\(\int (a+b x)^3 \sqrt {a^2-b^2 x^2} \, dx\) [778]

Optimal result

Integrand size = 24, antiderivative size = 140 \[ \int (a+b x)^3 \sqrt {a^2-b^2 x^2} \, dx=\frac {7}{8} a^3 x \sqrt {a^2-b^2 x^2}-\frac {7 a^2 \left (a^2-b^2 x^2\right )^{3/2}}{12 b}-\frac {7 a (a+b x) \left (a^2-b^2 x^2\right )^{3/2}}{20 b}-\frac {(a+b x)^2 \left (a^2-b^2 x^2\right )^{3/2}}{5 b}+\frac {7 a^5 \arctan \left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{8 b} \]

[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] (verified)

Time = 0.04 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {685, 655, 201, 223, 209} \[ \int (a+b x)^3 \sqrt {a^2-b^2 x^2} \, dx=-\frac {7 a^2 \left (a^2-b^2 x^2\right )^{3/2}}{12 b}-\frac {7 a (a+b x) \left (a^2-b^2 x^2\right )^{3/2}}{20 b}-\frac {(a+b x)^2 \left (a^2-b^2 x^2\right )^{3/2}}{5 b}+\frac {7 a^5 \arctan \left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{8 b}+\frac {7}{8} a^3 x \sqrt {a^2-b^2 x^2} \]

[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 201

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 209

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

Rule 223

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 655

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 685

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*} \text {integral}& = -\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 ) \text {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] (verified)

Time = 0.43 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.79 \[ \int (a+b x)^3 \sqrt {a^2-b^2 x^2} \, dx=\frac {\sqrt {a^2-b^2 x^2} \left (-136 a^4+15 a^3 b x+112 a^2 b^2 x^2+90 a b^3 x^3+24 b^4 x^4\right )}{120 b}-\frac {7 a^5 \log \left (-\sqrt {-b^2} x+\sqrt {a^2-b^2 x^2}\right )}{8 \sqrt {-b^2}} \]

[In]

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

[Out]

(Sqrt[a^2 - b^2*x^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))/(120*b) - (7*a^5*L
og[-(Sqrt[-b^2]*x) + Sqrt[a^2 - b^2*x^2]])/(8*Sqrt[-b^2])

Maple [A] (verified)

Time = 2.49 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.67

method	result	size
risch	\(-\frac {\left (-24 b^{4} x^{4}-90 a \,b^{3} x^{3}-112 a^{2} b^{2} x^{2}-15 a^{3} b x +136 a^{4}\right ) \sqrt {-b^{2} x^{2}+a^{2}}}{120 b}+\frac {7 a^{5} \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-b^{2} x^{2}+a^{2}}}\right )}{8 \sqrt {b^{2}}}\)	\(94\)
default	\(a^{3} \left (\frac {x \sqrt {-b^{2} x^{2}+a^{2}}}{2}+\frac {a^{2} \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-b^{2} x^{2}+a^{2}}}\right )}{2 \sqrt {b^{2}}}\right )+b^{3} \left (-\frac {x^{2} \left (-b^{2} x^{2}+a^{2}\right )^{\frac {3}{2}}}{5 b^{2}}-\frac {2 a^{2} \left (-b^{2} x^{2}+a^{2}\right )^{\frac {3}{2}}}{15 b^{4}}\right )+3 a \,b^{2} \left (-\frac {x \left (-b^{2} x^{2}+a^{2}\right )^{\frac {3}{2}}}{4 b^{2}}+\frac {a^{2} \left (\frac {x \sqrt {-b^{2} x^{2}+a^{2}}}{2}+\frac {a^{2} \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-b^{2} x^{2}+a^{2}}}\right )}{2 \sqrt {b^{2}}}\right )}{4 b^{2}}\right )-\frac {a^{2} \left (-b^{2} x^{2}+a^{2}\right )^{\frac {3}{2}}}{b}\)	\(212\)

[In]

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

[Out]

-1/120*(-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)/b*(-b^2*x^2+a^2)^(1/2)+7/8*a^5/(b^2)^(1/2
)*arctan((b^2)^(1/2)*x/(-b^2*x^2+a^2)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.68 \[ \int (a+b x)^3 \sqrt {a^2-b^2 x^2} \, dx=-\frac {210 \, a^{5} \arctan \left (-\frac {a - \sqrt {-b^{2} x^{2} + a^{2}}}{b x}\right ) - {\left (24 \, b^{4} x^{4} + 90 \, a b^{3} x^{3} + 112 \, a^{2} b^{2} x^{2} + 15 \, a^{3} b x - 136 \, a^{4}\right )} \sqrt {-b^{2} x^{2} + a^{2}}}{120 \, b} \]

[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

Sympy [A] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.07 \[ \int (a+b x)^3 \sqrt {a^2-b^2 x^2} \, dx=\begin {cases} \frac {7 a^{5} \left (\begin {cases} \frac {\log {\left (- 2 b^{2} x + 2 \sqrt {- b^{2}} \sqrt {a^{2} - b^{2} x^{2}} \right )}}{\sqrt {- b^{2}}} & \text {for}\: a^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- b^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{8} + \sqrt {a^{2} - b^{2} x^{2}} \left (- \frac {17 a^{4}}{15 b} + \frac {a^{3} x}{8} + \frac {14 a^{2} b x^{2}}{15} + \frac {3 a b^{2} x^{3}}{4} + \frac {b^{3} x^{4}}{5}\right ) & \text {for}\: b^{2} \neq 0 \\\sqrt {a^{2}} \left (\begin {cases} a^{3} x & \text {for}\: b = 0 \\\frac {\left (a + b x\right )^{4}}{4 b} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((7*a**5*Piecewise((log(-2*b**2*x + 2*sqrt(-b**2)*sqrt(a**2 - b**2*x**2))/sqrt(-b**2), Ne(a**2, 0)),
(x*log(x)/sqrt(-b**2*x**2), True))/8 + sqrt(a**2 - b**2*x**2)*(-17*a**4/(15*b) + a**3*x/8 + 14*a**2*b*x**2/15
+ 3*a*b**2*x**3/4 + b**3*x**4/5), Ne(b**2, 0)), (sqrt(a**2)*Piecewise((a**3*x, Eq(b, 0)), ((a + b*x)**4/(4*b),
True)), True))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.69 \[ \int (a+b x)^3 \sqrt {a^2-b^2 x^2} \, dx=\frac {7 \, a^{5} \arcsin \left (\frac {b x}{a}\right )}{8 \, b} + \frac {7}{8} \, \sqrt {-b^{2} x^{2} + a^{2}} a^{3} x - \frac {1}{5} \, {\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}} b x^{2} - \frac {3}{4} \, {\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}} a x - \frac {17 \, {\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}} a^{2}}{15 \, b} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.58 \[ \int (a+b x)^3 \sqrt {a^2-b^2 x^2} \, dx=\frac {7 \, a^{5} \arcsin \left (\frac {b x}{a}\right ) \mathrm {sgn}\left (a\right ) \mathrm {sgn}\left (b\right )}{8 \, {\left | b \right |}} - \frac {1}{120} \, \sqrt {-b^{2} x^{2} + a^{2}} {\left (\frac {136 \, a^{4}}{b} - {\left (15 \, a^{3} + 2 \, {\left (56 \, a^{2} b + 3 \, {\left (4 \, b^{3} x + 15 \, a b^{2}\right )} x\right )} x\right )} x\right )} \]

[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)

Mupad [F(-1)]

Timed out. \[ \int (a+b x)^3 \sqrt {a^2-b^2 x^2} \, dx=\int \sqrt {a^2-b^2\,x^2}\,{\left (a+b\,x\right )}^3 \,d x \]

[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)