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

Optimal. Leaf size=90 \[ \frac{x (b c-a d)^2}{a b^2 \sqrt{a+b x^2}}+\frac{d (4 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{5/2}}+\frac{d^2 x \sqrt{a+b x^2}}{2 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0620589, antiderivative size = 105, normalized size of antiderivative = 1.17, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {413, 388, 217, 206} \[ -\frac{d x \sqrt{a+b x^2} (2 b c-3 a d)}{2 a b^2}+\frac{d (4 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{5/2}}+\frac{x \left (c+d x^2\right ) (b c-a d)}{a b \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [C]  time = 2.36876, size = 160, normalized size = 1.78 \[ \frac{x \sqrt{\frac{b x^2}{a}+1} \left (-6 b x^2 \left (c+d x^2\right )^2 \text{HypergeometricPFQ}\left (\left \{\frac{3}{2},2,\frac{5}{2}\right \},\left \{1,\frac{9}{2}\right \},-\frac{b x^2}{a}\right )-12 b x^2 \left (2 c^2+3 c d x^2+d^2 x^4\right ) \, _2F_1\left (\frac{3}{2},\frac{5}{2};\frac{9}{2};-\frac{b x^2}{a}\right )+7 a \left (15 c^2+10 c d x^2+3 d^2 x^4\right ) \, _2F_1\left (\frac{1}{2},\frac{3}{2};\frac{7}{2};-\frac{b x^2}{a}\right )\right )}{105 a^2 \sqrt{a+b x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*Sqrt[1 + (b*x^2)/a]*(7*a*(15*c^2 + 10*c*d*x^2 + 3*d^2*x^4)*Hypergeometric2F1[1/2, 3/2, 7/2, -((b*x^2)/a)] -
12*b*x^2*(2*c^2 + 3*c*d*x^2 + d^2*x^4)*Hypergeometric2F1[3/2, 5/2, 9/2, -((b*x^2)/a)] - 6*b*x^2*(c + d*x^2)^2
*HypergeometricPFQ[{3/2, 2, 5/2}, {1, 9/2}, -((b*x^2)/a)]))/(105*a^2*Sqrt[a + b*x^2])

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Maple [A]  time = 0.006, size = 123, normalized size = 1.4 \begin{align*}{\frac{{d}^{2}{x}^{3}}{2\,b}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{3\,a{d}^{2}x}{2\,{b}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{\frac{3\,a{d}^{2}}{2}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{5}{2}}}}-2\,{\frac{cdx}{b\sqrt{b{x}^{2}+a}}}+2\,{\frac{cd\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ) }{{b}^{3/2}}}+{\frac{{c}^{2}x}{a}{\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)^2/(b*x^2+a)^(3/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.63845, size = 597, normalized size = 6.63 \begin{align*} \left [-\frac{{\left (4 \, a^{2} b c d - 3 \, a^{3} d^{2} +{\left (4 \, a b^{2} c d - 3 \, a^{2} b d^{2}\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 (a b^{2} d^{2} x^{3} +{\left (2 \, b^{3} c^{2} - 4 \, a b^{2} c d + 3 \, a^{2} b d^{2}\right )} x\right )} \sqrt{b x^{2} + a}}{4 \,{\left (a b^{4} x^{2} + a^{2} b^{3}\right )}}, -\frac{{\left (4 \, a^{2} b c d - 3 \, a^{3} d^{2} +{\left (4 \, a b^{2} c d - 3 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) -{\left (a b^{2} d^{2} x^{3} +{\left (2 \, b^{3} c^{2} - 4 \, a b^{2} c d + 3 \, a^{2} b d^{2}\right )} x\right )} \sqrt{b x^{2} + a}}{2 \,{\left (a b^{4} x^{2} + a^{2} b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*((4*a^2*b*c*d - 3*a^3*d^2 + (4*a*b^2*c*d - 3*a^2*b*d^2)*x^2)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sq
rt(b)*x - a) - 2*(a*b^2*d^2*x^3 + (2*b^3*c^2 - 4*a*b^2*c*d + 3*a^2*b*d^2)*x)*sqrt(b*x^2 + a))/(a*b^4*x^2 + a^2
*b^3), -1/2*((4*a^2*b*c*d - 3*a^3*d^2 + (4*a*b^2*c*d - 3*a^2*b*d^2)*x^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2
+ a)) - (a*b^2*d^2*x^3 + (2*b^3*c^2 - 4*a*b^2*c*d + 3*a^2*b*d^2)*x)*sqrt(b*x^2 + a))/(a*b^4*x^2 + a^2*b^3)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.11688, size = 124, normalized size = 1.38 \begin{align*} \frac{{\left (\frac{d^{2} x^{2}}{b} + \frac{2 \, b^{3} c^{2} - 4 \, a b^{2} c d + 3 \, a^{2} b d^{2}}{a b^{3}}\right )} x}{2 \, \sqrt{b x^{2} + a}} - \frac{{\left (4 \, b c d - 3 \, a d^{2}\right )} \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{2 \, b^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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