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

Optimal. Leaf size=197 \[ -\frac{x^3 \sqrt{c+d x^2} \left (24 a^2 d^2-60 a b c d+35 b^2 c^2\right )}{24 c d^3}+\frac{x \sqrt{c+d x^2} \left (24 a^2 d^2-60 a b c d+35 b^2 c^2\right )}{16 d^4}-\frac{c \left (24 a^2 d^2-60 a b c d+35 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{16 d^{9/2}}+\frac{x^5 (b c-a d)^2}{c d^2 \sqrt{c+d x^2}}+\frac{b^2 x^5 \sqrt{c+d x^2}}{6 d^2} \]

[Out]

((b*c - a*d)^2*x^5)/(c*d^2*Sqrt[c + d*x^2]) + ((35*b^2*c^2 - 60*a*b*c*d + 24*a^2*d^2)*x*Sqrt[c + d*x^2])/(16*d
^4) - ((35*b^2*c^2 - 60*a*b*c*d + 24*a^2*d^2)*x^3*Sqrt[c + d*x^2])/(24*c*d^3) + (b^2*x^5*Sqrt[c + d*x^2])/(6*d
^2) - (c*(35*b^2*c^2 - 60*a*b*c*d + 24*a^2*d^2)*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(16*d^(9/2))

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Rubi [A]  time = 0.149479, antiderivative size = 197, normalized size of antiderivative = 1., 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 = {463, 459, 321, 217, 206} \[ -\frac{x^3 \sqrt{c+d x^2} \left (24 a^2 d^2-60 a b c d+35 b^2 c^2\right )}{24 c d^3}+\frac{x \sqrt{c+d x^2} \left (24 a^2 d^2-60 a b c d+35 b^2 c^2\right )}{16 d^4}-\frac{c \left (24 a^2 d^2-60 a b c d+35 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{16 d^{9/2}}+\frac{x^5 (b c-a d)^2}{c d^2 \sqrt{c+d x^2}}+\frac{b^2 x^5 \sqrt{c+d x^2}}{6 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)^2*x^5)/(c*d^2*Sqrt[c + d*x^2]) + ((35*b^2*c^2 - 60*a*b*c*d + 24*a^2*d^2)*x*Sqrt[c + d*x^2])/(16*d
^4) - ((35*b^2*c^2 - 60*a*b*c*d + 24*a^2*d^2)*x^3*Sqrt[c + d*x^2])/(24*c*d^3) + (b^2*x^5*Sqrt[c + d*x^2])/(6*d
^2) - (c*(35*b^2*c^2 - 60*a*b*c*d + 24*a^2*d^2)*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(16*d^(9/2))

Rule 463

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

Rule 459

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

Rule 321

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

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 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])

Rubi steps

\begin{align*} \int \frac{x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx &=\frac{(b c-a d)^2 x^5}{c d^2 \sqrt{c+d x^2}}-\frac{\int \frac{x^4 \left (-a^2 d^2+5 (b c-a d)^2-b^2 c d x^2\right )}{\sqrt{c+d x^2}} \, dx}{c d^2}\\ &=\frac{(b c-a d)^2 x^5}{c d^2 \sqrt{c+d x^2}}+\frac{b^2 x^5 \sqrt{c+d x^2}}{6 d^2}-\frac{\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) \int \frac{x^4}{\sqrt{c+d x^2}} \, dx}{6 c d^2}\\ &=\frac{(b c-a d)^2 x^5}{c d^2 \sqrt{c+d x^2}}-\frac{\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) x^3 \sqrt{c+d x^2}}{24 c d^3}+\frac{b^2 x^5 \sqrt{c+d x^2}}{6 d^2}+\frac{\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) \int \frac{x^2}{\sqrt{c+d x^2}} \, dx}{8 d^3}\\ &=\frac{(b c-a d)^2 x^5}{c d^2 \sqrt{c+d x^2}}+\frac{\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) x \sqrt{c+d x^2}}{16 d^4}-\frac{\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) x^3 \sqrt{c+d x^2}}{24 c d^3}+\frac{b^2 x^5 \sqrt{c+d x^2}}{6 d^2}-\frac{\left (c \left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right )\right ) \int \frac{1}{\sqrt{c+d x^2}} \, dx}{16 d^4}\\ &=\frac{(b c-a d)^2 x^5}{c d^2 \sqrt{c+d x^2}}+\frac{\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) x \sqrt{c+d x^2}}{16 d^4}-\frac{\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) x^3 \sqrt{c+d x^2}}{24 c d^3}+\frac{b^2 x^5 \sqrt{c+d x^2}}{6 d^2}-\frac{\left (c \left (35 b^2 c^2-60 a b c d+24 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 )}{16 d^4}\\ &=\frac{(b c-a d)^2 x^5}{c d^2 \sqrt{c+d x^2}}+\frac{\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) x \sqrt{c+d x^2}}{16 d^4}-\frac{\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) x^3 \sqrt{c+d x^2}}{24 c d^3}+\frac{b^2 x^5 \sqrt{c+d x^2}}{6 d^2}-\frac{c \left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{16 d^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.145878, size = 158, normalized size = 0.8 \[ \sqrt{c+d x^2} \left (\frac{x \left (8 a^2 d^2-28 a b c d+19 b^2 c^2\right )}{16 d^4}-\frac{b x^3 (11 b c-12 a d)}{24 d^3}+\frac{c x (b c-a d)^2}{d^4 \left (c+d x^2\right )}+\frac{b^2 x^5}{6 d^2}\right )-\frac{c \left (24 a^2 d^2-60 a b c d+35 b^2 c^2\right ) \log \left (\sqrt{d} \sqrt{c+d x^2}+d x\right )}{16 d^{9/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[c + d*x^2]*(((19*b^2*c^2 - 28*a*b*c*d + 8*a^2*d^2)*x)/(16*d^4) - (b*(11*b*c - 12*a*d)*x^3)/(24*d^3) + (b^
2*x^5)/(6*d^2) + (c*(b*c - a*d)^2*x)/(d^4*(c + d*x^2))) - (c*(35*b^2*c^2 - 60*a*b*c*d + 24*a^2*d^2)*Log[d*x +
Sqrt[d]*Sqrt[c + d*x^2]])/(16*d^(9/2))

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Maple [A]  time = 0.015, size = 263, normalized size = 1.3 \begin{align*}{\frac{{b}^{2}{x}^{7}}{6\,d}{\frac{1}{\sqrt{d{x}^{2}+c}}}}-{\frac{7\,{b}^{2}c{x}^{5}}{24\,{d}^{2}}{\frac{1}{\sqrt{d{x}^{2}+c}}}}+{\frac{35\,{x}^{3}{b}^{2}{c}^{2}}{48\,{d}^{3}}{\frac{1}{\sqrt{d{x}^{2}+c}}}}+{\frac{35\,{b}^{2}{c}^{3}x}{16\,{d}^{4}}{\frac{1}{\sqrt{d{x}^{2}+c}}}}-{\frac{35\,{b}^{2}{c}^{3}}{16}\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ){d}^{-{\frac{9}{2}}}}+{\frac{ab{x}^{5}}{2\,d}{\frac{1}{\sqrt{d{x}^{2}+c}}}}-{\frac{5\,abc{x}^{3}}{4\,{d}^{2}}{\frac{1}{\sqrt{d{x}^{2}+c}}}}-{\frac{15\,ab{c}^{2}x}{4\,{d}^{3}}{\frac{1}{\sqrt{d{x}^{2}+c}}}}+{\frac{15\,ab{c}^{2}}{4}\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ){d}^{-{\frac{7}{2}}}}+{\frac{{a}^{2}{x}^{3}}{2\,d}{\frac{1}{\sqrt{d{x}^{2}+c}}}}+{\frac{3\,{a}^{2}cx}{2\,{d}^{2}}{\frac{1}{\sqrt{d{x}^{2}+c}}}}-{\frac{3\,{a}^{2}c}{2}\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ){d}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b^2*x^7/d/(d*x^2+c)^(1/2)-7/24*b^2*c/d^2*x^5/(d*x^2+c)^(1/2)+35/48*b^2*c^2/d^3*x^3/(d*x^2+c)^(1/2)+35/16*b
^2*c^3/d^4*x/(d*x^2+c)^(1/2)-35/16*b^2*c^3/d^(9/2)*ln(x*d^(1/2)+(d*x^2+c)^(1/2))+1/2*a*b*x^5/d/(d*x^2+c)^(1/2)
-5/4*a*b*c/d^2*x^3/(d*x^2+c)^(1/2)-15/4*a*b*c^2/d^3*x/(d*x^2+c)^(1/2)+15/4*a*b*c^2/d^(7/2)*ln(x*d^(1/2)+(d*x^2
+c)^(1/2))+1/2*a^2*x^3/d/(d*x^2+c)^(1/2)+3/2*a^2*c/d^2*x/(d*x^2+c)^(1/2)-3/2*a^2*c/d^(5/2)*ln(x*d^(1/2)+(d*x^2
+c)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.69173, size = 950, normalized size = 4.82 \begin{align*} \left [\frac{3 \,{\left (35 \, b^{2} c^{4} - 60 \, a b c^{3} d + 24 \, a^{2} c^{2} d^{2} +{\left (35 \, b^{2} c^{3} d - 60 \, a b c^{2} d^{2} + 24 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt{d} \log \left (-2 \, d x^{2} + 2 \, \sqrt{d x^{2} + c} \sqrt{d} x - c\right ) + 2 \,{\left (8 \, b^{2} d^{4} x^{7} - 2 \,{\left (7 \, b^{2} c d^{3} - 12 \, a b d^{4}\right )} x^{5} +{\left (35 \, b^{2} c^{2} d^{2} - 60 \, a b c d^{3} + 24 \, a^{2} d^{4}\right )} x^{3} + 3 \,{\left (35 \, b^{2} c^{3} d - 60 \, a b c^{2} d^{2} + 24 \, a^{2} c d^{3}\right )} x\right )} \sqrt{d x^{2} + c}}{96 \,{\left (d^{6} x^{2} + c d^{5}\right )}}, \frac{3 \,{\left (35 \, b^{2} c^{4} - 60 \, a b c^{3} d + 24 \, a^{2} c^{2} d^{2} +{\left (35 \, b^{2} c^{3} d - 60 \, a b c^{2} d^{2} + 24 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x}{\sqrt{d x^{2} + c}}\right ) +{\left (8 \, b^{2} d^{4} x^{7} - 2 \,{\left (7 \, b^{2} c d^{3} - 12 \, a b d^{4}\right )} x^{5} +{\left (35 \, b^{2} c^{2} d^{2} - 60 \, a b c d^{3} + 24 \, a^{2} d^{4}\right )} x^{3} + 3 \,{\left (35 \, b^{2} c^{3} d - 60 \, a b c^{2} d^{2} + 24 \, a^{2} c d^{3}\right )} x\right )} \sqrt{d x^{2} + c}}{48 \,{\left (d^{6} x^{2} + c d^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(3*(35*b^2*c^4 - 60*a*b*c^3*d + 24*a^2*c^2*d^2 + (35*b^2*c^3*d - 60*a*b*c^2*d^2 + 24*a^2*c*d^3)*x^2)*sqr
t(d)*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*(8*b^2*d^4*x^7 - 2*(7*b^2*c*d^3 - 12*a*b*d^4)*x^5 + (
35*b^2*c^2*d^2 - 60*a*b*c*d^3 + 24*a^2*d^4)*x^3 + 3*(35*b^2*c^3*d - 60*a*b*c^2*d^2 + 24*a^2*c*d^3)*x)*sqrt(d*x
^2 + c))/(d^6*x^2 + c*d^5), 1/48*(3*(35*b^2*c^4 - 60*a*b*c^3*d + 24*a^2*c^2*d^2 + (35*b^2*c^3*d - 60*a*b*c^2*d
^2 + 24*a^2*c*d^3)*x^2)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + (8*b^2*d^4*x^7 - 2*(7*b^2*c*d^3 - 12*a*b
*d^4)*x^5 + (35*b^2*c^2*d^2 - 60*a*b*c*d^3 + 24*a^2*d^4)*x^3 + 3*(35*b^2*c^3*d - 60*a*b*c^2*d^2 + 24*a^2*c*d^3
)*x)*sqrt(d*x^2 + c))/(d^6*x^2 + c*d^5)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.13322, size = 236, normalized size = 1.2 \begin{align*} \frac{{\left ({\left (2 \,{\left (\frac{4 \, b^{2} x^{2}}{d} - \frac{7 \, b^{2} c d^{5} - 12 \, a b d^{6}}{d^{7}}\right )} x^{2} + \frac{35 \, b^{2} c^{2} d^{4} - 60 \, a b c d^{5} + 24 \, a^{2} d^{6}}{d^{7}}\right )} x^{2} + \frac{3 \,{\left (35 \, b^{2} c^{3} d^{3} - 60 \, a b c^{2} d^{4} + 24 \, a^{2} c d^{5}\right )}}{d^{7}}\right )} x}{48 \, \sqrt{d x^{2} + c}} + \frac{{\left (35 \, b^{2} c^{3} - 60 \, a b c^{2} d + 24 \, a^{2} c d^{2}\right )} \log \left ({\left | -\sqrt{d} x + \sqrt{d x^{2} + c} \right |}\right )}{16 \, d^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*((2*(4*b^2*x^2/d - (7*b^2*c*d^5 - 12*a*b*d^6)/d^7)*x^2 + (35*b^2*c^2*d^4 - 60*a*b*c*d^5 + 24*a^2*d^6)/d^7
)*x^2 + 3*(35*b^2*c^3*d^3 - 60*a*b*c^2*d^4 + 24*a^2*c*d^5)/d^7)*x/sqrt(d*x^2 + c) + 1/16*(35*b^2*c^3 - 60*a*b*
c^2*d + 24*a^2*c*d^2)*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))/d^(9/2)