\(\int \frac {(c+d x^2)^{3/2}}{x^4 (a+b x^2)} \, dx\) [693]

Optimal result

Integrand size = 24, antiderivative size = 102 \[ \int \frac {\left (c+d x^2\right )^{3/2}}{x^4 \left (a+b x^2\right )} \, dx=-\frac {c \sqrt {c+d x^2}}{3 a x^3}+\frac {(3 b c-4 a d) \sqrt {c+d x^2}}{3 a^2 x}+\frac {(b c-a d)^{3/2} \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{5/2}} \]

[Out]

(-a*d+b*c)^(3/2)*arctan(x*(-a*d+b*c)^(1/2)/a^(1/2)/(d*x^2+c)^(1/2))/a^(5/2)-1/3*c*(d*x^2+c)^(1/2)/a/x^3+1/3*(-
4*a*d+3*b*c)*(d*x^2+c)^(1/2)/a^2/x

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {485, 597, 12, 385, 211} \[ \int \frac {\left (c+d x^2\right )^{3/2}}{x^4 \left (a+b x^2\right )} \, dx=\frac {(b c-a d)^{3/2} \arctan \left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{5/2}}+\frac {\sqrt {c+d x^2} (3 b c-4 a d)}{3 a^2 x}-\frac {c \sqrt {c+d x^2}}{3 a x^3} \]

[In]

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

[Out]

-1/3*(c*Sqrt[c + d*x^2])/(a*x^3) + ((3*b*c - 4*a*d)*Sqrt[c + d*x^2])/(3*a^2*x) + ((b*c - a*d)^(3/2)*ArcTan[(Sq
rt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/a^(5/2)

Rule 12

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

Rule 211

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

Rule 385

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 485

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

Rule 597

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(343\) vs. \(2(102)=204\).

Time = 1.17 (sec) , antiderivative size = 343, normalized size of antiderivative = 3.36 \[ \int \frac {\left (c+d x^2\right )^{3/2}}{x^4 \left (a+b x^2\right )} \, dx=\frac {\frac {a^{3/2} \sqrt {c+d x^2} \left (3 b c x^2-a \left (c+4 d x^2\right )\right )}{x^3}+\frac {3 (b c-a d) \sqrt {2 b c-a d-2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} \left (b c-a d+\sqrt {b} \sqrt {c} \sqrt {b c-a d}\right ) \arctan \left (\frac {\sqrt {2 b c-a d-2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} x}{\sqrt {a} \left (\sqrt {c}-\sqrt {c+d x^2}\right )}\right )}{d}+\frac {3 (b c-a d) \left (-b c+a d+\sqrt {b} \sqrt {c} \sqrt {b c-a d}\right ) \sqrt {2 b c-a d+2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} \arctan \left (\frac {\sqrt {2 b c-a d+2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} x}{\sqrt {a} \left (-\sqrt {c}+\sqrt {c+d x^2}\right )}\right )}{d}}{3 a^{7/2}} \]

[In]

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

[Out]

((a^(3/2)*Sqrt[c + d*x^2]*(3*b*c*x^2 - a*(c + 4*d*x^2)))/x^3 + (3*(b*c - a*d)*Sqrt[2*b*c - a*d - 2*Sqrt[b]*Sqr
t[c]*Sqrt[b*c - a*d]]*(b*c - a*d + Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d])*ArcTan[(Sqrt[2*b*c - a*d - 2*Sqrt[b]*Sqrt[
c]*Sqrt[b*c - a*d]]*x)/(Sqrt[a]*(Sqrt[c] - Sqrt[c + d*x^2]))])/d + (3*(b*c - a*d)*(-(b*c) + a*d + Sqrt[b]*Sqrt
[c]*Sqrt[b*c - a*d])*Sqrt[2*b*c - a*d + 2*Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d]]*ArcTan[(Sqrt[2*b*c - a*d + 2*Sqrt[b
]*Sqrt[c]*Sqrt[b*c - a*d]]*x)/(Sqrt[a]*(-Sqrt[c] + Sqrt[c + d*x^2]))])/d)/(3*a^(7/2))

Maple [A] (verified)

Time = 3.00 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.02

method	result	size
pseudoelliptic	\(-\frac {-3 x^{3} \left (a d -b c \right )^{2} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}\, a}{x \sqrt {\left (a d -b c \right ) a}}\right )+\sqrt {\left (a d -b c \right ) a}\, \left (\left (4 d \,x^{2}+c \right ) a -3 c b \,x^{2}\right ) \sqrt {d \,x^{2}+c}}{3 \sqrt {\left (a d -b c \right ) a}\, a^{2} x^{3}}\)	\(104\)
risch	\(-\frac {\sqrt {d \,x^{2}+c}\, \left (4 a d \,x^{2}-3 c b \,x^{2}+a c \right )}{3 a^{2} x^{3}}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (-\frac {\ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}}+\frac {\ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}}\right )}{a^{2}}\)	\(367\)
default	\(\text {Expression too large to display}\)	\(1439\)

[In]

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

[Out]

-1/3*(-3*x^3*(a*d-b*c)^2*arctanh((d*x^2+c)^(1/2)/x*a/((a*d-b*c)*a)^(1/2))+((a*d-b*c)*a)^(1/2)*((4*d*x^2+c)*a-3
*c*b*x^2)*(d*x^2+c)^(1/2))/((a*d-b*c)*a)^(1/2)/a^2/x^3

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 331, normalized size of antiderivative = 3.25 \[ \int \frac {\left (c+d x^2\right )^{3/2}}{x^4 \left (a+b x^2\right )} \, dx=\left [-\frac {3 \, {\left (b c - a d\right )} x^{3} \sqrt {-\frac {b c - a d}{a}} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} + 4 \, {\left (a^{2} c x - {\left (a b c - 2 \, a^{2} d\right )} x^{3}\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{a}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \, {\left ({\left (3 \, b c - 4 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c}}{12 \, a^{2} x^{3}}, \frac {3 \, {\left (b c - a d\right )} x^{3} \sqrt {\frac {b c - a d}{a}} \arctan \left (\frac {{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{a}}}{2 \, {\left ({\left (b c d - a d^{2}\right )} x^{3} + {\left (b c^{2} - a c d\right )} x\right )}}\right ) + 2 \, {\left ({\left (3 \, b c - 4 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c}}{6 \, a^{2} x^{3}}\right ] \]

[In]

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

[Out]

[-1/12*(3*(b*c - a*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*((3*b*c - 4*a*d)*x^2 - a*c)*sqrt(d*x^2 + c))/(a^2*x^3), 1/6*(3*(b*c - a*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*((3*b*c - 4*a*d)*x^2 - a*c)*sqrt(d*x^2 + c))/(a^2*x^3)]

Sympy [F]

\[ \int \frac {\left (c+d x^2\right )^{3/2}}{x^4 \left (a+b x^2\right )} \, dx=\int \frac {\left (c + d x^{2}\right )^{\frac {3}{2}}}{x^{4} \left (a + b x^{2}\right )}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\left (c+d x^2\right )^{3/2}}{x^4 \left (a+b x^2\right )} \, dx=\int { \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}}}{{\left (b x^{2} + a\right )} x^{4}} \,d x } \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 256 vs. \(2 (84) = 168\).

Time = 1.10 (sec) , antiderivative size = 256, normalized size of antiderivative = 2.51 \[ \int \frac {\left (c+d x^2\right )^{3/2}}{x^4 \left (a+b x^2\right )} \, dx=-\frac {{\left (b^{2} c^{2} \sqrt {d} - 2 \, a b c d^{\frac {3}{2}} + a^{2} d^{\frac {5}{2}}\right )} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{\sqrt {a b c d - a^{2} d^{2}} a^{2}} - \frac {2 \, {\left (3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b c^{2} \sqrt {d} - 6 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a c d^{\frac {3}{2}} - 6 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c^{3} \sqrt {d} + 6 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a c^{2} d^{\frac {3}{2}} + 3 \, b c^{4} \sqrt {d} - 4 \, a c^{3} d^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{3} a^{2}} \]

[In]

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

[Out]

-(b^2*c^2*sqrt(d) - 2*a*b*c*d^(3/2) + 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^2) - 2/3*(3*(sqrt(d)*x - sqrt(d*x^2 + c))^4*b*c^2*sqrt(d
) - 6*(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) + 6*(sqrt(
d)*x - sqrt(d*x^2 + c))^2*a*c^2*d^(3/2) + 3*b*c^4*sqrt(d) - 4*a*c^3*d^(3/2))/(((sqrt(d)*x - sqrt(d*x^2 + c))^2
- c)^3*a^2)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^2\right )^{3/2}}{x^4 \left (a+b x^2\right )} \, dx=\int \frac {{\left (d\,x^2+c\right )}^{3/2}}{x^4\,\left (b\,x^2+a\right )} \,d x \]

[In]

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

[Out]

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