3.956 \(\int \frac{x (a+b x^2)^{5/2}}{\sqrt{c+d x^2}} \, dx\)

Optimal. Leaf size=164 \[ \frac{5 \sqrt{a+b x^2} \sqrt{c+d x^2} (b c-a d)^2}{16 d^3}-\frac{5 \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2} (b c-a d)}{24 d^2}-\frac{5 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b} \sqrt{c+d x^2}}\right )}{16 \sqrt{b} d^{7/2}}+\frac{\left (a+b x^2\right )^{5/2} \sqrt{c+d x^2}}{6 d} \]

[Out]

(5*(b*c - a*d)^2*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(16*d^3) - (5*(b*c - a*d)*(a + b*x^2)^(3/2)*Sqrt[c + d*x^2])
/(24*d^2) + ((a + b*x^2)^(5/2)*Sqrt[c + d*x^2])/(6*d) - (5*(b*c - a*d)^3*ArcTanh[(Sqrt[d]*Sqrt[a + b*x^2])/(Sq
rt[b]*Sqrt[c + d*x^2])])/(16*Sqrt[b]*d^(7/2))

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Rubi [A]  time = 0.140962, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {444, 50, 63, 217, 206} \[ \frac{5 \sqrt{a+b x^2} \sqrt{c+d x^2} (b c-a d)^2}{16 d^3}-\frac{5 \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2} (b c-a d)}{24 d^2}-\frac{5 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b} \sqrt{c+d x^2}}\right )}{16 \sqrt{b} d^{7/2}}+\frac{\left (a+b x^2\right )^{5/2} \sqrt{c+d x^2}}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*(b*c - a*d)^2*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(16*d^3) - (5*(b*c - a*d)*(a + b*x^2)^(3/2)*Sqrt[c + d*x^2])
/(24*d^2) + ((a + b*x^2)^(5/2)*Sqrt[c + d*x^2])/(6*d) - (5*(b*c - a*d)^3*ArcTanh[(Sqrt[d]*Sqrt[a + b*x^2])/(Sq
rt[b]*Sqrt[c + d*x^2])])/(16*Sqrt[b]*d^(7/2))

Rule 444

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, 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 \left (a+b x^2\right )^{5/2}}{\sqrt{c+d x^2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^{5/2}}{\sqrt{c+d x}} \, dx,x,x^2\right )\\ &=\frac{\left (a+b x^2\right )^{5/2} \sqrt{c+d x^2}}{6 d}-\frac{(5 (b c-a d)) \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{\sqrt{c+d x}} \, dx,x,x^2\right )}{12 d}\\ &=-\frac{5 (b c-a d) \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{24 d^2}+\frac{\left (a+b x^2\right )^{5/2} \sqrt{c+d x^2}}{6 d}+\frac{\left (5 (b c-a d)^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{\sqrt{c+d x}} \, dx,x,x^2\right )}{16 d^2}\\ &=\frac{5 (b c-a d)^2 \sqrt{a+b x^2} \sqrt{c+d x^2}}{16 d^3}-\frac{5 (b c-a d) \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{24 d^2}+\frac{\left (a+b x^2\right )^{5/2} \sqrt{c+d x^2}}{6 d}-\frac{\left (5 (b c-a d)^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx,x,x^2\right )}{32 d^3}\\ &=\frac{5 (b c-a d)^2 \sqrt{a+b x^2} \sqrt{c+d x^2}}{16 d^3}-\frac{5 (b c-a d) \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{24 d^2}+\frac{\left (a+b x^2\right )^{5/2} \sqrt{c+d x^2}}{6 d}-\frac{\left (5 (b c-a d)^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{a d}{b}+\frac{d x^2}{b}}} \, dx,x,\sqrt{a+b x^2}\right )}{16 b d^3}\\ &=\frac{5 (b c-a d)^2 \sqrt{a+b x^2} \sqrt{c+d x^2}}{16 d^3}-\frac{5 (b c-a d) \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{24 d^2}+\frac{\left (a+b x^2\right )^{5/2} \sqrt{c+d x^2}}{6 d}-\frac{\left (5 (b c-a d)^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{d x^2}{b}} \, dx,x,\frac{\sqrt{a+b x^2}}{\sqrt{c+d x^2}}\right )}{16 b d^3}\\ &=\frac{5 (b c-a d)^2 \sqrt{a+b x^2} \sqrt{c+d x^2}}{16 d^3}-\frac{5 (b c-a d) \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{24 d^2}+\frac{\left (a+b x^2\right )^{5/2} \sqrt{c+d x^2}}{6 d}-\frac{5 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b} \sqrt{c+d x^2}}\right )}{16 \sqrt{b} d^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.514228, size = 164, normalized size = 1. \[ \frac{\sqrt{d} \sqrt{a+b x^2} \left (c+d x^2\right ) \left (33 a^2 d^2+2 a b d \left (13 d x^2-20 c\right )+b^2 \left (15 c^2-10 c d x^2+8 d^2 x^4\right )\right )-\frac{15 (b c-a d)^{7/2} \sqrt{\frac{b \left (c+d x^2\right )}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b c-a d}}\right )}{b}}{48 d^{7/2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d]*Sqrt[a + b*x^2]*(c + d*x^2)*(33*a^2*d^2 + 2*a*b*d*(-20*c + 13*d*x^2) + b^2*(15*c^2 - 10*c*d*x^2 + 8*d
^2*x^4)) - (15*(b*c - a*d)^(7/2)*Sqrt[(b*(c + d*x^2))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x^2])/Sqrt[b*c
- a*d]])/b)/(48*d^(7/2)*Sqrt[c + d*x^2])

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Maple [B]  time = 0.014, size = 529, normalized size = 3.2 \begin{align*}{\frac{1}{96\,{d}^{3}}\sqrt{b{x}^{2}+a}\sqrt{d{x}^{2}+c} \left ( 16\,{x}^{4}{b}^{2}{d}^{2}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+52\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}{x}^{2}ab{d}^{2}\sqrt{bd}-20\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}{x}^{2}c{b}^{2}d\sqrt{bd}+15\,{d}^{3}\ln \left ( 1/2\,{\frac{2\,d{x}^{2}b+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{3}-45\,\ln \left ( 1/2\,{\frac{2\,d{x}^{2}b+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{2}cb{d}^{2}+45\,\ln \left ( 1/2\,{\frac{2\,d{x}^{2}b+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){c}^{2}a{b}^{2}d-15\,{b}^{3}\ln \left ( 1/2\,{\frac{2\,d{x}^{2}b+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){c}^{3}+66\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}{a}^{2}{d}^{2}\sqrt{bd}-80\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}acbd\sqrt{bd}+30\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}{c}^{2}{b}^{2}\sqrt{bd} \right ){\frac{1}{\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}}}{\frac{1}{\sqrt{bd}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(b*x^2+a)^(5/2)/(d*x^2+c)^(1/2),x)

[Out]

1/96*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)*(16*x^4*b^2*d^2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)+52*(b*d*x
^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*x^2*a*b*d^2*(b*d)^(1/2)-20*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*x^2*c*b^2*d*(b*d)
^(1/2)+15*d^3*ln(1/2*(2*d*x^2*b+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^3-45
*ln(1/2*(2*d*x^2*b+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^2*c*b*d^2+45*ln(1
/2*(2*d*x^2*b+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*c^2*a*b^2*d-15*b^3*ln(1/
2*(2*d*x^2*b+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*c^3+66*(b*d*x^4+a*d*x^2+b
*c*x^2+a*c)^(1/2)*a^2*d^2*(b*d)^(1/2)-80*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*a*c*b*d*(b*d)^(1/2)+30*(b*d*x^4+a
*d*x^2+b*c*x^2+a*c)^(1/2)*c^2*b^2*(b*d)^(1/2))/d^3/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)/(b*d)^(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*(b*x^2+a)^(5/2)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.98681, size = 971, normalized size = 5.92 \begin{align*} \left [-\frac{15 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{b d} \log \left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \,{\left (b^{2} c d + a b d^{2}\right )} x^{2} + 4 \,{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt{b x^{2} + a} \sqrt{d x^{2} + c} \sqrt{b d}\right ) - 4 \,{\left (8 \, b^{3} d^{3} x^{4} + 15 \, b^{3} c^{2} d - 40 \, a b^{2} c d^{2} + 33 \, a^{2} b d^{3} - 2 \,{\left (5 \, b^{3} c d^{2} - 13 \, a b^{2} d^{3}\right )} x^{2}\right )} \sqrt{b x^{2} + a} \sqrt{d x^{2} + c}}{192 \, b d^{4}}, \frac{15 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{-b d} \arctan \left (\frac{{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt{b x^{2} + a} \sqrt{d x^{2} + c} \sqrt{-b d}}{2 \,{\left (b^{2} d^{2} x^{4} + a b c d +{\left (b^{2} c d + a b d^{2}\right )} x^{2}\right )}}\right ) + 2 \,{\left (8 \, b^{3} d^{3} x^{4} + 15 \, b^{3} c^{2} d - 40 \, a b^{2} c d^{2} + 33 \, a^{2} b d^{3} - 2 \,{\left (5 \, b^{3} c d^{2} - 13 \, a b^{2} d^{3}\right )} x^{2}\right )} \sqrt{b x^{2} + a} \sqrt{d x^{2} + c}}{96 \, b d^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x^2+a)^(5/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

[-1/192*(15*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(b*d)*log(8*b^2*d^2*x^4 + b^2*c^2 + 6*a*b*
c*d + a^2*d^2 + 8*(b^2*c*d + a*b*d^2)*x^2 + 4*(2*b*d*x^2 + b*c + a*d)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(b*d
)) - 4*(8*b^3*d^3*x^4 + 15*b^3*c^2*d - 40*a*b^2*c*d^2 + 33*a^2*b*d^3 - 2*(5*b^3*c*d^2 - 13*a*b^2*d^3)*x^2)*sqr
t(b*x^2 + a)*sqrt(d*x^2 + c))/(b*d^4), 1/96*(15*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(-b*d)
*arctan(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-b*d)/(b^2*d^2*x^4 + a*b*c*d + (b^2*c
*d + a*b*d^2)*x^2)) + 2*(8*b^3*d^3*x^4 + 15*b^3*c^2*d - 40*a*b^2*c*d^2 + 33*a^2*b*d^3 - 2*(5*b^3*c*d^2 - 13*a*
b^2*d^3)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(b*d^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \left (a + b x^{2}\right )^{\frac{5}{2}}}{\sqrt{c + d x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x**2+a)**(5/2)/(d*x**2+c)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.23526, size = 284, normalized size = 1.73 \begin{align*} \frac{{\left (\sqrt{b^{2} c +{\left (b x^{2} + a\right )} b d - a b d} \sqrt{b x^{2} + a}{\left (2 \,{\left (b x^{2} + a\right )}{\left (\frac{4 \,{\left (b x^{2} + a\right )}}{b d} - \frac{5 \,{\left (b c d^{3} - a d^{4}\right )}}{b d^{5}}\right )} + \frac{15 \,{\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )}}{b d^{5}}\right )} + \frac{15 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left ({\left | -\sqrt{b x^{2} + a} \sqrt{b d} + \sqrt{b^{2} c +{\left (b x^{2} + a\right )} b d - a b d} \right |}\right )}{\sqrt{b d} d^{3}}\right )} b}{48 \,{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x^2+a)^(5/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

1/48*(sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d)*sqrt(b*x^2 + a)*(2*(b*x^2 + a)*(4*(b*x^2 + a)/(b*d) - 5*(b*c*d^3 -
a*d^4)/(b*d^5)) + 15*(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)/(b*d^5)) + 15*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c
*d^2 - a^3*d^3)*log(abs(-sqrt(b*x^2 + a)*sqrt(b*d) + sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d)))/(sqrt(b*d)*d^3))*
b/abs(b)