\(\int (a+b x)^4 \sqrt {a^2-b^2 x^2} \, dx\) [777]

Optimal result

Integrand size = 24, antiderivative size = 173 \[ \int (a+b x)^4 \sqrt {a^2-b^2 x^2} \, dx=\frac {21}{16} a^4 x \sqrt {a^2-b^2 x^2}-\frac {7 a^3 \left (a^2-b^2 x^2\right )^{3/2}}{8 b}-\frac {21 a^2 (a+b x) \left (a^2-b^2 x^2\right )^{3/2}}{40 b}-\frac {3 a (a+b x)^2 \left (a^2-b^2 x^2\right )^{3/2}}{10 b}-\frac {(a+b x)^3 \left (a^2-b^2 x^2\right )^{3/2}}{6 b}+\frac {21 a^6 \arctan \left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{16 b} \]

[Out]

-7/8*a^3*(-b^2*x^2+a^2)^(3/2)/b-21/40*a^2*(b*x+a)*(-b^2*x^2+a^2)^(3/2)/b-3/10*a*(b*x+a)^2*(-b^2*x^2+a^2)^(3/2)
/b-1/6*(b*x+a)^3*(-b^2*x^2+a^2)^(3/2)/b+21/16*a^6*arctan(b*x/(-b^2*x^2+a^2)^(1/2))/b+21/16*a^4*x*(-b^2*x^2+a^2
)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {685, 655, 201, 223, 209} \[ \int (a+b x)^4 \sqrt {a^2-b^2 x^2} \, dx=-\frac {21 a^2 (a+b x) \left (a^2-b^2 x^2\right )^{3/2}}{40 b}-\frac {3 a (a+b x)^2 \left (a^2-b^2 x^2\right )^{3/2}}{10 b}-\frac {(a+b x)^3 \left (a^2-b^2 x^2\right )^{3/2}}{6 b}+\frac {21 a^6 \arctan \left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{16 b}+\frac {21}{16} a^4 x \sqrt {a^2-b^2 x^2}-\frac {7 a^3 \left (a^2-b^2 x^2\right )^{3/2}}{8 b} \]

[In]

Int[(a + b*x)^4*Sqrt[a^2 - b^2*x^2],x]

[Out]

(21*a^4*x*Sqrt[a^2 - b^2*x^2])/16 - (7*a^3*(a^2 - b^2*x^2)^(3/2))/(8*b) - (21*a^2*(a + b*x)*(a^2 - b^2*x^2)^(3
/2))/(40*b) - (3*a*(a + b*x)^2*(a^2 - b^2*x^2)^(3/2))/(10*b) - ((a + b*x)^3*(a^2 - b^2*x^2)^(3/2))/(6*b) + (21
*a^6*ArcTan[(b*x)/Sqrt[a^2 - b^2*x^2]])/(16*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)^3 \left (a^2-b^2 x^2\right )^{3/2}}{6 b}+\frac {1}{2} (3 a) \int (a+b x)^3 \sqrt {a^2-b^2 x^2} \, dx \\ & = -\frac {3 a (a+b x)^2 \left (a^2-b^2 x^2\right )^{3/2}}{10 b}-\frac {(a+b x)^3 \left (a^2-b^2 x^2\right )^{3/2}}{6 b}+\frac {1}{10} \left (21 a^2\right ) \int (a+b x)^2 \sqrt {a^2-b^2 x^2} \, dx \\ & = -\frac {21 a^2 (a+b x) \left (a^2-b^2 x^2\right )^{3/2}}{40 b}-\frac {3 a (a+b x)^2 \left (a^2-b^2 x^2\right )^{3/2}}{10 b}-\frac {(a+b x)^3 \left (a^2-b^2 x^2\right )^{3/2}}{6 b}+\frac {1}{8} \left (21 a^3\right ) \int (a+b x) \sqrt {a^2-b^2 x^2} \, dx \\ & = -\frac {7 a^3 \left (a^2-b^2 x^2\right )^{3/2}}{8 b}-\frac {21 a^2 (a+b x) \left (a^2-b^2 x^2\right )^{3/2}}{40 b}-\frac {3 a (a+b x)^2 \left (a^2-b^2 x^2\right )^{3/2}}{10 b}-\frac {(a+b x)^3 \left (a^2-b^2 x^2\right )^{3/2}}{6 b}+\frac {1}{8} \left (21 a^4\right ) \int \sqrt {a^2-b^2 x^2} \, dx \\ & = \frac {21}{16} a^4 x \sqrt {a^2-b^2 x^2}-\frac {7 a^3 \left (a^2-b^2 x^2\right )^{3/2}}{8 b}-\frac {21 a^2 (a+b x) \left (a^2-b^2 x^2\right )^{3/2}}{40 b}-\frac {3 a (a+b x)^2 \left (a^2-b^2 x^2\right )^{3/2}}{10 b}-\frac {(a+b x)^3 \left (a^2-b^2 x^2\right )^{3/2}}{6 b}+\frac {1}{16} \left (21 a^6\right ) \int \frac {1}{\sqrt {a^2-b^2 x^2}} \, dx \\ & = \frac {21}{16} a^4 x \sqrt {a^2-b^2 x^2}-\frac {7 a^3 \left (a^2-b^2 x^2\right )^{3/2}}{8 b}-\frac {21 a^2 (a+b x) \left (a^2-b^2 x^2\right )^{3/2}}{40 b}-\frac {3 a (a+b x)^2 \left (a^2-b^2 x^2\right )^{3/2}}{10 b}-\frac {(a+b x)^3 \left (a^2-b^2 x^2\right )^{3/2}}{6 b}+\frac {1}{16} \left (21 a^6\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 {21}{16} a^4 x \sqrt {a^2-b^2 x^2}-\frac {7 a^3 \left (a^2-b^2 x^2\right )^{3/2}}{8 b}-\frac {21 a^2 (a+b x) \left (a^2-b^2 x^2\right )^{3/2}}{40 b}-\frac {3 a (a+b x)^2 \left (a^2-b^2 x^2\right )^{3/2}}{10 b}-\frac {(a+b x)^3 \left (a^2-b^2 x^2\right )^{3/2}}{6 b}+\frac {21 a^6 \tan ^{-1}\left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{16 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.66 \[ \int (a+b x)^4 \sqrt {a^2-b^2 x^2} \, dx=\frac {\sqrt {a^2-b^2 x^2} \left (-448 a^5-75 a^4 b x+256 a^3 b^2 x^2+350 a^2 b^3 x^3+192 a b^4 x^4+40 b^5 x^5\right )-630 a^6 \arctan \left (\frac {b x}{\sqrt {a^2}-\sqrt {a^2-b^2 x^2}}\right )}{240 b} \]

[In]

Integrate[(a + b*x)^4*Sqrt[a^2 - b^2*x^2],x]

[Out]

(Sqrt[a^2 - b^2*x^2]*(-448*a^5 - 75*a^4*b*x + 256*a^3*b^2*x^2 + 350*a^2*b^3*x^3 + 192*a*b^4*x^4 + 40*b^5*x^5)
- 630*a^6*ArcTan[(b*x)/(Sqrt[a^2] - Sqrt[a^2 - b^2*x^2])])/(240*b)

Maple [A] (verified)

Time = 2.30 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.61

method	result	size
risch	\(-\frac {\left (-40 b^{5} x^{5}-192 a \,b^{4} x^{4}-350 a^{2} b^{3} x^{3}-256 a^{3} b^{2} x^{2}+75 a^{4} b x +448 a^{5}\right ) \sqrt {-b^{2} x^{2}+a^{2}}}{240 b}+\frac {21 a^{6} \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-b^{2} x^{2}+a^{2}}}\right )}{16 \sqrt {b^{2}}}\)	\(105\)
default	\(a^{4} \left (\frac {x \sqrt {-b^{2} x^{2}+a^{2}}}{2}+\frac {a^{2} \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-b^{2} x^{2}+a^{2}}}\right )}{2 \sqrt {b^{2}}}\right )+b^{4} \left (-\frac {x^{3} \left (-b^{2} x^{2}+a^{2}\right )^{\frac {3}{2}}}{6 b^{2}}+\frac {a^{2} \left (-\frac {x \left (-b^{2} x^{2}+a^{2}\right )^{\frac {3}{2}}}{4 b^{2}}+\frac {a^{2} \left (\frac {x \sqrt {-b^{2} x^{2}+a^{2}}}{2}+\frac {a^{2} \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-b^{2} x^{2}+a^{2}}}\right )}{2 \sqrt {b^{2}}}\right )}{4 b^{2}}\right )}{2 b^{2}}\right )+4 a \,b^{3} \left (-\frac {x^{2} \left (-b^{2} x^{2}+a^{2}\right )^{\frac {3}{2}}}{5 b^{2}}-\frac {2 a^{2} \left (-b^{2} x^{2}+a^{2}\right )^{\frac {3}{2}}}{15 b^{4}}\right )+6 a^{2} b^{2} \left (-\frac {x \left (-b^{2} x^{2}+a^{2}\right )^{\frac {3}{2}}}{4 b^{2}}+\frac {a^{2} \left (\frac {x \sqrt {-b^{2} x^{2}+a^{2}}}{2}+\frac {a^{2} \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-b^{2} x^{2}+a^{2}}}\right )}{2 \sqrt {b^{2}}}\right )}{4 b^{2}}\right )-\frac {4 a^{3} \left (-b^{2} x^{2}+a^{2}\right )^{\frac {3}{2}}}{3 b}\)	\(330\)

[In]

int((b*x+a)^4*(-b^2*x^2+a^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/240*(-40*b^5*x^5-192*a*b^4*x^4-350*a^2*b^3*x^3-256*a^3*b^2*x^2+75*a^4*b*x+448*a^5)/b*(-b^2*x^2+a^2)^(1/2)+2
1/16*a^6/(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 = 0.28 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.61 \[ \int (a+b x)^4 \sqrt {a^2-b^2 x^2} \, dx=-\frac {630 \, a^{6} \arctan \left (-\frac {a - \sqrt {-b^{2} x^{2} + a^{2}}}{b x}\right ) - {\left (40 \, b^{5} x^{5} + 192 \, a b^{4} x^{4} + 350 \, a^{2} b^{3} x^{3} + 256 \, a^{3} b^{2} x^{2} - 75 \, a^{4} b x - 448 \, a^{5}\right )} \sqrt {-b^{2} x^{2} + a^{2}}}{240 \, b} \]

[In]

integrate((b*x+a)^4*(-b^2*x^2+a^2)^(1/2),x, algorithm="fricas")

[Out]

-1/240*(630*a^6*arctan(-(a - sqrt(-b^2*x^2 + a^2))/(b*x)) - (40*b^5*x^5 + 192*a*b^4*x^4 + 350*a^2*b^3*x^3 + 25
6*a^3*b^2*x^2 - 75*a^4*b*x - 448*a^5)*sqrt(-b^2*x^2 + a^2))/b

Sympy [A] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.95 \[ \int (a+b x)^4 \sqrt {a^2-b^2 x^2} \, dx=\begin {cases} \frac {21 a^{6} \left (\begin {cases} \frac {\log {\left (- 2 b^{2} x + 2 \sqrt {- b^{2}} \sqrt {a^{2} - b^{2} x^{2}} \right )}}{\sqrt {- b^{2}}} & \text {for}\: a^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- b^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{16} + \sqrt {a^{2} - b^{2} x^{2}} \left (- \frac {28 a^{5}}{15 b} - \frac {5 a^{4} x}{16} + \frac {16 a^{3} b x^{2}}{15} + \frac {35 a^{2} b^{2} x^{3}}{24} + \frac {4 a b^{3} x^{4}}{5} + \frac {b^{4} x^{5}}{6}\right ) & \text {for}\: b^{2} \neq 0 \\\sqrt {a^{2}} \left (\begin {cases} a^{4} x & \text {for}\: b = 0 \\\frac {\left (a + b x\right )^{5}}{5 b} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate((b*x+a)**4*(-b**2*x**2+a**2)**(1/2),x)

[Out]

Piecewise((21*a**6*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))/16 + sqrt(a**2 - b**2*x**2)*(-28*a**5/(15*b) - 5*a**4*x/16 + 16*a**3*b*x**
2/15 + 35*a**2*b**2*x**3/24 + 4*a*b**3*x**4/5 + b**4*x**5/6), Ne(b**2, 0)), (sqrt(a**2)*Piecewise((a**4*x, Eq(
b, 0)), ((a + b*x)**5/(5*b), True)), True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.70 \[ \int (a+b x)^4 \sqrt {a^2-b^2 x^2} \, dx=-\frac {1}{6} \, {\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}} b^{2} x^{3} + \frac {21 \, a^{6} \arcsin \left (\frac {b x}{a}\right )}{16 \, b} + \frac {21}{16} \, \sqrt {-b^{2} x^{2} + a^{2}} a^{4} x - \frac {4}{5} \, {\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}} a b x^{2} - \frac {13}{8} \, {\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}} a^{2} x - \frac {28 \, {\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}} a^{3}}{15 \, b} \]

[In]

integrate((b*x+a)^4*(-b^2*x^2+a^2)^(1/2),x, algorithm="maxima")

[Out]

-1/6*(-b^2*x^2 + a^2)^(3/2)*b^2*x^3 + 21/16*a^6*arcsin(b*x/a)/b + 21/16*sqrt(-b^2*x^2 + a^2)*a^4*x - 4/5*(-b^2
*x^2 + a^2)^(3/2)*a*b*x^2 - 13/8*(-b^2*x^2 + a^2)^(3/2)*a^2*x - 28/15*(-b^2*x^2 + a^2)^(3/2)*a^3/b

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.53 \[ \int (a+b x)^4 \sqrt {a^2-b^2 x^2} \, dx=\frac {21 \, a^{6} \arcsin \left (\frac {b x}{a}\right ) \mathrm {sgn}\left (a\right ) \mathrm {sgn}\left (b\right )}{16 \, {\left | b \right |}} - \frac {1}{240} \, {\left (\frac {448 \, a^{5}}{b} + {\left (75 \, a^{4} - 2 \, {\left (128 \, a^{3} b + {\left (175 \, a^{2} b^{2} + 4 \, {\left (5 \, b^{4} x + 24 \, a b^{3}\right )} x\right )} x\right )} x\right )} x\right )} \sqrt {-b^{2} x^{2} + a^{2}} \]

[In]

integrate((b*x+a)^4*(-b^2*x^2+a^2)^(1/2),x, algorithm="giac")

[Out]

21/16*a^6*arcsin(b*x/a)*sgn(a)*sgn(b)/abs(b) - 1/240*(448*a^5/b + (75*a^4 - 2*(128*a^3*b + (175*a^2*b^2 + 4*(5
*b^4*x + 24*a*b^3)*x)*x)*x)*x)*sqrt(-b^2*x^2 + a^2)

Mupad [F(-1)]

Timed out. \[ \int (a+b x)^4 \sqrt {a^2-b^2 x^2} \, dx=\int \sqrt {a^2-b^2\,x^2}\,{\left (a+b\,x\right )}^4 \,d x \]

[In]

int((a^2 - b^2*x^2)^(1/2)*(a + b*x)^4,x)

[Out]

int((a^2 - b^2*x^2)^(1/2)*(a + b*x)^4, x)