3.987 \(\int \frac{x^3}{(a+b x^2)^{7/2} \sqrt{c+d x^2}} \, dx\)

Optimal. Leaf size=138 \[ \frac{a \sqrt{c+d x^2}}{5 b \left (a+b x^2\right )^{5/2} (b c-a d)}+\frac{2 d \sqrt{c+d x^2} (5 b c-a d)}{15 b \sqrt{a+b x^2} (b c-a d)^3}-\frac{\sqrt{c+d x^2} (5 b c-a d)}{15 b \left (a+b x^2\right )^{3/2} (b c-a d)^2} \]

[Out]

(a*Sqrt[c + d*x^2])/(5*b*(b*c - a*d)*(a + b*x^2)^(5/2)) - ((5*b*c - a*d)*Sqrt[c + d*x^2])/(15*b*(b*c - a*d)^2*
(a + b*x^2)^(3/2)) + (2*d*(5*b*c - a*d)*Sqrt[c + d*x^2])/(15*b*(b*c - a*d)^3*Sqrt[a + b*x^2])

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Rubi [A]  time = 0.100783, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {446, 78, 45, 37} \[ \frac{a \sqrt{c+d x^2}}{5 b \left (a+b x^2\right )^{5/2} (b c-a d)}+\frac{2 d \sqrt{c+d x^2} (5 b c-a d)}{15 b \sqrt{a+b x^2} (b c-a d)^3}-\frac{\sqrt{c+d x^2} (5 b c-a d)}{15 b \left (a+b x^2\right )^{3/2} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Int[x^3/((a + b*x^2)^(7/2)*Sqrt[c + d*x^2]),x]

[Out]

(a*Sqrt[c + d*x^2])/(5*b*(b*c - a*d)*(a + b*x^2)^(5/2)) - ((5*b*c - a*d)*Sqrt[c + d*x^2])/(15*b*(b*c - a*d)^2*
(a + b*x^2)^(3/2)) + (2*d*(5*b*c - a*d)*Sqrt[c + d*x^2])/(15*b*(b*c - a*d)^3*Sqrt[a + b*x^2])

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]

Rule 78

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

Rule 45

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

Rule 37

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

Rubi steps

\begin{align*} \int \frac{x^3}{\left (a+b x^2\right )^{7/2} \sqrt{c+d x^2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x}{(a+b x)^{7/2} \sqrt{c+d x}} \, dx,x,x^2\right )\\ &=\frac{a \sqrt{c+d x^2}}{5 b (b c-a d) \left (a+b x^2\right )^{5/2}}+\frac{(5 b c-a d) \operatorname{Subst}\left (\int \frac{1}{(a+b x)^{5/2} \sqrt{c+d x}} \, dx,x,x^2\right )}{10 b (b c-a d)}\\ &=\frac{a \sqrt{c+d x^2}}{5 b (b c-a d) \left (a+b x^2\right )^{5/2}}-\frac{(5 b c-a d) \sqrt{c+d x^2}}{15 b (b c-a d)^2 \left (a+b x^2\right )^{3/2}}-\frac{(d (5 b c-a d)) \operatorname{Subst}\left (\int \frac{1}{(a+b x)^{3/2} \sqrt{c+d x}} \, dx,x,x^2\right )}{15 b (b c-a d)^2}\\ &=\frac{a \sqrt{c+d x^2}}{5 b (b c-a d) \left (a+b x^2\right )^{5/2}}-\frac{(5 b c-a d) \sqrt{c+d x^2}}{15 b (b c-a d)^2 \left (a+b x^2\right )^{3/2}}+\frac{2 d (5 b c-a d) \sqrt{c+d x^2}}{15 b (b c-a d)^3 \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0353247, size = 91, normalized size = 0.66 \[ \frac{\sqrt{c+d x^2} \left (-5 a^2 d \left (d x^2-2 c\right )-2 a b \left (c^2-13 c d x^2+d^2 x^4\right )-5 b^2 c x^2 \left (c-2 d x^2\right )\right )}{15 \left (a+b x^2\right )^{5/2} (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/((a + b*x^2)^(7/2)*Sqrt[c + d*x^2]),x]

[Out]

(Sqrt[c + d*x^2]*(-5*b^2*c*x^2*(c - 2*d*x^2) - 5*a^2*d*(-2*c + d*x^2) - 2*a*b*(c^2 - 13*c*d*x^2 + d^2*x^4)))/(
15*(b*c - a*d)^3*(a + b*x^2)^(5/2))

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Maple [A]  time = 0.007, size = 125, normalized size = 0.9 \begin{align*} -{\frac{-2\,ab{d}^{2}{x}^{4}+10\,{b}^{2}cd{x}^{4}-5\,{a}^{2}{d}^{2}{x}^{2}+26\,abcd{x}^{2}-5\,{b}^{2}{c}^{2}{x}^{2}+10\,{a}^{2}cd-2\,ab{c}^{2}}{15\,{a}^{3}{d}^{3}-45\,{a}^{2}c{d}^{2}b+45\,a{c}^{2}d{b}^{2}-15\,{c}^{3}{b}^{3}}\sqrt{d{x}^{2}+c} \left ( b{x}^{2}+a \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(b*x^2+a)^(7/2)/(d*x^2+c)^(1/2),x)

[Out]

-1/15*(d*x^2+c)^(1/2)*(-2*a*b*d^2*x^4+10*b^2*c*d*x^4-5*a^2*d^2*x^2+26*a*b*c*d*x^2-5*b^2*c^2*x^2+10*a^2*c*d-2*a
*b*c^2)/(b*x^2+a)^(5/2)/(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)

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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^3/(b*x^2+a)^(7/2)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 4.32376, size = 537, normalized size = 3.89 \begin{align*} \frac{{\left (2 \,{\left (5 \, b^{2} c d - a b d^{2}\right )} x^{4} - 2 \, a b c^{2} + 10 \, a^{2} c d -{\left (5 \, b^{2} c^{2} - 26 \, a b c d + 5 \, a^{2} d^{2}\right )} x^{2}\right )} \sqrt{b x^{2} + a} \sqrt{d x^{2} + c}}{15 \,{\left (a^{3} b^{3} c^{3} - 3 \, a^{4} b^{2} c^{2} d + 3 \, a^{5} b c d^{2} - a^{6} d^{3} +{\left (b^{6} c^{3} - 3 \, a b^{5} c^{2} d + 3 \, a^{2} b^{4} c d^{2} - a^{3} b^{3} d^{3}\right )} x^{6} + 3 \,{\left (a b^{5} c^{3} - 3 \, a^{2} b^{4} c^{2} d + 3 \, a^{3} b^{3} c d^{2} - a^{4} b^{2} d^{3}\right )} x^{4} + 3 \,{\left (a^{2} b^{4} c^{3} - 3 \, a^{3} b^{3} c^{2} d + 3 \, a^{4} b^{2} c d^{2} - a^{5} b d^{3}\right )} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^2+a)^(7/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

1/15*(2*(5*b^2*c*d - a*b*d^2)*x^4 - 2*a*b*c^2 + 10*a^2*c*d - (5*b^2*c^2 - 26*a*b*c*d + 5*a^2*d^2)*x^2)*sqrt(b*
x^2 + a)*sqrt(d*x^2 + c)/(a^3*b^3*c^3 - 3*a^4*b^2*c^2*d + 3*a^5*b*c*d^2 - a^6*d^3 + (b^6*c^3 - 3*a*b^5*c^2*d +
3*a^2*b^4*c*d^2 - a^3*b^3*d^3)*x^6 + 3*(a*b^5*c^3 - 3*a^2*b^4*c^2*d + 3*a^3*b^3*c*d^2 - a^4*b^2*d^3)*x^4 + 3*
(a^2*b^4*c^3 - 3*a^3*b^3*c^2*d + 3*a^4*b^2*c*d^2 - a^5*b*d^3)*x^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\left (a + b x^{2}\right )^{\frac{7}{2}} \sqrt{c + d x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(b*x**2+a)**(7/2)/(d*x**2+c)**(1/2),x)

[Out]

Integral(x**3/((a + b*x**2)**(7/2)*sqrt(c + d*x**2)), x)

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Giac [B]  time = 1.31265, size = 637, normalized size = 4.62 \begin{align*} \frac{4 \,{\left (5 \, \sqrt{b d} b^{8} c^{3} d - 11 \, \sqrt{b d} a b^{7} c^{2} d^{2} + 7 \, \sqrt{b d} a^{2} b^{6} c d^{3} - \sqrt{b d} a^{3} b^{5} d^{4} - 25 \, \sqrt{b d}{\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 )}^{2} b^{6} c^{2} d + 30 \, \sqrt{b d}{\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 )}^{2} a b^{5} c d^{2} - 5 \, \sqrt{b d}{\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 )}^{2} a^{2} b^{4} d^{3} + 35 \, \sqrt{b d}{\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 )}^{4} b^{4} c d + 5 \, \sqrt{b d}{\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 )}^{4} a b^{3} d^{2} - 15 \, \sqrt{b d}{\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 )}^{6} b^{2} d\right )}}{15 \,{\left (b^{2} c - a b d -{\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 )}^{2}\right )}^{5} b{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^2+a)^(7/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

4/15*(5*sqrt(b*d)*b^8*c^3*d - 11*sqrt(b*d)*a*b^7*c^2*d^2 + 7*sqrt(b*d)*a^2*b^6*c*d^3 - sqrt(b*d)*a^3*b^5*d^4 -
25*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*b^6*c^2*d + 30*sqrt(b*d)*(
sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*a*b^5*c*d^2 - 5*sqrt(b*d)*(sqrt(b*x^2 + a
)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*a^2*b^4*d^3 + 35*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) -
sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^4*b^4*c*d + 5*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x
^2 + a)*b*d - a*b*d))^4*a*b^3*d^2 - 15*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a
*b*d))^6*b^2*d)/((b^2*c - a*b*d - (sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2)^5*b*a
bs(b))