3.90 \(\int \frac{(c+d x^2)^3}{(a+b x^2)^{5/2}} \, dx\)

Optimal. Leaf size=172 \[ -\frac{d x \sqrt{a+b x^2} \left (-15 a^2 d^2+8 a b c d+4 b^2 c^2\right )}{6 a^2 b^3}+\frac{x \left (c+d x^2\right ) (b c-a d) (5 a d+2 b c)}{3 a^2 b^2 \sqrt{a+b x^2}}+\frac{d^2 (6 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{7/2}}+\frac{x \left (c+d x^2\right )^2 (b c-a d)}{3 a b \left (a+b x^2\right )^{3/2}} \]

[Out]

-(d*(4*b^2*c^2 + 8*a*b*c*d - 15*a^2*d^2)*x*Sqrt[a + b*x^2])/(6*a^2*b^3) + ((b*c - a*d)*(2*b*c + 5*a*d)*x*(c +
d*x^2))/(3*a^2*b^2*Sqrt[a + b*x^2]) + ((b*c - a*d)*x*(c + d*x^2)^2)/(3*a*b*(a + b*x^2)^(3/2)) + (d^2*(6*b*c -
5*a*d)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*b^(7/2))

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Rubi [A]  time = 0.156522, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {413, 526, 388, 217, 206} \[ -\frac{d x \sqrt{a+b x^2} \left (-15 a^2 d^2+8 a b c d+4 b^2 c^2\right )}{6 a^2 b^3}+\frac{x \left (c+d x^2\right ) (b c-a d) (5 a d+2 b c)}{3 a^2 b^2 \sqrt{a+b x^2}}+\frac{d^2 (6 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{7/2}}+\frac{x \left (c+d x^2\right )^2 (b c-a d)}{3 a b \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)^3/(a + b*x^2)^(5/2),x]

[Out]

-(d*(4*b^2*c^2 + 8*a*b*c*d - 15*a^2*d^2)*x*Sqrt[a + b*x^2])/(6*a^2*b^3) + ((b*c - a*d)*(2*b*c + 5*a*d)*x*(c +
d*x^2))/(3*a^2*b^2*Sqrt[a + b*x^2]) + ((b*c - a*d)*x*(c + d*x^2)^2)/(3*a*b*(a + b*x^2)^(3/2)) + (d^2*(6*b*c -
5*a*d)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*b^(7/2))

Rule 413

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

Rule 526

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

Rule 388

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

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{\left (c+d x^2\right )^3}{\left (a+b x^2\right )^{5/2}} \, dx &=\frac{(b c-a d) x \left (c+d x^2\right )^2}{3 a b \left (a+b x^2\right )^{3/2}}+\frac{\int \frac{\left (c+d x^2\right ) \left (c (2 b c+a d)-d (2 b c-5 a d) x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx}{3 a b}\\ &=\frac{(b c-a d) (2 b c+5 a d) x \left (c+d x^2\right )}{3 a^2 b^2 \sqrt{a+b x^2}}+\frac{(b c-a d) x \left (c+d x^2\right )^2}{3 a b \left (a+b x^2\right )^{3/2}}-\frac{\int \frac{a c d (2 b c-5 a d)+d \left (4 b^2 c^2+8 a b c d-15 a^2 d^2\right ) x^2}{\sqrt{a+b x^2}} \, dx}{3 a^2 b^2}\\ &=-\frac{d \left (4 b^2 c^2+8 a b c d-15 a^2 d^2\right ) x \sqrt{a+b x^2}}{6 a^2 b^3}+\frac{(b c-a d) (2 b c+5 a d) x \left (c+d x^2\right )}{3 a^2 b^2 \sqrt{a+b x^2}}+\frac{(b c-a d) x \left (c+d x^2\right )^2}{3 a b \left (a+b x^2\right )^{3/2}}+\frac{\left (d^2 (6 b c-5 a d)\right ) \int \frac{1}{\sqrt{a+b x^2}} \, dx}{2 b^3}\\ &=-\frac{d \left (4 b^2 c^2+8 a b c d-15 a^2 d^2\right ) x \sqrt{a+b x^2}}{6 a^2 b^3}+\frac{(b c-a d) (2 b c+5 a d) x \left (c+d x^2\right )}{3 a^2 b^2 \sqrt{a+b x^2}}+\frac{(b c-a d) x \left (c+d x^2\right )^2}{3 a b \left (a+b x^2\right )^{3/2}}+\frac{\left (d^2 (6 b c-5 a d)\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )}{2 b^3}\\ &=-\frac{d \left (4 b^2 c^2+8 a b c d-15 a^2 d^2\right ) x \sqrt{a+b x^2}}{6 a^2 b^3}+\frac{(b c-a d) (2 b c+5 a d) x \left (c+d x^2\right )}{3 a^2 b^2 \sqrt{a+b x^2}}+\frac{(b c-a d) x \left (c+d x^2\right )^2}{3 a b \left (a+b x^2\right )^{3/2}}+\frac{d^2 (6 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{7/2}}\\ \end{align*}

Mathematica [A]  time = 5.09688, size = 125, normalized size = 0.73 \[ \frac{x \left (3 a^2 d^3 \left (a+b x^2\right )^2+2 \left (a+b x^2\right ) (b c-a d)^2 (7 a d+2 b c)+2 a (b c-a d)^3\right )}{6 a^2 b^3 \left (a+b x^2\right )^{3/2}}+\frac{d^2 (6 b c-5 a d) \log \left (\sqrt{b} \sqrt{a+b x^2}+b x\right )}{2 b^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)^3/(a + b*x^2)^(5/2),x]

[Out]

(x*(2*a*(b*c - a*d)^3 + 2*(b*c - a*d)^2*(2*b*c + 7*a*d)*(a + b*x^2) + 3*a^2*d^3*(a + b*x^2)^2))/(6*a^2*b^3*(a
+ b*x^2)^(3/2)) + (d^2*(6*b*c - 5*a*d)*Log[b*x + Sqrt[b]*Sqrt[a + b*x^2]])/(2*b^(7/2))

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Maple [A]  time = 0.007, size = 228, normalized size = 1.3 \begin{align*}{\frac{{d}^{3}{x}^{5}}{2\,b} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{5\,a{d}^{3}{x}^{3}}{6\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{5\,a{d}^{3}x}{2\,{b}^{3}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{\frac{5\,a{d}^{3}}{2}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{7}{2}}}}-{\frac{c{d}^{2}{x}^{3}}{b} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-3\,{\frac{c{d}^{2}x}{{b}^{2}\sqrt{b{x}^{2}+a}}}+3\,{\frac{c{d}^{2}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ) }{{b}^{5/2}}}-{\frac{{c}^{2}dx}{b} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{{c}^{2}dx}{ab}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{{c}^{3}x}{3\,a} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,{c}^{3}x}{3\,{a}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)^3/(b*x^2+a)^(5/2),x)

[Out]

1/2*d^3*x^5/b/(b*x^2+a)^(3/2)+5/6*d^3/b^2*a*x^3/(b*x^2+a)^(3/2)+5/2*d^3/b^3*a*x/(b*x^2+a)^(1/2)-5/2*d^3/b^(7/2
)*a*ln(x*b^(1/2)+(b*x^2+a)^(1/2))-c*d^2*x^3/b/(b*x^2+a)^(3/2)-3*c*d^2/b^2*x/(b*x^2+a)^(1/2)+3*c*d^2/b^(5/2)*ln
(x*b^(1/2)+(b*x^2+a)^(1/2))-c^2*d/b*x/(b*x^2+a)^(3/2)+c^2*d/b/a*x/(b*x^2+a)^(1/2)+1/3*c^3*x/a/(b*x^2+a)^(3/2)+
2/3*c^3/a^2*x/(b*x^2+a)^(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((d*x^2+c)^3/(b*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.96421, size = 1008, normalized size = 5.86 \begin{align*} \left [-\frac{3 \,{\left (6 \, a^{4} b c d^{2} - 5 \, a^{5} d^{3} +{\left (6 \, a^{2} b^{3} c d^{2} - 5 \, a^{3} b^{2} d^{3}\right )} x^{4} + 2 \,{\left (6 \, a^{3} b^{2} c d^{2} - 5 \, a^{4} b d^{3}\right )} x^{2}\right )} \sqrt{b} \log \left (-2 \, b x^{2} + 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) - 2 \,{\left (3 \, a^{2} b^{3} d^{3} x^{5} + 2 \,{\left (2 \, b^{5} c^{3} + 3 \, a b^{4} c^{2} d - 12 \, a^{2} b^{3} c d^{2} + 10 \, a^{3} b^{2} d^{3}\right )} x^{3} + 3 \,{\left (2 \, a b^{4} c^{3} - 6 \, a^{3} b^{2} c d^{2} + 5 \, a^{4} b d^{3}\right )} x\right )} \sqrt{b x^{2} + a}}{12 \,{\left (a^{2} b^{6} x^{4} + 2 \, a^{3} b^{5} x^{2} + a^{4} b^{4}\right )}}, -\frac{3 \,{\left (6 \, a^{4} b c d^{2} - 5 \, a^{5} d^{3} +{\left (6 \, a^{2} b^{3} c d^{2} - 5 \, a^{3} b^{2} d^{3}\right )} x^{4} + 2 \,{\left (6 \, a^{3} b^{2} c d^{2} - 5 \, a^{4} b d^{3}\right )} x^{2}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) -{\left (3 \, a^{2} b^{3} d^{3} x^{5} + 2 \,{\left (2 \, b^{5} c^{3} + 3 \, a b^{4} c^{2} d - 12 \, a^{2} b^{3} c d^{2} + 10 \, a^{3} b^{2} d^{3}\right )} x^{3} + 3 \,{\left (2 \, a b^{4} c^{3} - 6 \, a^{3} b^{2} c d^{2} + 5 \, a^{4} b d^{3}\right )} x\right )} \sqrt{b x^{2} + a}}{6 \,{\left (a^{2} b^{6} x^{4} + 2 \, a^{3} b^{5} x^{2} + a^{4} b^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^3/(b*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

[-1/12*(3*(6*a^4*b*c*d^2 - 5*a^5*d^3 + (6*a^2*b^3*c*d^2 - 5*a^3*b^2*d^3)*x^4 + 2*(6*a^3*b^2*c*d^2 - 5*a^4*b*d^
3)*x^2)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(3*a^2*b^3*d^3*x^5 + 2*(2*b^5*c^3 + 3*a*b^
4*c^2*d - 12*a^2*b^3*c*d^2 + 10*a^3*b^2*d^3)*x^3 + 3*(2*a*b^4*c^3 - 6*a^3*b^2*c*d^2 + 5*a^4*b*d^3)*x)*sqrt(b*x
^2 + a))/(a^2*b^6*x^4 + 2*a^3*b^5*x^2 + a^4*b^4), -1/6*(3*(6*a^4*b*c*d^2 - 5*a^5*d^3 + (6*a^2*b^3*c*d^2 - 5*a^
3*b^2*d^3)*x^4 + 2*(6*a^3*b^2*c*d^2 - 5*a^4*b*d^3)*x^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (3*a^2*b
^3*d^3*x^5 + 2*(2*b^5*c^3 + 3*a*b^4*c^2*d - 12*a^2*b^3*c*d^2 + 10*a^3*b^2*d^3)*x^3 + 3*(2*a*b^4*c^3 - 6*a^3*b^
2*c*d^2 + 5*a^4*b*d^3)*x)*sqrt(b*x^2 + a))/(a^2*b^6*x^4 + 2*a^3*b^5*x^2 + a^4*b^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c + d x^{2}\right )^{3}}{\left (a + b x^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)**3/(b*x**2+a)**(5/2),x)

[Out]

Integral((c + d*x**2)**3/(a + b*x**2)**(5/2), x)

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Giac [A]  time = 1.16924, size = 213, normalized size = 1.24 \begin{align*} \frac{{\left ({\left (\frac{3 \, d^{3} x^{2}}{b} + \frac{2 \,{\left (2 \, b^{6} c^{3} + 3 \, a b^{5} c^{2} d - 12 \, a^{2} b^{4} c d^{2} + 10 \, a^{3} b^{3} d^{3}\right )}}{a^{2} b^{5}}\right )} x^{2} + \frac{3 \,{\left (2 \, a b^{5} c^{3} - 6 \, a^{3} b^{3} c d^{2} + 5 \, a^{4} b^{2} d^{3}\right )}}{a^{2} b^{5}}\right )} x}{6 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}}} - \frac{{\left (6 \, b c d^{2} - 5 \, a d^{3}\right )} \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{2 \, b^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^3/(b*x^2+a)^(5/2),x, algorithm="giac")

[Out]

1/6*((3*d^3*x^2/b + 2*(2*b^6*c^3 + 3*a*b^5*c^2*d - 12*a^2*b^4*c*d^2 + 10*a^3*b^3*d^3)/(a^2*b^5))*x^2 + 3*(2*a*
b^5*c^3 - 6*a^3*b^3*c*d^2 + 5*a^4*b^2*d^3)/(a^2*b^5))*x/(b*x^2 + a)^(3/2) - 1/2*(6*b*c*d^2 - 5*a*d^3)*log(abs(
-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)