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]

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

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Rubi [A]  time = 0.03, antiderivative size = 87, normalized size of antiderivative = 1.00, 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 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 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])

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 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]

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*}

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Mathematica [A]  time = 0.08, 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)

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fricas [A]  time = 1.02, size = 176, normalized size = 2.02 \[ \left [\frac {15 \, A a^{2} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (8 \, B b^{2} x^{4} + 10 \, A b^{2} x^{3} + 16 \, B a b x^{2} + 25 \, A a b x + 8 \, B a^{2}\right )} \sqrt {b x^{2} + a}}{80 \, b}, -\frac {15 \, A a^{2} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (8 \, B b^{2} x^{4} + 10 \, A b^{2} x^{3} + 16 \, B a b x^{2} + 25 \, A a b x + 8 \, B a^{2}\right )} \sqrt {b x^{2} + a}}{40 \, b}\right ] \]

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]

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giac [A]  time = 0.44, size = 76, normalized size = 0.87 \[ -\frac {3 \, A a^{2} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, \sqrt {b}} + \frac {1}{40} \, \sqrt {b x^{2} + a} {\left (\frac {8 \, B a^{2}}{b} + {\left (25 \, A a + 2 \, {\left (8 \, B a + {\left (4 \, B b x + 5 \, A b\right )} x\right )} x\right )} x\right )} \]

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)

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maple [A]  time = 0.00, size = 69, normalized size = 0.79 \[ \frac {3 A \,a^{2} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8 \sqrt {b}}+\frac {3 \sqrt {b \,x^{2}+a}\, A a x}{8}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} A x}{4}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} B}{5 b} \]

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(b^(1/2)*x+(b*x^
2+a)^(1/2))

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maxima [A]  time = 1.41, size = 61, normalized size = 0.70 \[ \frac {1}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A x + \frac {3}{8} \, \sqrt {b x^{2} + a} A a x + \frac {3 \, A a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {b}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B}{5 \, b} \]

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]

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

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mupad [B]  time = 1.18, size = 54, normalized size = 0.62 \[ \frac {B\,{\left (b\,x^2+a\right )}^{5/2}}{5\,b}+\frac {A\,x\,{\left (b\,x^2+a\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(3/2)*(A + B*x),x)

[Out]

(B*(a + b*x^2)^(5/2))/(5*b) + (A*x*(a + b*x^2)^(3/2)*hypergeom([-3/2, 1/2], 3/2, -(b*x^2)/a))/((b*x^2)/a + 1)^
(3/2)

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sympy [A]  time = 12.91, size = 219, normalized size = 2.52 \[ \frac {A a^{\frac {3}{2}} x \sqrt {1 + \frac {b x^{2}}{a}}}{2} + \frac {A a^{\frac {3}{2}} x}{8 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 A \sqrt {a} b x^{3}}{8 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 A a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{8 \sqrt {b}} + \frac {A b^{2} x^{5}}{4 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} + B a \left (\begin {cases} \frac {\sqrt {a} x^{2}}{2} & \text {for}\: b = 0 \\\frac {\left (a + b x^{2}\right )^{\frac {3}{2}}}{3 b} & \text {otherwise} \end {cases}\right ) + B b \left (\begin {cases} - \frac {2 a^{2} \sqrt {a + b x^{2}}}{15 b^{2}} + \frac {a x^{2} \sqrt {a + b x^{2}}}{15 b} + \frac {x^{4} \sqrt {a + b x^{2}}}{5} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{4}}{4} & \text {otherwise} \end {cases}\right ) \]

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))

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