3.7.100 \(\int \frac {(c+d x^2)^{5/2}}{x^2 (a+b x^2)} \, dx\) [700]

Optimal. Leaf size=145 \[ \frac {d (2 b c+a d) x \sqrt {c+d x^2}}{2 a b}-\frac {c \left (c+d x^2\right )^{3/2}}{a x}-\frac {(b c-a d)^{5/2} \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{3/2} b^2}+\frac {d^{3/2} (5 b c-2 a d) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 b^2} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {485, 542, 537, 223, 212, 385, 211} \begin {gather*} -\frac {(b c-a d)^{5/2} \text {ArcTan}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{3/2} b^2}+\frac {d^{3/2} (5 b c-2 a d) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 b^2}+\frac {d x \sqrt {c+d x^2} (a d+2 b c)}{2 a b}-\frac {c \left (c+d x^2\right )^{3/2}}{a x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(2*b*c + a*d)*x*Sqrt[c + d*x^2])/(2*a*b) - (c*(c + d*x^2)^(3/2))/(a*x) - ((b*c - a*d)^(5/2)*ArcTan[(Sqrt[b*
c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(a^(3/2)*b^2) + (d^(3/2)*(5*b*c - 2*a*d)*ArcTanh[(Sqrt[d]*x)/Sqrt[c +
d*x^2]])/(2*b^2)

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 212

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

Rule 223

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 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 537

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 542

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

Rubi steps

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

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Mathematica [A]
time = 0.34, size = 148, normalized size = 1.02 \begin {gather*} \frac {\frac {b \sqrt {c+d x^2} \left (-2 b c^2+a d^2 x^2\right )}{a x}+\frac {2 (b c-a d)^{5/2} \tan ^{-1}\left (\frac {a \sqrt {d}+b x \left (\sqrt {d} x-\sqrt {c+d x^2}\right )}{\sqrt {a} \sqrt {b c-a d}}\right )}{a^{3/2}}+d^{3/2} (-5 b c+2 a d) \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*Sqrt[c + d*x^2]*(-2*b*c^2 + a*d^2*x^2))/(a*x) + (2*(b*c - a*d)^(5/2)*ArcTan[(a*Sqrt[d] + b*x*(Sqrt[d]*x -
Sqrt[c + d*x^2]))/(Sqrt[a]*Sqrt[b*c - a*d])])/a^(3/2) + d^(3/2)*(-5*b*c + 2*a*d)*Log[-(Sqrt[d]*x) + Sqrt[c + d
*x^2]])/(2*b^2)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(2176\) vs. \(2(121)=242\).
time = 0.13, size = 2177, normalized size = 15.01

method	result	size
risch	\(\frac {\sqrt {d \,x^{2}+c}\, \left (a \,d^{2} x^{2}-2 b \,c^{2}\right )}{2 b a x}-\frac {a \,d^{\frac {5}{2}} \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{b^{2}}+\frac {5 d^{\frac {3}{2}} \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right ) c}{2 b}-\frac {a^{2} \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 ) d^{3}}{2 b^{2} \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}}+\frac {3 a \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 ) c \,d^{2}}{2 b \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}}-\frac {3 \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 ) c^{2} d}{2 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}}+\frac {b \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 ) c^{3}}{2 a \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}}+\frac {a^{2} \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 ) d^{3}}{2 b^{2} \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}}-\frac {3 a \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 ) c \,d^{2}}{2 b \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}}+\frac {3 \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 ) c^{2} d}{2 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}}-\frac {b \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 ) c^{3}}{2 a \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}}\)	\(1359\)
default	\(\text {Expression too large to display}\)	\(2177\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b/a/(-a*b)^(1/2)*(1/5*(d*(x-1/b*(-a*b)^(1/2))^2+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(5/2
)+d*(-a*b)^(1/2)/b*(1/8*(2*d*(x-1/b*(-a*b)^(1/2))+2*d*(-a*b)^(1/2)/b)/d*(d*(x-1/b*(-a*b)^(1/2))^2+2*d*(-a*b)^(
1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(3/2)+3/16*(-4*d*(a*d-b*c)/b+4*d^2*a/b)/d*(1/4*(2*d*(x-1/b*(-a*b)^(1/
2))+2*d*(-a*b)^(1/2)/b)/d*(d*(x-1/b*(-a*b)^(1/2))^2+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)
+1/8*(-4*d*(a*d-b*c)/b+4*d^2*a/b)/d^(3/2)*ln((d*(-a*b)^(1/2)/b+d*(x-1/b*(-a*b)^(1/2)))/d^(1/2)+(d*(x-1/b*(-a*b
)^(1/2))^2+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))))-(a*d-b*c)/b*(1/3*(d*(x-1/b*(-a*b)^(1/
2))^2+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(3/2)+d*(-a*b)^(1/2)/b*(1/4*(2*d*(x-1/b*(-a*b)^(1/2
))+2*d*(-a*b)^(1/2)/b)/d*(d*(x-1/b*(-a*b)^(1/2))^2+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)+
1/8*(-4*d*(a*d-b*c)/b+4*d^2*a/b)/d^(3/2)*ln((d*(-a*b)^(1/2)/b+d*(x-1/b*(-a*b)^(1/2)))/d^(1/2)+(d*(x-1/b*(-a*b)
^(1/2))^2+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)))-(a*d-b*c)/b*((d*(x-1/b*(-a*b)^(1/2))^2+
2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)+d^(1/2)*(-a*b)^(1/2)/b*ln((d*(-a*b)^(1/2)/b+d*(x-1/
b*(-a*b)^(1/2)))/d^(1/2)+(d*(x-1/b*(-a*b)^(1/2))^2+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))
+(a*d-b*c)/b/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^
(1/2)*(d*(x-1/b*(-a*b)^(1/2))^2+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x-1/b*(-a*b)^(1/2
))))))+1/2*b/a/(-a*b)^(1/2)*(1/5*(d*(x+1/b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b
)^(5/2)-d*(-a*b)^(1/2)/b*(1/8*(2*d*(x+1/b*(-a*b)^(1/2))-2*d*(-a*b)^(1/2)/b)/d*(d*(x+1/b*(-a*b)^(1/2))^2-2*d*(-
a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(3/2)+3/16*(-4*d*(a*d-b*c)/b+4*d^2*a/b)/d*(1/4*(2*d*(x+1/b*(-a*
b)^(1/2))-2*d*(-a*b)^(1/2)/b)/d*(d*(x+1/b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)
^(1/2)+1/8*(-4*d*(a*d-b*c)/b+4*d^2*a/b)/d^(3/2)*ln((-d*(-a*b)^(1/2)/b+d*(x+1/b*(-a*b)^(1/2)))/d^(1/2)+(d*(x+1/
b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))))-(a*d-b*c)/b*(1/3*(d*(x+1/b*(-a
*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(3/2)-d*(-a*b)^(1/2)/b*(1/4*(2*d*(x+1/b*(-a*
b)^(1/2))-2*d*(-a*b)^(1/2)/b)/d*(d*(x+1/b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)
^(1/2)+1/8*(-4*d*(a*d-b*c)/b+4*d^2*a/b)/d^(3/2)*ln((-d*(-a*b)^(1/2)/b+d*(x+1/b*(-a*b)^(1/2)))/d^(1/2)+(d*(x+1/
b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)))-(a*d-b*c)/b*((d*(x+1/b*(-a*b)^(
1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)-d^(1/2)*(-a*b)^(1/2)/b*ln((-d*(-a*b)^(1/2)/
b+d*(x+1/b*(-a*b)^(1/2)))/d^(1/2)+(d*(x+1/b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/
b)^(1/2))+(a*d-b*c)/b/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))+2*(-(a*d
-b*c)/b)^(1/2)*(d*(x+1/b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x+1/b*(-
a*b)^(1/2))))))+1/a*(-1/c/x*(d*x^2+c)^(7/2)+6*d/c*(1/6*x*(d*x^2+c)^(5/2)+5/6*c*(1/4*x*(d*x^2+c)^(3/2)+3/4*c*(1
/2*x*(d*x^2+c)^(1/2)+1/2*c/d^(1/2)*ln(x*d^(1/2)+(d*x^2+c)^(1/2))))))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.03, size = 887, normalized size = 6.12 \begin {gather*} \left [-\frac {{\left (5 \, a b c d - 2 \, a^{2} d^{2}\right )} \sqrt {d} x \log \left (-2 \, d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) - {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x \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 ) - 2 \, {\left (a b d^{2} x^{2} - 2 \, b^{2} c^{2}\right )} \sqrt {d x^{2} + c}}{4 \, a b^{2} x}, -\frac {2 \, {\left (5 \, a b c d - 2 \, a^{2} d^{2}\right )} \sqrt {-d} x \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x \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 ) - 2 \, {\left (a b d^{2} x^{2} - 2 \, b^{2} c^{2}\right )} \sqrt {d x^{2} + c}}{4 \, a b^{2} x}, -\frac {2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x \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 ) + {\left (5 \, a b c d - 2 \, a^{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 (a b d^{2} x^{2} - 2 \, b^{2} c^{2}\right )} \sqrt {d x^{2} + c}}{4 \, a b^{2} x}, -\frac {{\left (5 \, a b c d - 2 \, a^{2} d^{2}\right )} \sqrt {-d} x \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x \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 ) - {\left (a b d^{2} x^{2} - 2 \, b^{2} c^{2}\right )} \sqrt {d x^{2} + c}}{2 \, a b^{2} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*((5*a*b*c*d - 2*a^2*d^2)*sqrt(d)*x*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) - (b^2*c^2 - 2*a*b*c*
d + a^2*d^2)*x*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)) - 2*(a*b*d^2*x^2 - 2*b^2*c^2)*sqrt(d*x^2 + c))/(a*b^2*x), -1/4*(2*(5*a*b*c*d - 2*a^2*d^2)*sqrt(-d)*x*arc
tan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x*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)) - 2*(a*b*d^2*x^2 - 2*b^2*c^2)*sqrt(d*x^2 + c))/(a*b
^2*x), -1/4*(2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x*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)) + (5*a*b*c*d - 2*a^2*d^2)*sqrt(d)*x
*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) - 2*(a*b*d^2*x^2 - 2*b^2*c^2)*sqrt(d*x^2 + c))/(a*b^2*x), -1/
2*((5*a*b*c*d - 2*a^2*d^2)*sqrt(-d)*x*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x*s
qrt((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)) - (a*b*d^2*x^2 - 2*b^2*c^2)*sqrt(d*x^2 + c))/(a*b^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 {5}{2}}}{x^{2} \left (a + b x^{2}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:index.cc index_m i_lex_is_greater Error: Bad Argument Value

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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