3.11 \(\int (A+B x) (a+b x^2)^{3/2} \, dx\)

Optimal. Leaf size=87 \[ \frac{3 a^2 A \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{8 \sqrt{b}}+\frac{1}{4} A x \left (a+b x^2\right )^{3/2}+\frac{3}{8} a A x \sqrt{a+b x^2}+\frac{B \left (a+b x^2\right )^{5/2}}{5 b} \]

[Out]

(3*a*A*x*Sqrt[a + b*x^2])/8 + (A*x*(a + b*x^2)^(3/2))/4 + (B*(a + b*x^2)^(5/2))/(5*b) + (3*a^2*A*ArcTanh[(Sqrt
[b]*x)/Sqrt[a + b*x^2]])/(8*Sqrt[b])

________________________________________________________________________________________

Rubi [A]  time = 0.0261586, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {641, 195, 217, 206} \[ \frac{3 a^2 A \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{8 \sqrt{b}}+\frac{1}{4} A x \left (a+b x^2\right )^{3/2}+\frac{3}{8} a A x \sqrt{a+b x^2}+\frac{B \left (a+b x^2\right )^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)*(a + b*x^2)^(3/2),x]

[Out]

(3*a*A*x*Sqrt[a + b*x^2])/8 + (A*x*(a + b*x^2)^(3/2))/4 + (B*(a + b*x^2)^(5/2))/(5*b) + (3*a^2*A*ArcTanh[(Sqrt
[b]*x)/Sqrt[a + b*x^2]])/(8*Sqrt[b])

Rule 641

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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

Mathematica [A]  time = 0.0696007, size = 88, normalized size = 1.01 \[ \frac{\sqrt{a+b x^2} \left (8 a^2 B+a b x (25 A+16 B x)+2 b^2 x^3 (5 A+4 B x)\right )+15 a^2 A \sqrt{b} \log \left (\sqrt{b} \sqrt{a+b x^2}+b x\right )}{40 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)*(a + b*x^2)^(3/2),x]

[Out]

(Sqrt[a + b*x^2]*(8*a^2*B + 2*b^2*x^3*(5*A + 4*B*x) + a*b*x*(25*A + 16*B*x)) + 15*a^2*A*Sqrt[b]*Log[b*x + Sqrt
[b]*Sqrt[a + b*x^2]])/(40*b)

________________________________________________________________________________________

Maple [A]  time = 0.004, size = 69, normalized size = 0.8 \begin{align*}{\frac{B}{5\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{Ax}{4} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,aAx}{8}\sqrt{b{x}^{2}+a}}+{\frac{3\,A{a}^{2}}{8}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(b*x^2+a)^(3/2),x)

[Out]

1/5*B*(b*x^2+a)^(5/2)/b+1/4*A*x*(b*x^2+a)^(3/2)+3/8*a*A*x*(b*x^2+a)^(1/2)+3/8*A*a^2/b^(1/2)*ln(x*b^(1/2)+(b*x^
2+a)^(1/2))

________________________________________________________________________________________

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((B*x+A)*(b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.63008, size = 425, normalized size = 4.89 \begin{align*} \left [\frac{15 \, A a^{2} \sqrt{b} \log \left (-2 \, b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) + 2 \,{\left (8 \, B b^{2} x^{4} + 10 \, A b^{2} x^{3} + 16 \, B a b x^{2} + 25 \, A a b x + 8 \, B a^{2}\right )} \sqrt{b x^{2} + a}}{80 \, b}, -\frac{15 \, A a^{2} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) -{\left (8 \, B b^{2} x^{4} + 10 \, A b^{2} x^{3} + 16 \, B a b x^{2} + 25 \, A a b x + 8 \, B a^{2}\right )} \sqrt{b x^{2} + a}}{40 \, b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/80*(15*A*a^2*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(8*B*b^2*x^4 + 10*A*b^2*x^3 + 16*B
*a*b*x^2 + 25*A*a*b*x + 8*B*a^2)*sqrt(b*x^2 + a))/b, -1/40*(15*A*a^2*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a
)) - (8*B*b^2*x^4 + 10*A*b^2*x^3 + 16*B*a*b*x^2 + 25*A*a*b*x + 8*B*a^2)*sqrt(b*x^2 + a))/b]

________________________________________________________________________________________

Sympy [A]  time = 6.30866, size = 219, normalized size = 2.52 \begin{align*} \frac{A a^{\frac{3}{2}} x \sqrt{1 + \frac{b x^{2}}{a}}}{2} + \frac{A a^{\frac{3}{2}} x}{8 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{3 A \sqrt{a} b x^{3}}{8 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{3 A a^{2} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{8 \sqrt{b}} + \frac{A b^{2} x^{5}}{4 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} + B a \left (\begin{cases} \frac{\sqrt{a} x^{2}}{2} & \text{for}\: b = 0 \\\frac{\left (a + b x^{2}\right )^{\frac{3}{2}}}{3 b} & \text{otherwise} \end{cases}\right ) + B b \left (\begin{cases} - \frac{2 a^{2} \sqrt{a + b x^{2}}}{15 b^{2}} + \frac{a x^{2} \sqrt{a + b x^{2}}}{15 b} + \frac{x^{4} \sqrt{a + b x^{2}}}{5} & \text{for}\: b \neq 0 \\\frac{\sqrt{a} x^{4}}{4} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x**2+a)**(3/2),x)

[Out]

A*a**(3/2)*x*sqrt(1 + b*x**2/a)/2 + A*a**(3/2)*x/(8*sqrt(1 + b*x**2/a)) + 3*A*sqrt(a)*b*x**3/(8*sqrt(1 + b*x**
2/a)) + 3*A*a**2*asinh(sqrt(b)*x/sqrt(a))/(8*sqrt(b)) + A*b**2*x**5/(4*sqrt(a)*sqrt(1 + b*x**2/a)) + B*a*Piece
wise((sqrt(a)*x**2/2, Eq(b, 0)), ((a + b*x**2)**(3/2)/(3*b), True)) + B*b*Piecewise((-2*a**2*sqrt(a + b*x**2)/
(15*b**2) + a*x**2*sqrt(a + b*x**2)/(15*b) + x**4*sqrt(a + b*x**2)/5, Ne(b, 0)), (sqrt(a)*x**4/4, True))

________________________________________________________________________________________

Giac [A]  time = 1.18071, size = 103, normalized size = 1.18 \begin{align*} -\frac{3 \, A a^{2} \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{8 \, \sqrt{b}} + \frac{1}{40} \, \sqrt{b x^{2} + a}{\left (\frac{8 \, B a^{2}}{b} +{\left (25 \, A a + 2 \,{\left (8 \, B a +{\left (4 \, B b x + 5 \, A b\right )} x\right )} x\right )} x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/8*A*a^2*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/sqrt(b) + 1/40*sqrt(b*x^2 + a)*(8*B*a^2/b + (25*A*a + 2*(8*B
*a + (4*B*b*x + 5*A*b)*x)*x)*x)