3.778 \(\int (a+b x)^3 \sqrt {a^2-b^2 x^2} \, dx\)

Optimal. Leaf size=140 \[ -\frac {7 a^2 \left (a^2-b^2 x^2\right )^{3/2}}{12 b}-\frac {7 a (a+b x) \left (a^2-b^2 x^2\right )^{3/2}}{20 b}-\frac {(a+b x)^2 \left (a^2-b^2 x^2\right )^{3/2}}{5 b}+\frac {7 a^5 \tan ^{-1}\left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{8 b}+\frac {7}{8} a^3 x \sqrt {a^2-b^2 x^2} \]

[Out]

-7/12*a^2*(-b^2*x^2+a^2)^(3/2)/b-7/20*a*(b*x+a)*(-b^2*x^2+a^2)^(3/2)/b-1/5*(b*x+a)^2*(-b^2*x^2+a^2)^(3/2)/b+7/
8*a^5*arctan(b*x/(-b^2*x^2+a^2)^(1/2))/b+7/8*a^3*x*(-b^2*x^2+a^2)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {671, 641, 195, 217, 203} \[ \frac {7}{8} a^3 x \sqrt {a^2-b^2 x^2}-\frac {7 a^2 \left (a^2-b^2 x^2\right )^{3/2}}{12 b}-\frac {7 a (a+b x) \left (a^2-b^2 x^2\right )^{3/2}}{20 b}-\frac {(a+b x)^2 \left (a^2-b^2 x^2\right )^{3/2}}{5 b}+\frac {7 a^5 \tan ^{-1}\left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^3*Sqrt[a^2 - b^2*x^2],x]

[Out]

(7*a^3*x*Sqrt[a^2 - b^2*x^2])/8 - (7*a^2*(a^2 - b^2*x^2)^(3/2))/(12*b) - (7*a*(a + b*x)*(a^2 - b^2*x^2)^(3/2))
/(20*b) - ((a + b*x)^2*(a^2 - b^2*x^2)^(3/2))/(5*b) + (7*a^5*ArcTan[(b*x)/Sqrt[a^2 - b^2*x^2]])/(8*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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[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]

Rule 671

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(m + 2*p + 1)), x] + Dist[(2*c*d*(m + p))/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p, x]
, x] /; FreeQ[{a, c, d, e, p}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p
]

Rubi steps

\begin {align*} \int (a+b x)^3 \sqrt {a^2-b^2 x^2} \, dx &=-\frac {(a+b x)^2 \left (a^2-b^2 x^2\right )^{3/2}}{5 b}+\frac {1}{5} (7 a) \int (a+b x)^2 \sqrt {a^2-b^2 x^2} \, dx\\ &=-\frac {7 a (a+b x) \left (a^2-b^2 x^2\right )^{3/2}}{20 b}-\frac {(a+b x)^2 \left (a^2-b^2 x^2\right )^{3/2}}{5 b}+\frac {1}{4} \left (7 a^2\right ) \int (a+b x) \sqrt {a^2-b^2 x^2} \, dx\\ &=-\frac {7 a^2 \left (a^2-b^2 x^2\right )^{3/2}}{12 b}-\frac {7 a (a+b x) \left (a^2-b^2 x^2\right )^{3/2}}{20 b}-\frac {(a+b x)^2 \left (a^2-b^2 x^2\right )^{3/2}}{5 b}+\frac {1}{4} \left (7 a^3\right ) \int \sqrt {a^2-b^2 x^2} \, dx\\ &=\frac {7}{8} a^3 x \sqrt {a^2-b^2 x^2}-\frac {7 a^2 \left (a^2-b^2 x^2\right )^{3/2}}{12 b}-\frac {7 a (a+b x) \left (a^2-b^2 x^2\right )^{3/2}}{20 b}-\frac {(a+b x)^2 \left (a^2-b^2 x^2\right )^{3/2}}{5 b}+\frac {1}{8} \left (7 a^5\right ) \int \frac {1}{\sqrt {a^2-b^2 x^2}} \, dx\\ &=\frac {7}{8} a^3 x \sqrt {a^2-b^2 x^2}-\frac {7 a^2 \left (a^2-b^2 x^2\right )^{3/2}}{12 b}-\frac {7 a (a+b x) \left (a^2-b^2 x^2\right )^{3/2}}{20 b}-\frac {(a+b x)^2 \left (a^2-b^2 x^2\right )^{3/2}}{5 b}+\frac {1}{8} \left (7 a^5\right ) \operatorname {Subst}\left (\int \frac {1}{1+b^2 x^2} \, dx,x,\frac {x}{\sqrt {a^2-b^2 x^2}}\right )\\ &=\frac {7}{8} a^3 x \sqrt {a^2-b^2 x^2}-\frac {7 a^2 \left (a^2-b^2 x^2\right )^{3/2}}{12 b}-\frac {7 a (a+b x) \left (a^2-b^2 x^2\right )^{3/2}}{20 b}-\frac {(a+b x)^2 \left (a^2-b^2 x^2\right )^{3/2}}{5 b}+\frac {7 a^5 \tan ^{-1}\left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{8 b}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.22, size = 112, normalized size = 0.80 \[ \frac {\sqrt {a^2-b^2 x^2} \left (105 a^4 \sin ^{-1}\left (\frac {b x}{a}\right )+\sqrt {1-\frac {b^2 x^2}{a^2}} \left (-136 a^4+15 a^3 b x+112 a^2 b^2 x^2+90 a b^3 x^3+24 b^4 x^4\right )\right )}{120 b \sqrt {1-\frac {b^2 x^2}{a^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^3*Sqrt[a^2 - b^2*x^2],x]

[Out]

(Sqrt[a^2 - b^2*x^2]*(Sqrt[1 - (b^2*x^2)/a^2]*(-136*a^4 + 15*a^3*b*x + 112*a^2*b^2*x^2 + 90*a*b^3*x^3 + 24*b^4
*x^4) + 105*a^4*ArcSin[(b*x)/a]))/(120*b*Sqrt[1 - (b^2*x^2)/a^2])

________________________________________________________________________________________

fricas [A]  time = 0.93, size = 95, normalized size = 0.68 \[ -\frac {210 \, a^{5} \arctan \left (-\frac {a - \sqrt {-b^{2} x^{2} + a^{2}}}{b x}\right ) - {\left (24 \, b^{4} x^{4} + 90 \, a b^{3} x^{3} + 112 \, a^{2} b^{2} x^{2} + 15 \, a^{3} b x - 136 \, a^{4}\right )} \sqrt {-b^{2} x^{2} + a^{2}}}{120 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(210*a^5*arctan(-(a - sqrt(-b^2*x^2 + a^2))/(b*x)) - (24*b^4*x^4 + 90*a*b^3*x^3 + 112*a^2*b^2*x^2 + 15*
a^3*b*x - 136*a^4)*sqrt(-b^2*x^2 + a^2))/b

________________________________________________________________________________________

giac [A]  time = 0.21, size = 81, normalized size = 0.58 \[ \frac {7 \, a^{5} \arcsin \left (\frac {b x}{a}\right ) \mathrm {sgn}\relax (a) \mathrm {sgn}\relax (b)}{8 \, {\left | b \right |}} - \frac {1}{120} \, \sqrt {-b^{2} x^{2} + a^{2}} {\left (\frac {136 \, a^{4}}{b} - {\left (15 \, a^{3} + 2 \, {\left (56 \, a^{2} b + 3 \, {\left (4 \, b^{3} x + 15 \, a b^{2}\right )} x\right )} x\right )} x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7/8*a^5*arcsin(b*x/a)*sgn(a)*sgn(b)/abs(b) - 1/120*sqrt(-b^2*x^2 + a^2)*(136*a^4/b - (15*a^3 + 2*(56*a^2*b + 3
*(4*b^3*x + 15*a*b^2)*x)*x)*x)

________________________________________________________________________________________

maple [A]  time = 0.06, size = 114, normalized size = 0.81 \[ \frac {7 a^{5} \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-b^{2} x^{2}+a^{2}}}\right )}{8 \sqrt {b^{2}}}+\frac {7 \sqrt {-b^{2} x^{2}+a^{2}}\, a^{3} x}{8}-\frac {\left (-b^{2} x^{2}+a^{2}\right )^{\frac {3}{2}} b \,x^{2}}{5}-\frac {3 \left (-b^{2} x^{2}+a^{2}\right )^{\frac {3}{2}} a x}{4}-\frac {17 \left (-b^{2} x^{2}+a^{2}\right )^{\frac {3}{2}} a^{2}}{15 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^3*(-b^2*x^2+a^2)^(1/2),x)

[Out]

-1/5*b*x^2*(-b^2*x^2+a^2)^(3/2)-17/15*a^2*(-b^2*x^2+a^2)^(3/2)/b-3/4*a*x*(-b^2*x^2+a^2)^(3/2)+7/8*a^3*x*(-b^2*
x^2+a^2)^(1/2)+7/8*a^5/(b^2)^(1/2)*arctan((b^2)^(1/2)/(-b^2*x^2+a^2)^(1/2)*x)

________________________________________________________________________________________

maxima [A]  time = 3.00, size = 96, normalized size = 0.69 \[ \frac {7 \, a^{5} \arcsin \left (\frac {b x}{a}\right )}{8 \, b} + \frac {7}{8} \, \sqrt {-b^{2} x^{2} + a^{2}} a^{3} x - \frac {1}{5} \, {\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}} b x^{2} - \frac {3}{4} \, {\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}} a x - \frac {17 \, {\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}} a^{2}}{15 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(-b^2*x^2+a^2)^(1/2),x, algorithm="maxima")

[Out]

7/8*a^5*arcsin(b*x/a)/b + 7/8*sqrt(-b^2*x^2 + a^2)*a^3*x - 1/5*(-b^2*x^2 + a^2)^(3/2)*b*x^2 - 3/4*(-b^2*x^2 +
a^2)^(3/2)*a*x - 17/15*(-b^2*x^2 + a^2)^(3/2)*a^2/b

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {a^2-b^2\,x^2}\,{\left (a+b\,x\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 - b^2*x^2)^(1/2)*(a + b*x)^3,x)

[Out]

int((a^2 - b^2*x^2)^(1/2)*(a + b*x)^3, x)

________________________________________________________________________________________

sympy [C]  time = 8.24, size = 439, normalized size = 3.14 \[ a^{3} \left (\begin {cases} - \frac {i a^{2} \operatorname {acosh}{\left (\frac {b x}{a} \right )}}{2 b} - \frac {i a x}{2 \sqrt {-1 + \frac {b^{2} x^{2}}{a^{2}}}} + \frac {i b^{2} x^{3}}{2 a \sqrt {-1 + \frac {b^{2} x^{2}}{a^{2}}}} & \text {for}\: \left |{\frac {b^{2} x^{2}}{a^{2}}}\right | > 1 \\\frac {a^{2} \operatorname {asin}{\left (\frac {b x}{a} \right )}}{2 b} + \frac {a x \sqrt {1 - \frac {b^{2} x^{2}}{a^{2}}}}{2} & \text {otherwise} \end {cases}\right ) + 3 a^{2} b \left (\begin {cases} \frac {x^{2} \sqrt {a^{2}}}{2} & \text {for}\: b^{2} = 0 \\- \frac {\left (a^{2} - b^{2} x^{2}\right )^{\frac {3}{2}}}{3 b^{2}} & \text {otherwise} \end {cases}\right ) + 3 a b^{2} \left (\begin {cases} - \frac {i a^{4} \operatorname {acosh}{\left (\frac {b x}{a} \right )}}{8 b^{3}} + \frac {i a^{3} x}{8 b^{2} \sqrt {-1 + \frac {b^{2} x^{2}}{a^{2}}}} - \frac {3 i a x^{3}}{8 \sqrt {-1 + \frac {b^{2} x^{2}}{a^{2}}}} + \frac {i b^{2} x^{5}}{4 a \sqrt {-1 + \frac {b^{2} x^{2}}{a^{2}}}} & \text {for}\: \left |{\frac {b^{2} x^{2}}{a^{2}}}\right | > 1 \\\frac {a^{4} \operatorname {asin}{\left (\frac {b x}{a} \right )}}{8 b^{3}} - \frac {a^{3} x}{8 b^{2} \sqrt {1 - \frac {b^{2} x^{2}}{a^{2}}}} + \frac {3 a x^{3}}{8 \sqrt {1 - \frac {b^{2} x^{2}}{a^{2}}}} - \frac {b^{2} x^{5}}{4 a \sqrt {1 - \frac {b^{2} x^{2}}{a^{2}}}} & \text {otherwise} \end {cases}\right ) + b^{3} \left (\begin {cases} - \frac {2 a^{4} \sqrt {a^{2} - b^{2} x^{2}}}{15 b^{4}} - \frac {a^{2} x^{2} \sqrt {a^{2} - b^{2} x^{2}}}{15 b^{2}} + \frac {x^{4} \sqrt {a^{2} - b^{2} x^{2}}}{5} & \text {for}\: b \neq 0 \\\frac {x^{4} \sqrt {a^{2}}}{4} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**3*(-b**2*x**2+a**2)**(1/2),x)

[Out]

a**3*Piecewise((-I*a**2*acosh(b*x/a)/(2*b) - I*a*x/(2*sqrt(-1 + b**2*x**2/a**2)) + I*b**2*x**3/(2*a*sqrt(-1 +
b**2*x**2/a**2)), Abs(b**2*x**2/a**2) > 1), (a**2*asin(b*x/a)/(2*b) + a*x*sqrt(1 - b**2*x**2/a**2)/2, True)) +
3*a**2*b*Piecewise((x**2*sqrt(a**2)/2, Eq(b**2, 0)), (-(a**2 - b**2*x**2)**(3/2)/(3*b**2), True)) + 3*a*b**2*
Piecewise((-I*a**4*acosh(b*x/a)/(8*b**3) + I*a**3*x/(8*b**2*sqrt(-1 + b**2*x**2/a**2)) - 3*I*a*x**3/(8*sqrt(-1
+ b**2*x**2/a**2)) + I*b**2*x**5/(4*a*sqrt(-1 + b**2*x**2/a**2)), Abs(b**2*x**2/a**2) > 1), (a**4*asin(b*x/a)
/(8*b**3) - a**3*x/(8*b**2*sqrt(1 - b**2*x**2/a**2)) + 3*a*x**3/(8*sqrt(1 - b**2*x**2/a**2)) - b**2*x**5/(4*a*
sqrt(1 - b**2*x**2/a**2)), True)) + b**3*Piecewise((-2*a**4*sqrt(a**2 - b**2*x**2)/(15*b**4) - a**2*x**2*sqrt(
a**2 - b**2*x**2)/(15*b**2) + x**4*sqrt(a**2 - b**2*x**2)/5, Ne(b, 0)), (x**4*sqrt(a**2)/4, True))

________________________________________________________________________________________