3.8.30 \(\int \frac {(c+d x^2)^{3/2}}{x^4 (a+b x^2)^2} \, dx\)

Optimal. Leaf size=166 \[ \frac {(5 b c-2 a d) \sqrt {b c-a d} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{7/2}}+\frac {\sqrt {c+d x^2} (15 b c-11 a d)}{6 a^3 x}-\frac {\sqrt {c+d x^2} (5 b c-3 a d)}{6 a^2 b x^3}+\frac {\sqrt {c+d x^2} (b c-a d)}{2 a b x^3 \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.24, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {468, 583, 12, 377, 205} \begin {gather*} \frac {\sqrt {c+d x^2} (15 b c-11 a d)}{6 a^3 x}-\frac {\sqrt {c+d x^2} (5 b c-3 a d)}{6 a^2 b x^3}+\frac {(5 b c-2 a d) \sqrt {b c-a d} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{7/2}}+\frac {\sqrt {c+d x^2} (b c-a d)}{2 a b x^3 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)^(3/2)/(x^4*(a + b*x^2)^2),x]

[Out]

-((5*b*c - 3*a*d)*Sqrt[c + d*x^2])/(6*a^2*b*x^3) + ((15*b*c - 11*a*d)*Sqrt[c + d*x^2])/(6*a^3*x) + ((b*c - a*d
)*Sqrt[c + d*x^2])/(2*a*b*x^3*(a + b*x^2)) + ((5*b*c - 2*a*d)*Sqrt[b*c - a*d]*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt
[a]*Sqrt[c + d*x^2])])/(2*a^(7/2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 377

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

Rule 468

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

Rule 583

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

Rubi steps

\begin {align*} \int \frac {\left (c+d x^2\right )^{3/2}}{x^4 \left (a+b x^2\right )^2} \, dx &=\frac {(b c-a d) \sqrt {c+d x^2}}{2 a b x^3 \left (a+b x^2\right )}-\frac {\int \frac {-c (5 b c-3 a d)-2 d (2 b c-a d) x^2}{x^4 \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{2 a b}\\ &=-\frac {(5 b c-3 a d) \sqrt {c+d x^2}}{6 a^2 b x^3}+\frac {(b c-a d) \sqrt {c+d x^2}}{2 a b x^3 \left (a+b x^2\right )}+\frac {\int \frac {-b c^2 (15 b c-11 a d)-2 b c d (5 b c-3 a d) x^2}{x^2 \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{6 a^2 b c}\\ &=-\frac {(5 b c-3 a d) \sqrt {c+d x^2}}{6 a^2 b x^3}+\frac {(15 b c-11 a d) \sqrt {c+d x^2}}{6 a^3 x}+\frac {(b c-a d) \sqrt {c+d x^2}}{2 a b x^3 \left (a+b x^2\right )}-\frac {\int -\frac {3 b c^2 (5 b c-2 a d) (b c-a d)}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{6 a^3 b c^2}\\ &=-\frac {(5 b c-3 a d) \sqrt {c+d x^2}}{6 a^2 b x^3}+\frac {(15 b c-11 a d) \sqrt {c+d x^2}}{6 a^3 x}+\frac {(b c-a d) \sqrt {c+d x^2}}{2 a b x^3 \left (a+b x^2\right )}+\frac {((5 b c-2 a d) (b c-a d)) \int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{2 a^3}\\ &=-\frac {(5 b c-3 a d) \sqrt {c+d x^2}}{6 a^2 b x^3}+\frac {(15 b c-11 a d) \sqrt {c+d x^2}}{6 a^3 x}+\frac {(b c-a d) \sqrt {c+d x^2}}{2 a b x^3 \left (a+b x^2\right )}+\frac {((5 b c-2 a d) (b c-a d)) \operatorname {Subst}\left (\int \frac {1}{a-(-b c+a d) x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{2 a^3}\\ &=-\frac {(5 b c-3 a d) \sqrt {c+d x^2}}{6 a^2 b x^3}+\frac {(15 b c-11 a d) \sqrt {c+d x^2}}{6 a^3 x}+\frac {(b c-a d) \sqrt {c+d x^2}}{2 a b x^3 \left (a+b x^2\right )}+\frac {(5 b c-2 a d) \sqrt {b c-a d} \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 5.16, size = 131, normalized size = 0.79 \begin {gather*} \frac {(5 b c-2 a d) \sqrt {b c-a d} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{7/2}}+\frac {\sqrt {c+d x^2} \left (-2 a^2 \left (c+4 d x^2\right )+a b x^2 \left (10 c-11 d x^2\right )+15 b^2 c x^4\right )}{6 a^3 x^3 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)^(3/2)/(x^4*(a + b*x^2)^2),x]

[Out]

(Sqrt[c + d*x^2]*(15*b^2*c*x^4 + a*b*x^2*(10*c - 11*d*x^2) - 2*a^2*(c + 4*d*x^2)))/(6*a^3*x^3*(a + b*x^2)) + (
(5*b*c - 2*a*d)*Sqrt[b*c - a*d]*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(2*a^(7/2))

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IntegrateAlgebraic [A]  time = 0.63, size = 154, normalized size = 0.93 \begin {gather*} \frac {\sqrt {c+d x^2} \left (-2 a^2 c-8 a^2 d x^2+10 a b c x^2-11 a b d x^4+15 b^2 c x^4\right )}{6 a^3 x^3 \left (a+b x^2\right )}-\frac {(5 b c-2 a d) \sqrt {b c-a d} \tan ^{-1}\left (\frac {a \sqrt {d}-b x \sqrt {c+d x^2}+b \sqrt {d} x^2}{\sqrt {a} \sqrt {b c-a d}}\right )}{2 a^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(c + d*x^2)^(3/2)/(x^4*(a + b*x^2)^2),x]

[Out]

(Sqrt[c + d*x^2]*(-2*a^2*c + 10*a*b*c*x^2 - 8*a^2*d*x^2 + 15*b^2*c*x^4 - 11*a*b*d*x^4))/(6*a^3*x^3*(a + b*x^2)
) - ((5*b*c - 2*a*d)*Sqrt[b*c - a*d]*ArcTan[(a*Sqrt[d] + b*Sqrt[d]*x^2 - b*x*Sqrt[c + d*x^2])/(Sqrt[a]*Sqrt[b*
c - a*d])])/(2*a^(7/2))

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fricas [A]  time = 1.15, size = 443, normalized size = 2.67 \begin {gather*} \left [-\frac {3 \, {\left ({\left (5 \, b^{2} c - 2 \, a b d\right )} x^{5} + {\left (5 \, a b c - 2 \, a^{2} d\right )} x^{3}\right )} \sqrt {-\frac {b c - a d}{a}} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} + 4 \, {\left (a^{2} c x - {\left (a b c - 2 \, a^{2} d\right )} x^{3}\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{a}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \, {\left ({\left (15 \, b^{2} c - 11 \, a b d\right )} x^{4} - 2 \, a^{2} c + 2 \, {\left (5 \, a b c - 4 \, a^{2} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{24 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}}, \frac {3 \, {\left ({\left (5 \, b^{2} c - 2 \, a b d\right )} x^{5} + {\left (5 \, a b c - 2 \, a^{2} d\right )} x^{3}\right )} \sqrt {\frac {b c - a d}{a}} \arctan \left (\frac {{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{a}}}{2 \, {\left ({\left (b c d - a d^{2}\right )} x^{3} + {\left (b c^{2} - a c d\right )} x\right )}}\right ) + 2 \, {\left ({\left (15 \, b^{2} c - 11 \, a b d\right )} x^{4} - 2 \, a^{2} c + 2 \, {\left (5 \, a b c - 4 \, a^{2} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{12 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^(3/2)/x^4/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

[-1/24*(3*((5*b^2*c - 2*a*b*d)*x^5 + (5*a*b*c - 2*a^2*d)*x^3)*sqrt(-(b*c - a*d)/a)*log(((b^2*c^2 - 8*a*b*c*d +
8*a^2*d^2)*x^4 + a^2*c^2 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^2 + 4*(a^2*c*x - (a*b*c - 2*a^2*d)*x^3)*sqrt(d*x^2 + c
)*sqrt(-(b*c - a*d)/a))/(b^2*x^4 + 2*a*b*x^2 + a^2)) - 4*((15*b^2*c - 11*a*b*d)*x^4 - 2*a^2*c + 2*(5*a*b*c - 4
*a^2*d)*x^2)*sqrt(d*x^2 + c))/(a^3*b*x^5 + a^4*x^3), 1/12*(3*((5*b^2*c - 2*a*b*d)*x^5 + (5*a*b*c - 2*a^2*d)*x^
3)*sqrt((b*c - a*d)/a)*arctan(1/2*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + c)*sqrt((b*c - a*d)/a)/((b*c*d - a*d^
2)*x^3 + (b*c^2 - a*c*d)*x)) + 2*((15*b^2*c - 11*a*b*d)*x^4 - 2*a^2*c + 2*(5*a*b*c - 4*a^2*d)*x^2)*sqrt(d*x^2
+ c))/(a^3*b*x^5 + a^4*x^3)]

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giac [B]  time = 4.27, size = 442, normalized size = 2.66 \begin {gather*} -\frac {{\left (5 \, b^{2} c^{2} \sqrt {d} - 7 \, a b c d^{\frac {3}{2}} + 2 \, a^{2} d^{\frac {5}{2}}\right )} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{2 \, \sqrt {a b c d - a^{2} d^{2}} a^{3}} - \frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b^{2} c^{2} \sqrt {d} - 3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b c d^{\frac {3}{2}} + 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} d^{\frac {5}{2}} - b^{2} c^{3} \sqrt {d} + a b c^{2} d^{\frac {3}{2}}}{{\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a d + b c^{2}\right )} a^{3}} - \frac {4 \, {\left (3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b c^{2} \sqrt {d} - 3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a c d^{\frac {3}{2}} - 6 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c^{3} \sqrt {d} + 3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a c^{2} d^{\frac {3}{2}} + 3 \, b c^{4} \sqrt {d} - 2 \, a c^{3} d^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{3} a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^(3/2)/x^4/(b*x^2+a)^2,x, algorithm="giac")

[Out]

-1/2*(5*b^2*c^2*sqrt(d) - 7*a*b*c*d^(3/2) + 2*a^2*d^(5/2))*arctan(1/2*((sqrt(d)*x - sqrt(d*x^2 + c))^2*b - b*c
+ 2*a*d)/sqrt(a*b*c*d - a^2*d^2))/(sqrt(a*b*c*d - a^2*d^2)*a^3) - ((sqrt(d)*x - sqrt(d*x^2 + c))^2*b^2*c^2*sq
rt(d) - 3*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b*c*d^(3/2) + 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a^2*d^(5/2) - b^2*
c^3*sqrt(d) + a*b*c^2*d^(3/2))/(((sqrt(d)*x - sqrt(d*x^2 + c))^4*b - 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c + 4
*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*d + b*c^2)*a^3) - 4/3*(3*(sqrt(d)*x - sqrt(d*x^2 + c))^4*b*c^2*sqrt(d) - 3*
(sqrt(d)*x - sqrt(d*x^2 + c))^4*a*c*d^(3/2) - 6*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c^3*sqrt(d) + 3*(sqrt(d)*x -
sqrt(d*x^2 + c))^2*a*c^2*d^(3/2) + 3*b*c^4*sqrt(d) - 2*a*c^3*d^(3/2))/(((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c)^
3*a^3)

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maple [B]  time = 0.02, size = 4908, normalized size = 29.57 \begin {gather*} \text {output too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)^(3/2)/x^4/(b*x^2+a)^2,x)

[Out]

1/a^2*d^(3/2)*ln(d^(1/2)*x+(d*x^2+c)^(1/2))-5/4/a^2*d^(3/2)*ln(((x-(-a*b)^(1/2)/b)*d+(-a*b)^(1/2)/b*d)/d^(1/2)
+((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))-5/4/a^2*d^(3/2)*ln(((x+(-a*
b)^(1/2)/b)*d-(-a*b)^(1/2)/b*d)/d^(1/2)+((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c
)/b)^(1/2))-1/4*b/a^3*(-a*b)^(1/2)*d/(a*d-b*c)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(
a*d-b*c)/b)^(3/2)-3/8*b/a^2*d^2/(a*d-b*c)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b
*c)/b)^(1/2)*x-9/8*b/a^2*d^(3/2)/(a*d-b*c)*ln(((x+(-a*b)^(1/2)/b)*d-(-a*b)^(1/2)/b*d)/d^(1/2)+((x+(-a*b)^(1/2)
/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))*c+1/4*b^2/a^3*d/(a*d-b*c)*((x+(-a*b)^(1/2)/b
)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(3/2)*x+3/8*b^2/a^3*d^(1/2)/(a*d-b*c)*c^2*ln(((x+(-a*
b)^(1/2)/b)*d-(-a*b)^(1/2)/b*d)/d^(1/2)+((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c
)/b)^(1/2))-5/4*b^2/a^3/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c
)/b+2*(-(a*d-b*c)/b)^(1/2)*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(
x-(-a*b)^(1/2)/b))*c^2-2*b/a^3*d/c*x*(d*x^2+c)^(3/2)+1/4*b/a^3*(-a*b)^(1/2)*d/(a*d-b*c)*((x-(-a*b)^(1/2)/b)^2*
d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(3/2)-3/8*b/a^2*d^2/(a*d-b*c)*((x-(-a*b)^(1/2)/b)^2*d+2*(
-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*x-9/8*b/a^2*d^(3/2)/(a*d-b*c)*ln(((x-(-a*b)^(1/2)/b)*d+(
-a*b)^(1/2)/b*d)/d^(1/2)+((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))*c+1
/4*b^2/a^3*d/(a*d-b*c)*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(3/2)*x+3/8*
b^2/a^3*d^(1/2)/(a*d-b*c)*c^2*ln(((x-(-a*b)^(1/2)/b)*d+(-a*b)^(1/2)/b*d)/d^(1/2)+((x-(-a*b)^(1/2)/b)^2*d+2*(-a
*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))+5/4*b^2/a^3/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(-a*
b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x
+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x+(-a*b)^(1/2)/b))*c^2-3/4*b/a^3*(-a*b)^(1/2)*d/(a*d-b*c)*((x+(-a*b)
^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*c+3/4/b/a*(-a*b)^(1/2)*d^3/(a*d-b*c)/(-
(a*d-b*c)/b)^(1/2)*ln((-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x+(-a*b)^
(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x+(-a*b)^(1/2)/b))-3/2/a^2*(-a*b)^(1/2
)*d^2/(a*d-b*c)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)
^(1/2)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x+(-a*b)^(1/2)/b))*c
+3/8*b^2/a^3*d/(a*d-b*c)*c*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*x+
5/2*b/a^2/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b
*c)/b)^(1/2)*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x-(-a*b)^(1/2)
/b))*d*c+3/4*b/a^3*(-a*b)^(1/2)*d/(a*d-b*c)*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d
-b*c)/b)^(1/2)*c-3/4/b/a*(-a*b)^(1/2)*d^3/(a*d-b*c)/(-(a*d-b*c)/b)^(1/2)*ln((2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)
/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b
*c)/b)^(1/2))/(x-(-a*b)^(1/2)/b))+3/2/a^2*(-a*b)^(1/2)*d^2/(a*d-b*c)/(-(a*d-b*c)/b)^(1/2)*ln((2*(-a*b)^(1/2)*(
x-(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1
/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x-(-a*b)^(1/2)/b))*c+5/12*b^2/a^3/(-a*b)^(1/2)*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b
)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(3/2)+3/4/a*d^(5/2)/(a*d-b*c)*ln(((x+(-a*b)^(1/2)/b)*d-(-a*b)^(1/2
)/b*d)/d^(1/2)+((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))+3/4/a*d^(5/2)
/(a*d-b*c)*ln(((x-(-a*b)^(1/2)/b)*d+(-a*b)^(1/2)/b*d)/d^(1/2)+((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)
^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))-5/12*b^2/a^3/(-a*b)^(1/2)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(
1/2)/b)/b*d-(a*d-b*c)/b)^(3/2)-1/3/a^2/c/x^3*(d*x^2+c)^(5/2)+15/8*b/a^3*d^(1/2)*ln(((x+(-a*b)^(1/2)/b)*d-(-a*b
)^(1/2)/b*d)/d^(1/2)+((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))*c+5/4*b
/a^2/(-a*b)^(1/2)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*d-5/4*b^2/a
^3/(-a*b)^(1/2)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*c+5/4/a/(-a*b
)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*(
(x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x+(-a*b)^(1/2)/b))*d^2+5/8*b
/a^3*d*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*x-1/4*b^2/a^3/(a*d-b*c
)/(x-(-a*b)^(1/2)/b)*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(5/2)-3/4/a^2*
(-a*b)^(1/2)*d^2/(a*d-b*c)*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)-1/
4*b^2/a^3/(a*d-b*c)/(x+(-a*b)^(1/2)/b)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)
/b)^(5/2)+3/4/a^2*(-a*b)^(1/2)*d^2/(a*d-b*c)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*
d-b*c)/b)^(1/2)+5/8*b/a^3*d*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*x
+15/8*b/a^3*d^(1/2)*ln(((x-(-a*b)^(1/2)/b)*d+(-a*b)^(1/2)/b*d)/d^(1/2)+((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*
(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))*c-5/4*b/a^2/(-a*b)^(1/2)*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-
(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*d+5/4*b^2/a^3/(-a*b)^(1/2)*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-
a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*c-5/4/a/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((2*(-a*b)^(1/2)*(x-(-a*b)^(1
/2)/b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-
(a*d-b*c)/b)^(1/2))/(x-(-a*b)^(1/2)/b))*d^2+2*b/a^3/c/x*(d*x^2+c)^(5/2)-3*b/a^3*d*x*(d*x^2+c)^(1/2)-3*b/a^3*d^
(1/2)*c*ln(d^(1/2)*x+(d*x^2+c)^(1/2))-2/3/a^2*d/c^2/x*(d*x^2+c)^(5/2)+2/3/a^2*d^2/c^2*x*(d*x^2+c)^(3/2)+1/a^2*
d^2/c*x*(d*x^2+c)^(1/2)+3/8*b^2/a^3*d/(a*d-b*c)*c*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*
d-(a*d-b*c)/b)^(1/2)*x-5/2*b/a^2/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-
2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b
)^(1/2))/(x+(-a*b)^(1/2)/b))*d*c-3/4*b/a^3*(-a*b)^(1/2)*d/(a*d-b*c)/(-(a*d-b*c)/b)^(1/2)*ln((2*(-a*b)^(1/2)*(x
-(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/
2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x-(-a*b)^(1/2)/b))*c^2+3/4*b/a^3*(-a*b)^(1/2)*d/(a*d-b*c)/(-(a*d-b*c)/b)^(1/2)*
ln((-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*
b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x+(-a*b)^(1/2)/b))*c^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}}}{{\left (b x^{2} + a\right )}^{2} x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^(3/2)/x^4/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)^(3/2)/((b*x^2 + a)^2*x^4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d\,x^2+c\right )}^{3/2}}{x^4\,{\left (b\,x^2+a\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^2)^(3/2)/(x^4*(a + b*x^2)^2),x)

[Out]

int((c + d*x^2)^(3/2)/(x^4*(a + b*x^2)^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c + d x^{2}\right )^{\frac {3}{2}}}{x^{4} \left (a + b x^{2}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)**(3/2)/x**4/(b*x**2+a)**2,x)

[Out]

Integral((c + d*x**2)**(3/2)/(x**4*(a + b*x**2)**2), x)

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