3.8.55 \(\int \frac {(c+d x^2)^{5/2}}{x^2 (a+b x^2)^2} \, dx\) [755]

Optimal. Leaf size=168 \[ -\frac {c (3 b c-a d) \sqrt {c+d x^2}}{2 a^2 b x}+\frac {(b c-a d) \left (c+d x^2\right )^{3/2}}{2 a b x \left (a+b x^2\right )}-\frac {(b c-a d)^{3/2} (3 b c+2 a d) \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{5/2} b^2}+\frac {d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{b^2} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 168, 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 = {479, 594, 537, 223, 212, 385, 211} \begin {gather*} -\frac {(b c-a d)^{3/2} (2 a d+3 b c) \text {ArcTan}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{5/2} b^2}-\frac {c \sqrt {c+d x^2} (3 b c-a d)}{2 a^2 b x}+\frac {\left (c+d x^2\right )^{3/2} (b c-a d)}{2 a b x \left (a+b x^2\right )}+\frac {d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(c*(3*b*c - a*d)*Sqrt[c + d*x^2])/(a^2*b*x) + ((b*c - a*d)*(c + d*x^2)^(3/2))/(2*a*b*x*(a + b*x^2)) - ((b
*c - a*d)^(3/2)*(3*b*c + 2*a*d)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(2*a^(5/2)*b^2) + (d^(5
/2)*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^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 479

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)),
Int[(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 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 594

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/(a*g*(m + 1))), x] - Dist[1/(a*g^n*(m + 1
)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f)*(m + 1) + e*n*(b*c*(p + 1) + a*d*q)
+ d*((b*e - a*f)*(m + 1) + b*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && IGtQ[n
, 0] && GtQ[q, 0] && LtQ[m, -1] &&  !(EqQ[q, 1] && SimplerQ[e + f*x^n, c + d*x^n])

Rubi steps

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

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Mathematica [A]
time = 0.60, size = 192, normalized size = 1.14 \begin {gather*} \frac {-\frac {b \sqrt {c+d x^2} \left (3 b^2 c^2 x^2+a^2 d^2 x^2+2 a b c \left (c-d x^2\right )\right )}{a^2 x \left (a+b x^2\right )}+\frac {\sqrt {b c-a d} \left (3 b^2 c^2-a b c d-2 a^2 d^2\right ) \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^{5/2}}-2 d^{5/2} \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)^2),x]

[Out]

(-((b*Sqrt[c + d*x^2]*(3*b^2*c^2*x^2 + a^2*d^2*x^2 + 2*a*b*c*(c - d*x^2)))/(a^2*x*(a + b*x^2))) + (Sqrt[b*c -
a*d]*(3*b^2*c^2 - a*b*c*d - 2*a^2*d^2)*ArcTan[(a*Sqrt[d] + b*x*(Sqrt[d]*x - Sqrt[c + d*x^2]))/(Sqrt[a]*Sqrt[b*
c - a*d])])/a^(5/2) - 2*d^(5/2)*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. \(5380\) vs. \(2(142)=284\).
time = 0.13, size = 5381, normalized size = 32.03

method	result	size
risch	\(\text {Expression too large to display}\)	\(3403\)
default	\(\text {Expression too large to display}\)	\(5381\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 1.72, size = 1184, normalized size = 7.05 \begin {gather*} \left [\frac {4 \, {\left (a^{2} b d^{2} x^{3} + a^{3} d^{2} x\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) - {\left ({\left (3 \, b^{3} c^{2} - a b^{2} c d - 2 \, a^{2} b d^{2}\right )} x^{3} + {\left (3 \, a b^{2} c^{2} - a^{2} b c d - 2 \, a^{3} d^{2}\right )} x\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 (2 \, a b^{2} c^{2} + {\left (3 \, b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{8 \, {\left (a^{2} b^{3} x^{3} + a^{3} b^{2} x\right )}}, -\frac {8 \, {\left (a^{2} b d^{2} x^{3} + a^{3} d^{2} x\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left ({\left (3 \, b^{3} c^{2} - a b^{2} c d - 2 \, a^{2} b d^{2}\right )} x^{3} + {\left (3 \, a b^{2} c^{2} - a^{2} b c d - 2 \, a^{3} d^{2}\right )} x\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 (2 \, a b^{2} c^{2} + {\left (3 \, b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{8 \, {\left (a^{2} b^{3} x^{3} + a^{3} b^{2} x\right )}}, -\frac {{\left ({\left (3 \, b^{3} c^{2} - a b^{2} c d - 2 \, a^{2} b d^{2}\right )} x^{3} + {\left (3 \, a b^{2} c^{2} - a^{2} b c d - 2 \, a^{3} d^{2}\right )} x\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 (a^{2} b d^{2} x^{3} + a^{3} d^{2} x\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (2 \, a b^{2} c^{2} + {\left (3 \, b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{4 \, {\left (a^{2} b^{3} x^{3} + a^{3} b^{2} x\right )}}, -\frac {4 \, {\left (a^{2} b d^{2} x^{3} + a^{3} d^{2} x\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left ({\left (3 \, b^{3} c^{2} - a b^{2} c d - 2 \, a^{2} b d^{2}\right )} x^{3} + {\left (3 \, a b^{2} c^{2} - a^{2} b c d - 2 \, a^{3} d^{2}\right )} x\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 (2 \, a b^{2} c^{2} + {\left (3 \, b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{4 \, {\left (a^{2} b^{3} x^{3} + a^{3} b^{2} x\right )}}\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)^2,x, algorithm="fricas")

[Out]

[1/8*(4*(a^2*b*d^2*x^3 + a^3*d^2*x)*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) - ((3*b^3*c^2 - a*
b^2*c*d - 2*a^2*b*d^2)*x^3 + (3*a*b^2*c^2 - a^2*b*c*d - 2*a^3*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)) - 4*(2*a*b^2*c^2 + (3*b^3*c^2 - 2*a*b^2*c*d + a^2
*b*d^2)*x^2)*sqrt(d*x^2 + c))/(a^2*b^3*x^3 + a^3*b^2*x), -1/8*(8*(a^2*b*d^2*x^3 + a^3*d^2*x)*sqrt(-d)*arctan(s
qrt(-d)*x/sqrt(d*x^2 + c)) + ((3*b^3*c^2 - a*b^2*c*d - 2*a^2*b*d^2)*x^3 + (3*a*b^2*c^2 - a^2*b*c*d - 2*a^3*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)) + 4*
(2*a*b^2*c^2 + (3*b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)*x^2)*sqrt(d*x^2 + c))/(a^2*b^3*x^3 + a^3*b^2*x), -1/4*(((
3*b^3*c^2 - a*b^2*c*d - 2*a^2*b*d^2)*x^3 + (3*a*b^2*c^2 - a^2*b*c*d - 2*a^3*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))
- 2*(a^2*b*d^2*x^3 + a^3*d^2*x)*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*(2*a*b^2*c^2 + (3*
b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)*x^2)*sqrt(d*x^2 + c))/(a^2*b^3*x^3 + a^3*b^2*x), -1/4*(4*(a^2*b*d^2*x^3 + a
^3*d^2*x)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + ((3*b^3*c^2 - a*b^2*c*d - 2*a^2*b*d^2)*x^3 + (3*a*b^2*
c^2 - a^2*b*c*d - 2*a^3*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)) + 2*(2*a*b^2*c^2 + (3*b^3*c^2 - 2*a*b^2*c*d + a^2*b*
d^2)*x^2)*sqrt(d*x^2 + c))/(a^2*b^3*x^3 + a^3*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 )^{2}}\, 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)**2,x)

[Out]

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

[Out]

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

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