^2*d^4*x^8 +
8*(17*b^2*c*d^3 + 16*a*b*d^4)*x^6 - 384*a^2*c^2*d^2 + 2*(59*b^2*c^2*d^2 + 208*a*b*c*d^3 + 48*a^2*d^4)*x^4 + 3
*(5*b^2*c^3*d + 176*a*b*c^2*d^2 + 144*a^2*c*d^3)*x^2)*sqrt(d*x^2 + c))/(d^2*x)]

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giac [A]  time = 0.56, size = 219, normalized size = 1.01 \[ \frac {2 \, a^{2} c^{3} \sqrt {d}}{{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c} + \frac {1}{384} \, {\left (2 \, {\left (4 \, {\left (6 \, b^{2} d^{2} x^{2} + \frac {17 \, b^{2} c d^{7} + 16 \, a b d^{8}}{d^{6}}\right )} x^{2} + \frac {59 \, b^{2} c^{2} d^{6} + 208 \, a b c d^{7} + 48 \, a^{2} d^{8}}{d^{6}}\right )} x^{2} + \frac {3 \, {\left (5 \, b^{2} c^{3} d^{5} + 176 \, a b c^{2} d^{6} + 144 \, a^{2} c d^{7}\right )}}{d^{6}}\right )} \sqrt {d x^{2} + c} x + \frac {5 \, {\left (b^{2} c^{4} \sqrt {d} - 16 \, a b c^{3} d^{\frac {3}{2}} - 48 \, a^{2} c^{2} d^{\frac {5}{2}}\right )} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right )}{256 \, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a^2*c^3*sqrt(d)/((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c) + 1/384*(2*(4*(6*b^2*d^2*x^2 + (17*b^2*c*d^7 + 16*a*b*
d^8)/d^6)*x^2 + (59*b^2*c^2*d^6 + 208*a*b*c*d^7 + 48*a^2*d^8)/d^6)*x^2 + 3*(5*b^2*c^3*d^5 + 176*a*b*c^2*d^6 +
144*a^2*c*d^7)/d^6)*sqrt(d*x^2 + c)*x + 5/256*(b^2*c^4*sqrt(d) - 16*a*b*c^3*d^(3/2) - 48*a^2*c^2*d^(5/2))*log(
(sqrt(d)*x - sqrt(d*x^2 + c))^2)/d^2

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maple [A]  time = 0.01, size = 278, normalized size = 1.28 \[ \frac {15 a^{2} c^{2} \sqrt {d}\, \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{8}+\frac {5 a b \,c^{3} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{8 \sqrt {d}}-\frac {5 b^{2} c^{4} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{128 d^{\frac {3}{2}}}+\frac {15 \sqrt {d \,x^{2}+c}\, a^{2} c d x}{8}+\frac {5 \sqrt {d \,x^{2}+c}\, a b \,c^{2} x}{8}-\frac {5 \sqrt {d \,x^{2}+c}\, b^{2} c^{3} x}{128 d}+\frac {5 \left (d \,x^{2}+c \right )^{\frac {3}{2}} a^{2} d x}{4}+\frac {5 \left (d \,x^{2}+c \right )^{\frac {3}{2}} a b c x}{12}-\frac {5 \left (d \,x^{2}+c \right )^{\frac {3}{2}} b^{2} c^{2} x}{192 d}+\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}} a^{2} d x}{c}+\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}} a b x}{3}-\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}} b^{2} c x}{48 d}+\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}} b^{2} x}{8 d}-\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}} a^{2}}{c x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*b^2*x*(d*x^2+c)^(7/2)/d-1/48*b^2*c/d*x*(d*x^2+c)^(5/2)-5/192*b^2*c^2/d*x*(d*x^2+c)^(3/2)-5/128*b^2*c^3/d*x
*(d*x^2+c)^(1/2)-5/128*b^2*c^4/d^(3/2)*ln(d^(1/2)*x+(d*x^2+c)^(1/2))+1/3*a*b*x*(d*x^2+c)^(5/2)+5/12*a*b*c*x*(d
*x^2+c)^(3/2)+5/8*a*b*c^2*x*(d*x^2+c)^(1/2)+5/8*a*b*c^3/d^(1/2)*ln(d^(1/2)*x+(d*x^2+c)^(1/2))-a^2*(d*x^2+c)^(7
/2)/c/x+a^2*d/c*x*(d*x^2+c)^(5/2)+5/4*a^2*d*x*(d*x^2+c)^(3/2)+15/8*a^2*d*c*x*(d*x^2+c)^(1/2)+15/8*a^2*d^(1/2)*
c^2*ln(d^(1/2)*x+(d*x^2+c)^(1/2))

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maxima [A]  time = 0.97, size = 235, normalized size = 1.08 \[ \frac {1}{3} \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b x + \frac {5}{12} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c x + \frac {5}{8} \, \sqrt {d x^{2} + c} a b c^{2} x + \frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2} x}{8 \, d} - \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} c x}{48 \, d} - \frac {5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{2} x}{192 \, d} - \frac {5 \, \sqrt {d x^{2} + c} b^{2} c^{3} x}{128 \, d} + \frac {5}{4} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d x + \frac {15}{8} \, \sqrt {d x^{2} + c} a^{2} c d x - \frac {5 \, b^{2} c^{4} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{128 \, d^{\frac {3}{2}}} + \frac {5 \, a b c^{3} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{8 \, \sqrt {d}} + \frac {15}{8} \, a^{2} c^{2} \sqrt {d} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right ) - \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(5/2)/x^2,x, algorithm="maxima")

[Out]

1/3*(d*x^2 + c)^(5/2)*a*b*x + 5/12*(d*x^2 + c)^(3/2)*a*b*c*x + 5/8*sqrt(d*x^2 + c)*a*b*c^2*x + 1/8*(d*x^2 + c)
^(7/2)*b^2*x/d - 1/48*(d*x^2 + c)^(5/2)*b^2*c*x/d - 5/192*(d*x^2 + c)^(3/2)*b^2*c^2*x/d - 5/128*sqrt(d*x^2 + c
)*b^2*c^3*x/d + 5/4*(d*x^2 + c)^(3/2)*a^2*d*x + 15/8*sqrt(d*x^2 + c)*a^2*c*d*x - 5/128*b^2*c^4*arcsinh(d*x/sqr
t(c*d))/d^(3/2) + 5/8*a*b*c^3*arcsinh(d*x/sqrt(c*d))/sqrt(d) + 15/8*a^2*c^2*sqrt(d)*arcsinh(d*x/sqrt(c*d)) - (
d*x^2 + c)^(5/2)*a^2/x

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^{5/2}}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 44.32, size = 496, normalized size = 2.29 \[ - \frac {a^{2} c^{\frac {5}{2}}}{x \sqrt {1 + \frac {d x^{2}}{c}}} + a^{2} c^{\frac {3}{2}} d x \sqrt {1 + \frac {d x^{2}}{c}} - \frac {7 a^{2} c^{\frac {3}{2}} d x}{8 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {3 a^{2} \sqrt {c} d^{2} x^{3}}{8 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {15 a^{2} c^{2} \sqrt {d} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{8} + \frac {a^{2} d^{3} x^{5}}{4 \sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} + a b c^{\frac {5}{2}} x \sqrt {1 + \frac {d x^{2}}{c}} + \frac {3 a b c^{\frac {5}{2}} x}{8 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {35 a b c^{\frac {3}{2}} d x^{3}}{24 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {17 a b \sqrt {c} d^{2} x^{5}}{12 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {5 a b c^{3} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{8 \sqrt {d}} + \frac {a b d^{3} x^{7}}{3 \sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {5 b^{2} c^{\frac {7}{2}} x}{128 d \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {133 b^{2} c^{\frac {5}{2}} x^{3}}{384 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {127 b^{2} c^{\frac {3}{2}} d x^{5}}{192 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {23 b^{2} \sqrt {c} d^{2} x^{7}}{48 \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {5 b^{2} c^{4} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{128 d^{\frac {3}{2}}} + \frac {b^{2} d^{3} x^{9}}{8 \sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(d*x**2+c)**(5/2)/x**2,x)

[Out]

-a**2*c**(5/2)/(x*sqrt(1 + d*x**2/c)) + a**2*c**(3/2)*d*x*sqrt(1 + d*x**2/c) - 7*a**2*c**(3/2)*d*x/(8*sqrt(1 +
d*x**2/c)) + 3*a**2*sqrt(c)*d**2*x**3/(8*sqrt(1 + d*x**2/c)) + 15*a**2*c**2*sqrt(d)*asinh(sqrt(d)*x/sqrt(c))/
8 + a**2*d**3*x**5/(4*sqrt(c)*sqrt(1 + d*x**2/c)) + a*b*c**(5/2)*x*sqrt(1 + d*x**2/c) + 3*a*b*c**(5/2)*x/(8*sq
rt(1 + d*x**2/c)) + 35*a*b*c**(3/2)*d*x**3/(24*sqrt(1 + d*x**2/c)) + 17*a*b*sqrt(c)*d**2*x**5/(12*sqrt(1 + d*x
**2/c)) + 5*a*b*c**3*asinh(sqrt(d)*x/sqrt(c))/(8*sqrt(d)) + a*b*d**3*x**7/(3*sqrt(c)*sqrt(1 + d*x**2/c)) + 5*b
**2*c**(7/2)*x/(128*d*sqrt(1 + d*x**2/c)) + 133*b**2*c**(5/2)*x**3/(384*sqrt(1 + d*x**2/c)) + 127*b**2*c**(3/2
)*d*x**5/(192*sqrt(1 + d*x**2/c)) + 23*b**2*sqrt(c)*d**2*x**7/(48*sqrt(1 + d*x**2/c)) - 5*b**2*c**4*asinh(sqrt
(d)*x/sqrt(c))/(128*d**(3/2)) + b**2*d**3*x**9/(8*sqrt(c)*sqrt(1 + d*x**2/c))

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