3.7.16 \(\int \frac {(a+b x^2)^2 (c+d x^2)^{5/2}}{x^6} \, dx\)

Optimal. Leaf size=228 \[ -\frac {a^2 \left (c+d x^2\right )^{7/2}}{5 c x^5}-\frac {\left (c+d x^2\right )^{5/2} \left (8 a d (a d+5 b c)+15 b^2 c^2\right )}{15 c^2 x}+\frac {d x \left (c+d x^2\right )^{3/2} \left (8 a d (a d+5 b c)+15 b^2 c^2\right )}{12 c^2}+\frac {d x \sqrt {c+d x^2} \left (8 a d (a d+5 b c)+15 b^2 c^2\right )}{8 c}+\frac {1}{8} \sqrt {d} \left (8 a d (a d+5 b c)+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )-\frac {2 a \left (c+d x^2\right )^{7/2} (a d+5 b c)}{15 c^2 x^3} \]

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Rubi [A]  time = 0.16, antiderivative size = 225, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {462, 453, 277, 195, 217, 206} \begin {gather*} -\frac {a^2 \left (c+d x^2\right )^{7/2}}{5 c x^5}-\frac {\left (c+d x^2\right )^{5/2} \left (\frac {8 a d (a d+5 b c)}{c^2}+15 b^2\right )}{15 x}+\frac {d x \left (c+d x^2\right )^{3/2} \left (8 a d (a d+5 b c)+15 b^2 c^2\right )}{12 c^2}+\frac {d x \sqrt {c+d x^2} \left (8 a d (a d+5 b c)+15 b^2 c^2\right )}{8 c}+\frac {1}{8} \sqrt {d} \left (8 a d (a d+5 b c)+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )-\frac {2 a \left (c+d x^2\right )^{7/2} (a d+5 b c)}{15 c^2 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^2*(c + d*x^2)^(5/2))/x^6,x]

[Out]

(d*(15*b^2*c^2 + 8*a*d*(5*b*c + a*d))*x*Sqrt[c + d*x^2])/(8*c) + (d*(15*b^2*c^2 + 8*a*d*(5*b*c + a*d))*x*(c +
d*x^2)^(3/2))/(12*c^2) - ((15*b^2 + (8*a*d*(5*b*c + a*d))/c^2)*(c + d*x^2)^(5/2))/(15*x) - (a^2*(c + d*x^2)^(7
/2))/(5*c*x^5) - (2*a*(5*b*c + a*d)*(c + d*x^2)^(7/2))/(15*c^2*x^3) + (Sqrt[d]*(15*b^2*c^2 + 8*a*d*(5*b*c + a*
d))*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/8

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 277

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

Rule 453

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

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^6} \, dx &=-\frac {a^2 \left (c+d x^2\right )^{7/2}}{5 c x^5}+\frac {\int \frac {\left (2 a (5 b c+a d)+5 b^2 c x^2\right ) \left (c+d x^2\right )^{5/2}}{x^4} \, dx}{5 c}\\ &=-\frac {a^2 \left (c+d x^2\right )^{7/2}}{5 c x^5}-\frac {2 a (5 b c+a d) \left (c+d x^2\right )^{7/2}}{15 c^2 x^3}-\frac {1}{15} \left (-15 b^2-\frac {8 a d (5 b c+a d)}{c^2}\right ) \int \frac {\left (c+d x^2\right )^{5/2}}{x^2} \, dx\\ &=-\frac {\left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}}{15 x}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{5 c x^5}-\frac {2 a (5 b c+a d) \left (c+d x^2\right )^{7/2}}{15 c^2 x^3}+\frac {1}{3} \left (d \left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right )\right ) \int \left (c+d x^2\right )^{3/2} \, dx\\ &=\frac {1}{12} d \left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right ) x \left (c+d x^2\right )^{3/2}-\frac {\left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}}{15 x}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{5 c x^5}-\frac {2 a (5 b c+a d) \left (c+d x^2\right )^{7/2}}{15 c^2 x^3}+\frac {1}{4} \left (c d \left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right )\right ) \int \sqrt {c+d x^2} \, dx\\ &=\frac {1}{8} c d \left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right ) x \sqrt {c+d x^2}+\frac {1}{12} d \left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right ) x \left (c+d x^2\right )^{3/2}-\frac {\left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}}{15 x}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{5 c x^5}-\frac {2 a (5 b c+a d) \left (c+d x^2\right )^{7/2}}{15 c^2 x^3}+\frac {1}{8} \left (d \left (15 b^2 c^2+40 a b c d+8 a^2 d^2\right )\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx\\ &=\frac {1}{8} c d \left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right ) x \sqrt {c+d x^2}+\frac {1}{12} d \left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right ) x \left (c+d x^2\right )^{3/2}-\frac {\left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}}{15 x}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{5 c x^5}-\frac {2 a (5 b c+a d) \left (c+d x^2\right )^{7/2}}{15 c^2 x^3}+\frac {1}{8} \left (d \left (15 b^2 c^2+40 a b c d+8 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )\\ &=\frac {1}{8} c d \left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right ) x \sqrt {c+d x^2}+\frac {1}{12} d \left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right ) x \left (c+d x^2\right )^{3/2}-\frac {\left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}}{15 x}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{5 c x^5}-\frac {2 a (5 b c+a d) \left (c+d x^2\right )^{7/2}}{15 c^2 x^3}+\frac {1}{8} \sqrt {d} \left (15 b^2 c^2+40 a b c d+8 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 158, normalized size = 0.69 \begin {gather*} \frac {1}{8} \sqrt {d} \left (8 a^2 d^2+40 a b c d+15 b^2 c^2\right ) \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )+\sqrt {c+d x^2} \left (\frac {-23 a^2 d^2-70 a b c d-15 b^2 c^2}{15 x}-\frac {a^2 c^2}{5 x^5}-\frac {a c (11 a d+10 b c)}{15 x^3}+\frac {1}{8} b d x (8 a d+9 b c)+\frac {1}{4} b^2 d^2 x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^2*(c + d*x^2)^(5/2))/x^6,x]

[Out]

Sqrt[c + d*x^2]*(-1/5*(a^2*c^2)/x^5 - (a*c*(10*b*c + 11*a*d))/(15*x^3) + (-15*b^2*c^2 - 70*a*b*c*d - 23*a^2*d^
2)/(15*x) + (b*d*(9*b*c + 8*a*d)*x)/8 + (b^2*d^2*x^3)/4) + (Sqrt[d]*(15*b^2*c^2 + 40*a*b*c*d + 8*a^2*d^2)*Log[
d*x + Sqrt[d]*Sqrt[c + d*x^2]])/8

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IntegrateAlgebraic [A]  time = 0.48, size = 169, normalized size = 0.74 \begin {gather*} \frac {1}{8} \left (-8 a^2 d^{5/2}-40 a b c d^{3/2}-15 b^2 c^2 \sqrt {d}\right ) \log \left (\sqrt {c+d x^2}-\sqrt {d} x\right )+\frac {\sqrt {c+d x^2} \left (-24 a^2 c^2-88 a^2 c d x^2-184 a^2 d^2 x^4-80 a b c^2 x^2-560 a b c d x^4+120 a b d^2 x^6-120 b^2 c^2 x^4+135 b^2 c d x^6+30 b^2 d^2 x^8\right )}{120 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b*x^2)^2*(c + d*x^2)^(5/2))/x^6,x]

[Out]

(Sqrt[c + d*x^2]*(-24*a^2*c^2 - 80*a*b*c^2*x^2 - 88*a^2*c*d*x^2 - 120*b^2*c^2*x^4 - 560*a*b*c*d*x^4 - 184*a^2*
d^2*x^4 + 135*b^2*c*d*x^6 + 120*a*b*d^2*x^6 + 30*b^2*d^2*x^8))/(120*x^5) + ((-15*b^2*c^2*Sqrt[d] - 40*a*b*c*d^
(3/2) - 8*a^2*d^(5/2))*Log[-(Sqrt[d]*x) + Sqrt[c + d*x^2]])/8

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fricas [A]  time = 1.05, size = 318, normalized size = 1.39 \begin {gather*} \left [\frac {15 \, {\left (15 \, b^{2} c^{2} + 40 \, a b c d + 8 \, a^{2} d^{2}\right )} \sqrt {d} x^{5} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (30 \, b^{2} d^{2} x^{8} + 15 \, {\left (9 \, b^{2} c d + 8 \, a b d^{2}\right )} x^{6} - 8 \, {\left (15 \, b^{2} c^{2} + 70 \, a b c d + 23 \, a^{2} d^{2}\right )} x^{4} - 24 \, a^{2} c^{2} - 8 \, {\left (10 \, a b c^{2} + 11 \, a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{240 \, x^{5}}, -\frac {15 \, {\left (15 \, b^{2} c^{2} + 40 \, a b c d + 8 \, a^{2} d^{2}\right )} \sqrt {-d} x^{5} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (30 \, b^{2} d^{2} x^{8} + 15 \, {\left (9 \, b^{2} c d + 8 \, a b d^{2}\right )} x^{6} - 8 \, {\left (15 \, b^{2} c^{2} + 70 \, a b c d + 23 \, a^{2} d^{2}\right )} x^{4} - 24 \, a^{2} c^{2} - 8 \, {\left (10 \, a b c^{2} + 11 \, a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{120 \, x^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(5/2)/x^6,x, algorithm="fricas")

[Out]

[1/240*(15*(15*b^2*c^2 + 40*a*b*c*d + 8*a^2*d^2)*sqrt(d)*x^5*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) +
2*(30*b^2*d^2*x^8 + 15*(9*b^2*c*d + 8*a*b*d^2)*x^6 - 8*(15*b^2*c^2 + 70*a*b*c*d + 23*a^2*d^2)*x^4 - 24*a^2*c^
2 - 8*(10*a*b*c^2 + 11*a^2*c*d)*x^2)*sqrt(d*x^2 + c))/x^5, -1/120*(15*(15*b^2*c^2 + 40*a*b*c*d + 8*a^2*d^2)*sq
rt(-d)*x^5*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (30*b^2*d^2*x^8 + 15*(9*b^2*c*d + 8*a*b*d^2)*x^6 - 8*(15*b^2*c
^2 + 70*a*b*c*d + 23*a^2*d^2)*x^4 - 24*a^2*c^2 - 8*(10*a*b*c^2 + 11*a^2*c*d)*x^2)*sqrt(d*x^2 + c))/x^5]

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giac [B]  time = 0.54, size = 510, normalized size = 2.24 \begin {gather*} \frac {1}{8} \, {\left (2 \, b^{2} d^{2} x^{2} + \frac {9 \, b^{2} c d^{3} + 8 \, a b d^{4}}{d^{2}}\right )} \sqrt {d x^{2} + c} x - \frac {1}{16} \, {\left (15 \, b^{2} c^{2} \sqrt {d} + 40 \, a b c d^{\frac {3}{2}} + 8 \, a^{2} d^{\frac {5}{2}}\right )} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right ) + \frac {2 \, {\left (15 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{8} b^{2} c^{3} \sqrt {d} + 90 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{8} a b c^{2} d^{\frac {3}{2}} + 45 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{8} a^{2} c d^{\frac {5}{2}} - 60 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{6} b^{2} c^{4} \sqrt {d} - 300 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{6} a b c^{3} d^{\frac {3}{2}} - 90 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{6} a^{2} c^{2} d^{\frac {5}{2}} + 90 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b^{2} c^{5} \sqrt {d} + 400 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a b c^{4} d^{\frac {3}{2}} + 140 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a^{2} c^{3} d^{\frac {5}{2}} - 60 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b^{2} c^{6} \sqrt {d} - 260 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b c^{5} d^{\frac {3}{2}} - 70 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} c^{4} d^{\frac {5}{2}} + 15 \, b^{2} c^{7} \sqrt {d} + 70 \, a b c^{6} d^{\frac {3}{2}} + 23 \, a^{2} c^{5} d^{\frac {5}{2}}\right )}}{15 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(5/2)/x^6,x, algorithm="giac")

[Out]

1/8*(2*b^2*d^2*x^2 + (9*b^2*c*d^3 + 8*a*b*d^4)/d^2)*sqrt(d*x^2 + c)*x - 1/16*(15*b^2*c^2*sqrt(d) + 40*a*b*c*d^
(3/2) + 8*a^2*d^(5/2))*log((sqrt(d)*x - sqrt(d*x^2 + c))^2) + 2/15*(15*(sqrt(d)*x - sqrt(d*x^2 + c))^8*b^2*c^3
*sqrt(d) + 90*(sqrt(d)*x - sqrt(d*x^2 + c))^8*a*b*c^2*d^(3/2) + 45*(sqrt(d)*x - sqrt(d*x^2 + c))^8*a^2*c*d^(5/
2) - 60*(sqrt(d)*x - sqrt(d*x^2 + c))^6*b^2*c^4*sqrt(d) - 300*(sqrt(d)*x - sqrt(d*x^2 + c))^6*a*b*c^3*d^(3/2)
- 90*(sqrt(d)*x - sqrt(d*x^2 + c))^6*a^2*c^2*d^(5/2) + 90*(sqrt(d)*x - sqrt(d*x^2 + c))^4*b^2*c^5*sqrt(d) + 40
0*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a*b*c^4*d^(3/2) + 140*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a^2*c^3*d^(5/2) - 60*(
sqrt(d)*x - sqrt(d*x^2 + c))^2*b^2*c^6*sqrt(d) - 260*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b*c^5*d^(3/2) - 70*(sqr
t(d)*x - sqrt(d*x^2 + c))^2*a^2*c^4*d^(5/2) + 15*b^2*c^7*sqrt(d) + 70*a*b*c^6*d^(3/2) + 23*a^2*c^5*d^(5/2))/((
sqrt(d)*x - sqrt(d*x^2 + c))^2 - c)^5

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maple [A]  time = 0.02, size = 369, normalized size = 1.62 \begin {gather*} a^{2} d^{\frac {5}{2}} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )+5 a b c \,d^{\frac {3}{2}} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )+\frac {15 b^{2} c^{2} \sqrt {d}\, \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{8}+\frac {\sqrt {d \,x^{2}+c}\, a^{2} d^{3} x}{c}+5 \sqrt {d \,x^{2}+c}\, a b \,d^{2} x +\frac {15 \sqrt {d \,x^{2}+c}\, b^{2} c d x}{8}+\frac {2 \left (d \,x^{2}+c \right )^{\frac {3}{2}} a^{2} d^{3} x}{3 c^{2}}+\frac {10 \left (d \,x^{2}+c \right )^{\frac {3}{2}} a b \,d^{2} x}{3 c}+\frac {5 \left (d \,x^{2}+c \right )^{\frac {3}{2}} b^{2} d x}{4}+\frac {8 \left (d \,x^{2}+c \right )^{\frac {5}{2}} a^{2} d^{3} x}{15 c^{3}}+\frac {8 \left (d \,x^{2}+c \right )^{\frac {5}{2}} a b \,d^{2} x}{3 c^{2}}+\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}} b^{2} d x}{c}-\frac {8 \left (d \,x^{2}+c \right )^{\frac {7}{2}} a^{2} d^{2}}{15 c^{3} x}-\frac {8 \left (d \,x^{2}+c \right )^{\frac {7}{2}} a b d}{3 c^{2} x}-\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}} b^{2}}{c x}-\frac {2 \left (d \,x^{2}+c \right )^{\frac {7}{2}} a^{2} d}{15 c^{2} x^{3}}-\frac {2 \left (d \,x^{2}+c \right )^{\frac {7}{2}} a b}{3 c \,x^{3}}-\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}} a^{2}}{5 c \,x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*(d*x^2+c)^(5/2)/x^6,x)

[Out]

-1/5*a^2*(d*x^2+c)^(7/2)/c/x^5-2/15*a^2*d/c^2/x^3*(d*x^2+c)^(7/2)-8/15*a^2*d^2/c^3/x*(d*x^2+c)^(7/2)+8/15*a^2*
d^3/c^3*x*(d*x^2+c)^(5/2)+2/3*a^2*d^3/c^2*x*(d*x^2+c)^(3/2)+a^2*d^3/c*x*(d*x^2+c)^(1/2)+a^2*d^(5/2)*ln(d^(1/2)
*x+(d*x^2+c)^(1/2))-b^2/c/x*(d*x^2+c)^(7/2)+b^2*d/c*x*(d*x^2+c)^(5/2)+5/4*b^2*d*x*(d*x^2+c)^(3/2)+15/8*b^2*d*c
*x*(d*x^2+c)^(1/2)+15/8*b^2*d^(1/2)*c^2*ln(d^(1/2)*x+(d*x^2+c)^(1/2))-2/3*a*b/c/x^3*(d*x^2+c)^(7/2)-8/3*a*b*d/
c^2/x*(d*x^2+c)^(7/2)+8/3*a*b*d^2/c^2*x*(d*x^2+c)^(5/2)+10/3*a*b*d^2/c*x*(d*x^2+c)^(3/2)+5*a*b*d^2*x*(d*x^2+c)
^(1/2)+5*a*b*d^(3/2)*c*ln(d^(1/2)*x+(d*x^2+c)^(1/2))

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maxima [A]  time = 0.96, size = 285, normalized size = 1.25 \begin {gather*} \frac {5}{4} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} d x + \frac {15}{8} \, \sqrt {d x^{2} + c} b^{2} c d x + 5 \, \sqrt {d x^{2} + c} a b d^{2} x + \frac {10 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b d^{2} x}{3 \, c} + \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{3} x}{3 \, c^{2}} + \frac {\sqrt {d x^{2} + c} a^{2} d^{3} x}{c} + \frac {15}{8} \, b^{2} c^{2} \sqrt {d} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right ) + 5 \, a b c d^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right ) + a^{2} d^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right ) - \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2}}{x} - \frac {8 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b d}{3 \, c x} - \frac {8 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} d^{2}}{15 \, c^{2} x} - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} a b}{3 \, c x^{3}} - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} a^{2} d}{15 \, c^{2} x^{3}} - \frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} a^{2}}{5 \, c x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(5/2)/x^6,x, algorithm="maxima")

[Out]

5/4*(d*x^2 + c)^(3/2)*b^2*d*x + 15/8*sqrt(d*x^2 + c)*b^2*c*d*x + 5*sqrt(d*x^2 + c)*a*b*d^2*x + 10/3*(d*x^2 + c
)^(3/2)*a*b*d^2*x/c + 2/3*(d*x^2 + c)^(3/2)*a^2*d^3*x/c^2 + sqrt(d*x^2 + c)*a^2*d^3*x/c + 15/8*b^2*c^2*sqrt(d)
*arcsinh(d*x/sqrt(c*d)) + 5*a*b*c*d^(3/2)*arcsinh(d*x/sqrt(c*d)) + a^2*d^(5/2)*arcsinh(d*x/sqrt(c*d)) - (d*x^2
+ c)^(5/2)*b^2/x - 8/3*(d*x^2 + c)^(5/2)*a*b*d/(c*x) - 8/15*(d*x^2 + c)^(5/2)*a^2*d^2/(c^2*x) - 2/3*(d*x^2 +
c)^(7/2)*a*b/(c*x^3) - 2/15*(d*x^2 + c)^(7/2)*a^2*d/(c^2*x^3) - 1/5*(d*x^2 + c)^(7/2)*a^2/(c*x^5)

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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 )}^2\,{\left (d\,x^2+c\right )}^{5/2}}{x^6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^2)^2*(c + d*x^2)^(5/2))/x^6,x)

[Out]

int(((a + b*x^2)^2*(c + d*x^2)^(5/2))/x^6, x)

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sympy [B]  time = 20.41, size = 474, normalized size = 2.08 \begin {gather*} - \frac {a^{2} \sqrt {c} d^{2}}{x \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {a^{2} c^{2} \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{5 x^{4}} - \frac {11 a^{2} c d^{\frac {3}{2}} \sqrt {\frac {c}{d x^{2}} + 1}}{15 x^{2}} - \frac {8 a^{2} d^{\frac {5}{2}} \sqrt {\frac {c}{d x^{2}} + 1}}{15} + a^{2} d^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )} - \frac {a^{2} d^{3} x}{\sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {4 a b c^{\frac {3}{2}} d}{x \sqrt {1 + \frac {d x^{2}}{c}}} + a b \sqrt {c} d^{2} x \sqrt {1 + \frac {d x^{2}}{c}} - \frac {4 a b \sqrt {c} d^{2} x}{\sqrt {1 + \frac {d x^{2}}{c}}} - \frac {2 a b c^{2} \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{3 x^{2}} - \frac {2 a b c d^{\frac {3}{2}} \sqrt {\frac {c}{d x^{2}} + 1}}{3} + 5 a b c d^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )} - \frac {b^{2} c^{\frac {5}{2}}}{x \sqrt {1 + \frac {d x^{2}}{c}}} + b^{2} c^{\frac {3}{2}} d x \sqrt {1 + \frac {d x^{2}}{c}} - \frac {7 b^{2} c^{\frac {3}{2}} d x}{8 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {3 b^{2} \sqrt {c} d^{2} x^{3}}{8 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {15 b^{2} c^{2} \sqrt {d} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{8} + \frac {b^{2} d^{3} x^{5}}{4 \sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(d*x**2+c)**(5/2)/x**6,x)

[Out]

-a**2*sqrt(c)*d**2/(x*sqrt(1 + d*x**2/c)) - a**2*c**2*sqrt(d)*sqrt(c/(d*x**2) + 1)/(5*x**4) - 11*a**2*c*d**(3/
2)*sqrt(c/(d*x**2) + 1)/(15*x**2) - 8*a**2*d**(5/2)*sqrt(c/(d*x**2) + 1)/15 + a**2*d**(5/2)*asinh(sqrt(d)*x/sq
rt(c)) - a**2*d**3*x/(sqrt(c)*sqrt(1 + d*x**2/c)) - 4*a*b*c**(3/2)*d/(x*sqrt(1 + d*x**2/c)) + a*b*sqrt(c)*d**2
*x*sqrt(1 + d*x**2/c) - 4*a*b*sqrt(c)*d**2*x/sqrt(1 + d*x**2/c) - 2*a*b*c**2*sqrt(d)*sqrt(c/(d*x**2) + 1)/(3*x
**2) - 2*a*b*c*d**(3/2)*sqrt(c/(d*x**2) + 1)/3 + 5*a*b*c*d**(3/2)*asinh(sqrt(d)*x/sqrt(c)) - b**2*c**(5/2)/(x*
sqrt(1 + d*x**2/c)) + b**2*c**(3/2)*d*x*sqrt(1 + d*x**2/c) - 7*b**2*c**(3/2)*d*x/(8*sqrt(1 + d*x**2/c)) + 3*b*
*2*sqrt(c)*d**2*x**3/(8*sqrt(1 + d*x**2/c)) + 15*b**2*c**2*sqrt(d)*asinh(sqrt(d)*x/sqrt(c))/8 + b**2*d**3*x**5
/(4*sqrt(c)*sqrt(1 + d*x**2/c))

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