3.7.14 \(\int \frac {(a+b x^2)^2 (c+d x^2)^{5/2}}{x^4} \, dx\)

Optimal. Leaf size=223 \[ -\frac {a^2 \left (c+d x^2\right )^{7/2}}{3 c x^3}+\frac {x \left (c+d x^2\right )^{5/2} \left (4 a d (2 a d+3 b c)+b^2 c^2\right )}{6 c^2}+\frac {5 x \left (c+d x^2\right )^{3/2} \left (4 a d (2 a d+3 b c)+b^2 c^2\right )}{24 c}+\frac {5}{16} x \sqrt {c+d x^2} \left (4 a d (2 a d+3 b c)+b^2 c^2\right )+\frac {5 c \left (4 a d (2 a d+3 b c)+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 \sqrt {d}}-\frac {2 a \left (c+d x^2\right )^{7/2} (2 a d+3 b c)}{3 c^2 x} \]

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Rubi [A]  time = 0.17, antiderivative size = 219, normalized size of antiderivative = 0.98, 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, 453, 195, 217, 206} \begin {gather*} -\frac {a^2 \left (c+d x^2\right )^{7/2}}{3 c x^3}+\frac {1}{6} x \left (c+d x^2\right )^{5/2} \left (\frac {4 a d (2 a d+3 b c)}{c^2}+b^2\right )+\frac {5 x \left (c+d x^2\right )^{3/2} \left (4 a d (2 a d+3 b c)+b^2 c^2\right )}{24 c}+\frac {5}{16} x \sqrt {c+d x^2} \left (4 a d (2 a d+3 b c)+b^2 c^2\right )+\frac {5 c \left (4 a d (2 a d+3 b c)+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 \sqrt {d}}-\frac {2 a \left (c+d x^2\right )^{7/2} (2 a d+3 b c)}{3 c^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^2*(c + d*x^2)^(5/2))/x^4,x]

[Out]

(5*(b^2*c^2 + 4*a*d*(3*b*c + 2*a*d))*x*Sqrt[c + d*x^2])/16 + (5*(b^2*c^2 + 4*a*d*(3*b*c + 2*a*d))*x*(c + d*x^2
)^(3/2))/(24*c) + ((b^2 + (4*a*d*(3*b*c + 2*a*d))/c^2)*x*(c + d*x^2)^(5/2))/6 - (a^2*(c + d*x^2)^(7/2))/(3*c*x
^3) - (2*a*(3*b*c + 2*a*d)*(c + d*x^2)^(7/2))/(3*c^2*x) + (5*c*(b^2*c^2 + 4*a*d*(3*b*c + 2*a*d))*ArcTanh[(Sqrt
[d]*x)/Sqrt[c + d*x^2]])/(16*Sqrt[d])

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 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^4} \, dx &=-\frac {a^2 \left (c+d x^2\right )^{7/2}}{3 c x^3}+\frac {\int \frac {\left (2 a (3 b c+2 a d)+3 b^2 c x^2\right ) \left (c+d x^2\right )^{5/2}}{x^2} \, dx}{3 c}\\ &=-\frac {a^2 \left (c+d x^2\right )^{7/2}}{3 c x^3}-\frac {2 a (3 b c+2 a d) \left (c+d x^2\right )^{7/2}}{3 c^2 x}+\left (b^2+\frac {4 a d (3 b c+2 a d)}{c^2}\right ) \int \left (c+d x^2\right )^{5/2} \, dx\\ &=\frac {1}{6} \left (b^2+\frac {4 a d (3 b c+2 a d)}{c^2}\right ) x \left (c+d x^2\right )^{5/2}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{3 c x^3}-\frac {2 a (3 b c+2 a d) \left (c+d x^2\right )^{7/2}}{3 c^2 x}+\frac {1}{6} \left (5 c \left (b^2+\frac {4 a d (3 b c+2 a d)}{c^2}\right )\right ) \int \left (c+d x^2\right )^{3/2} \, dx\\ &=\frac {5}{24} c \left (b^2+\frac {4 a d (3 b c+2 a d)}{c^2}\right ) x \left (c+d x^2\right )^{3/2}+\frac {1}{6} \left (b^2+\frac {4 a d (3 b c+2 a d)}{c^2}\right ) x \left (c+d x^2\right )^{5/2}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{3 c x^3}-\frac {2 a (3 b c+2 a d) \left (c+d x^2\right )^{7/2}}{3 c^2 x}+\frac {1}{8} \left (5 \left (b^2 c^2+12 a b c d+8 a^2 d^2\right )\right ) \int \sqrt {c+d x^2} \, dx\\ &=\frac {5}{16} \left (b^2 c^2+12 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}+\frac {5}{24} c \left (b^2+\frac {4 a d (3 b c+2 a d)}{c^2}\right ) x \left (c+d x^2\right )^{3/2}+\frac {1}{6} \left (b^2+\frac {4 a d (3 b c+2 a d)}{c^2}\right ) x \left (c+d x^2\right )^{5/2}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{3 c x^3}-\frac {2 a (3 b c+2 a d) \left (c+d x^2\right )^{7/2}}{3 c^2 x}+\frac {1}{16} \left (5 c \left (b^2 c^2+12 a b c d+8 a^2 d^2\right )\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx\\ &=\frac {5}{16} \left (b^2 c^2+12 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}+\frac {5}{24} c \left (b^2+\frac {4 a d (3 b c+2 a d)}{c^2}\right ) x \left (c+d x^2\right )^{3/2}+\frac {1}{6} \left (b^2+\frac {4 a d (3 b c+2 a d)}{c^2}\right ) x \left (c+d x^2\right )^{5/2}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{3 c x^3}-\frac {2 a (3 b c+2 a d) \left (c+d x^2\right )^{7/2}}{3 c^2 x}+\frac {1}{16} \left (5 c \left (b^2 c^2+12 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 {5}{16} \left (b^2 c^2+12 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}+\frac {5}{24} c \left (b^2+\frac {4 a d (3 b c+2 a d)}{c^2}\right ) x \left (c+d x^2\right )^{3/2}+\frac {1}{6} \left (b^2+\frac {4 a d (3 b c+2 a d)}{c^2}\right ) x \left (c+d x^2\right )^{5/2}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{3 c x^3}-\frac {2 a (3 b c+2 a d) \left (c+d x^2\right )^{7/2}}{3 c^2 x}+\frac {5 c \left (b^2 c^2+12 a b c d+8 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 \sqrt {d}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 155, normalized size = 0.70 \begin {gather*} \frac {1}{48} \left (\frac {15 c \left (8 a^2 d^2+12 a b c d+b^2 c^2\right ) \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )}{\sqrt {d}}+\frac {\sqrt {c+d x^2} \left (-8 a^2 \left (2 c^2+14 c d x^2-3 d^2 x^4\right )+12 a b x^2 \left (-8 c^2+9 c d x^2+2 d^2 x^4\right )+b^2 x^4 \left (33 c^2+26 c d x^2+8 d^2 x^4\right )\right )}{x^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^2*(c + d*x^2)^(5/2))/x^4,x]

[Out]

((Sqrt[c + d*x^2]*(-8*a^2*(2*c^2 + 14*c*d*x^2 - 3*d^2*x^4) + 12*a*b*x^2*(-8*c^2 + 9*c*d*x^2 + 2*d^2*x^4) + b^2
*x^4*(33*c^2 + 26*c*d*x^2 + 8*d^2*x^4)))/x^3 + (15*c*(b^2*c^2 + 12*a*b*c*d + 8*a^2*d^2)*Log[d*x + Sqrt[d]*Sqrt
[c + d*x^2]])/Sqrt[d])/48

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IntegrateAlgebraic [A]  time = 0.38, size = 165, normalized size = 0.74 \begin {gather*} \frac {\sqrt {c+d x^2} \left (-16 a^2 c^2-112 a^2 c d x^2+24 a^2 d^2 x^4-96 a b c^2 x^2+108 a b c d x^4+24 a b d^2 x^6+33 b^2 c^2 x^4+26 b^2 c d x^6+8 b^2 d^2 x^8\right )}{48 x^3}-\frac {5 \left (8 a^2 c d^2+12 a b c^2 d+b^2 c^3\right ) \log \left (\sqrt {c+d x^2}-\sqrt {d} x\right )}{16 \sqrt {d}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b*x^2)^2*(c + d*x^2)^(5/2))/x^4,x]

[Out]

(Sqrt[c + d*x^2]*(-16*a^2*c^2 - 96*a*b*c^2*x^2 - 112*a^2*c*d*x^2 + 33*b^2*c^2*x^4 + 108*a*b*c*d*x^4 + 24*a^2*d
^2*x^4 + 26*b^2*c*d*x^6 + 24*a*b*d^2*x^6 + 8*b^2*d^2*x^8))/(48*x^3) - (5*(b^2*c^3 + 12*a*b*c^2*d + 8*a^2*c*d^2
)*Log[-(Sqrt[d]*x) + Sqrt[c + d*x^2]])/(16*Sqrt[d])

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fricas [A]  time = 1.58, size = 346, normalized size = 1.55 \begin {gather*} \left [\frac {15 \, {\left (b^{2} c^{3} + 12 \, a b c^{2} d + 8 \, a^{2} c d^{2}\right )} \sqrt {d} x^{3} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (8 \, b^{2} d^{3} x^{8} + 2 \, {\left (13 \, b^{2} c d^{2} + 12 \, a b d^{3}\right )} x^{6} - 16 \, a^{2} c^{2} d + 3 \, {\left (11 \, b^{2} c^{2} d + 36 \, a b c d^{2} + 8 \, a^{2} d^{3}\right )} x^{4} - 16 \, {\left (6 \, a b c^{2} d + 7 \, a^{2} c d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{96 \, d x^{3}}, -\frac {15 \, {\left (b^{2} c^{3} + 12 \, a b c^{2} d + 8 \, a^{2} c d^{2}\right )} \sqrt {-d} x^{3} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (8 \, b^{2} d^{3} x^{8} + 2 \, {\left (13 \, b^{2} c d^{2} + 12 \, a b d^{3}\right )} x^{6} - 16 \, a^{2} c^{2} d + 3 \, {\left (11 \, b^{2} c^{2} d + 36 \, a b c d^{2} + 8 \, a^{2} d^{3}\right )} x^{4} - 16 \, {\left (6 \, a b c^{2} d + 7 \, a^{2} c d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{48 \, d x^{3}}\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^4,x, algorithm="fricas")

[Out]

[1/96*(15*(b^2*c^3 + 12*a*b*c^2*d + 8*a^2*c*d^2)*sqrt(d)*x^3*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) +
2*(8*b^2*d^3*x^8 + 2*(13*b^2*c*d^2 + 12*a*b*d^3)*x^6 - 16*a^2*c^2*d + 3*(11*b^2*c^2*d + 36*a*b*c*d^2 + 8*a^2*
d^3)*x^4 - 16*(6*a*b*c^2*d + 7*a^2*c*d^2)*x^2)*sqrt(d*x^2 + c))/(d*x^3), -1/48*(15*(b^2*c^3 + 12*a*b*c^2*d + 8
*a^2*c*d^2)*sqrt(-d)*x^3*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (8*b^2*d^3*x^8 + 2*(13*b^2*c*d^2 + 12*a*b*d^3)*x
^6 - 16*a^2*c^2*d + 3*(11*b^2*c^2*d + 36*a*b*c*d^2 + 8*a^2*d^3)*x^4 - 16*(6*a*b*c^2*d + 7*a^2*c*d^2)*x^2)*sqrt
(d*x^2 + c))/(d*x^3)]

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giac [A]  time = 0.60, size = 307, normalized size = 1.38 \begin {gather*} \frac {1}{48} \, {\left (2 \, {\left (4 \, b^{2} d^{2} x^{2} + \frac {13 \, b^{2} c d^{5} + 12 \, a b d^{6}}{d^{4}}\right )} x^{2} + \frac {3 \, {\left (11 \, b^{2} c^{2} d^{4} + 36 \, a b c d^{5} + 8 \, a^{2} d^{6}\right )}}{d^{4}}\right )} \sqrt {d x^{2} + c} x - \frac {5 \, {\left (b^{2} c^{3} \sqrt {d} + 12 \, a b c^{2} d^{\frac {3}{2}} + 8 \, a^{2} c d^{\frac {5}{2}}\right )} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right )}{32 \, d} + \frac {2 \, {\left (6 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a b c^{3} \sqrt {d} + 9 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a^{2} c^{2} d^{\frac {3}{2}} - 12 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b c^{4} \sqrt {d} - 12 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} c^{3} d^{\frac {3}{2}} + 6 \, a b c^{5} \sqrt {d} + 7 \, a^{2} c^{4} d^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{3}} \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^4,x, algorithm="giac")

[Out]

1/48*(2*(4*b^2*d^2*x^2 + (13*b^2*c*d^5 + 12*a*b*d^6)/d^4)*x^2 + 3*(11*b^2*c^2*d^4 + 36*a*b*c*d^5 + 8*a^2*d^6)/
d^4)*sqrt(d*x^2 + c)*x - 5/32*(b^2*c^3*sqrt(d) + 12*a*b*c^2*d^(3/2) + 8*a^2*c*d^(5/2))*log((sqrt(d)*x - sqrt(d
*x^2 + c))^2)/d + 2/3*(6*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a*b*c^3*sqrt(d) + 9*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a
^2*c^2*d^(3/2) - 12*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b*c^4*sqrt(d) - 12*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a^2*c
^3*d^(3/2) + 6*a*b*c^5*sqrt(d) + 7*a^2*c^4*d^(3/2))/((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c)^3

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maple [A]  time = 0.02, size = 298, normalized size = 1.34 \begin {gather*} \frac {5 a^{2} c \,d^{\frac {3}{2}} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{2}+\frac {15 a b \,c^{2} \sqrt {d}\, \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{4}+\frac {5 b^{2} c^{3} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{16 \sqrt {d}}+\frac {5 \sqrt {d \,x^{2}+c}\, a^{2} d^{2} x}{2}+\frac {15 \sqrt {d \,x^{2}+c}\, a b c d x}{4}+\frac {5 \sqrt {d \,x^{2}+c}\, b^{2} c^{2} x}{16}+\frac {5 \left (d \,x^{2}+c \right )^{\frac {3}{2}} a^{2} d^{2} x}{3 c}+\frac {5 \left (d \,x^{2}+c \right )^{\frac {3}{2}} a b d x}{2}+\frac {5 \left (d \,x^{2}+c \right )^{\frac {3}{2}} b^{2} c x}{24}+\frac {4 \left (d \,x^{2}+c \right )^{\frac {5}{2}} a^{2} d^{2} x}{3 c^{2}}+\frac {2 \left (d \,x^{2}+c \right )^{\frac {5}{2}} a b d x}{c}+\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}} b^{2} x}{6}-\frac {4 \left (d \,x^{2}+c \right )^{\frac {7}{2}} a^{2} d}{3 c^{2} x}-\frac {2 \left (d \,x^{2}+c \right )^{\frac {7}{2}} a b}{c x}-\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}} a^{2}}{3 c \,x^{3}} \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^4,x)

[Out]

1/6*x*b^2*(d*x^2+c)^(5/2)+5/24*b^2*c*x*(d*x^2+c)^(3/2)+5/16*b^2*c^2*x*(d*x^2+c)^(1/2)+5/16*b^2*c^3/d^(1/2)*ln(
d^(1/2)*x+(d*x^2+c)^(1/2))-2*a*b/c/x*(d*x^2+c)^(7/2)+2*a*b*d/c*x*(d*x^2+c)^(5/2)+5/2*a*b*d*x*(d*x^2+c)^(3/2)+1
5/4*a*b*d*c*x*(d*x^2+c)^(1/2)+15/4*a*b*d^(1/2)*c^2*ln(d^(1/2)*x+(d*x^2+c)^(1/2))-1/3*a^2*(d*x^2+c)^(7/2)/c/x^3
-4/3*a^2*d/c^2/x*(d*x^2+c)^(7/2)+4/3*a^2*d^2/c^2*x*(d*x^2+c)^(5/2)+5/3*a^2*d^2/c*x*(d*x^2+c)^(3/2)+5/2*a^2*d^2
*x*(d*x^2+c)^(1/2)+5/2*a^2*d^(3/2)*c*ln(d^(1/2)*x+(d*x^2+c)^(1/2))

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maxima [A]  time = 0.99, size = 234, normalized size = 1.05 \begin {gather*} \frac {1}{6} \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} x + \frac {5}{24} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c x + \frac {5}{16} \, \sqrt {d x^{2} + c} b^{2} c^{2} x + \frac {5}{2} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b d x + \frac {15}{4} \, \sqrt {d x^{2} + c} a b c d x + \frac {5}{2} \, \sqrt {d x^{2} + c} a^{2} d^{2} x + \frac {5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{2} x}{3 \, c} + \frac {5 \, b^{2} c^{3} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{16 \, \sqrt {d}} + \frac {15}{4} \, a b c^{2} \sqrt {d} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right ) + \frac {5}{2} \, a^{2} c d^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right ) - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b}{x} - \frac {4 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} d}{3 \, c x} - \frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} a^{2}}{3 \, c x^{3}} \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^4,x, algorithm="maxima")

[Out]

1/6*(d*x^2 + c)^(5/2)*b^2*x + 5/24*(d*x^2 + c)^(3/2)*b^2*c*x + 5/16*sqrt(d*x^2 + c)*b^2*c^2*x + 5/2*(d*x^2 + c
)^(3/2)*a*b*d*x + 15/4*sqrt(d*x^2 + c)*a*b*c*d*x + 5/2*sqrt(d*x^2 + c)*a^2*d^2*x + 5/3*(d*x^2 + c)^(3/2)*a^2*d
^2*x/c + 5/16*b^2*c^3*arcsinh(d*x/sqrt(c*d))/sqrt(d) + 15/4*a*b*c^2*sqrt(d)*arcsinh(d*x/sqrt(c*d)) + 5/2*a^2*c
*d^(3/2)*arcsinh(d*x/sqrt(c*d)) - 2*(d*x^2 + c)^(5/2)*a*b/x - 4/3*(d*x^2 + c)^(5/2)*a^2*d/(c*x) - 1/3*(d*x^2 +
c)^(7/2)*a^2/(c*x^3)

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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^4} \,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^4,x)

[Out]

int(((a + b*x^2)^2*(c + d*x^2)^(5/2))/x^4, x)

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sympy [B]  time = 24.42, size = 490, normalized size = 2.20 \begin {gather*} - \frac {2 a^{2} c^{\frac {3}{2}} d}{x \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {a^{2} \sqrt {c} d^{2} x \sqrt {1 + \frac {d x^{2}}{c}}}{2} - \frac {2 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}}{3 x^{2}} - \frac {a^{2} c d^{\frac {3}{2}} \sqrt {\frac {c}{d x^{2}} + 1}}{3} + \frac {5 a^{2} c d^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{2} - \frac {2 a b c^{\frac {5}{2}}}{x \sqrt {1 + \frac {d x^{2}}{c}}} + 2 a b c^{\frac {3}{2}} d x \sqrt {1 + \frac {d x^{2}}{c}} - \frac {7 a b c^{\frac {3}{2}} d x}{4 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {3 a b \sqrt {c} d^{2} x^{3}}{4 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {15 a b c^{2} \sqrt {d} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{4} + \frac {a b d^{3} x^{5}}{2 \sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {b^{2} c^{\frac {5}{2}} x \sqrt {1 + \frac {d x^{2}}{c}}}{2} + \frac {3 b^{2} c^{\frac {5}{2}} x}{16 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {35 b^{2} c^{\frac {3}{2}} d x^{3}}{48 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {17 b^{2} \sqrt {c} d^{2} x^{5}}{24 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {5 b^{2} c^{3} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{16 \sqrt {d}} + \frac {b^{2} d^{3} x^{7}}{6 \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**4,x)

[Out]

-2*a**2*c**(3/2)*d/(x*sqrt(1 + d*x**2/c)) + a**2*sqrt(c)*d**2*x*sqrt(1 + d*x**2/c)/2 - 2*a**2*sqrt(c)*d**2*x/s
qrt(1 + d*x**2/c) - a**2*c**2*sqrt(d)*sqrt(c/(d*x**2) + 1)/(3*x**2) - a**2*c*d**(3/2)*sqrt(c/(d*x**2) + 1)/3 +
5*a**2*c*d**(3/2)*asinh(sqrt(d)*x/sqrt(c))/2 - 2*a*b*c**(5/2)/(x*sqrt(1 + d*x**2/c)) + 2*a*b*c**(3/2)*d*x*sqr
t(1 + d*x**2/c) - 7*a*b*c**(3/2)*d*x/(4*sqrt(1 + d*x**2/c)) + 3*a*b*sqrt(c)*d**2*x**3/(4*sqrt(1 + d*x**2/c)) +
15*a*b*c**2*sqrt(d)*asinh(sqrt(d)*x/sqrt(c))/4 + a*b*d**3*x**5/(2*sqrt(c)*sqrt(1 + d*x**2/c)) + b**2*c**(5/2)
*x*sqrt(1 + d*x**2/c)/2 + 3*b**2*c**(5/2)*x/(16*sqrt(1 + d*x**2/c)) + 35*b**2*c**(3/2)*d*x**3/(48*sqrt(1 + d*x
**2/c)) + 17*b**2*sqrt(c)*d**2*x**5/(24*sqrt(1 + d*x**2/c)) + 5*b**2*c**3*asinh(sqrt(d)*x/sqrt(c))/(16*sqrt(d)
) + b**2*d**3*x**7/(6*sqrt(c)*sqrt(1 + d*x**2/c))

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