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

Optimal. Leaf size=173 \[ -\frac {21 a^2 (a+b x) \left (a^2-b^2 x^2\right )^{3/2}}{40 b}-\frac {3 a (a+b x)^2 \left (a^2-b^2 x^2\right )^{3/2}}{10 b}-\frac {(a+b x)^3 \left (a^2-b^2 x^2\right )^{3/2}}{6 b}+\frac {21 a^6 \tan ^{-1}\left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{16 b}+\frac {21}{16} a^4 x \sqrt {a^2-b^2 x^2}-\frac {7 a^3 \left (a^2-b^2 x^2\right )^{3/2}}{8 b} \]

[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)

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Rubi [A]  time = 0.08, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ \frac {21}{16} a^4 x \sqrt {a^2-b^2 x^2}-\frac {7 a^3 \left (a^2-b^2 x^2\right )^{3/2}}{8 b}-\frac {21 a^2 (a+b x) \left (a^2-b^2 x^2\right )^{3/2}}{40 b}-\frac {3 a (a+b x)^2 \left (a^2-b^2 x^2\right )^{3/2}}{10 b}-\frac {(a+b x)^3 \left (a^2-b^2 x^2\right )^{3/2}}{6 b}+\frac {21 a^6 \tan ^{-1}\left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{16 b} \]

Antiderivative was successfully verified.

[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 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)^4 \sqrt {a^2-b^2 x^2} \, dx &=-\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 ) \operatorname {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*}

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Mathematica [A]  time = 0.21, size = 123, normalized size = 0.71 \[ \frac {\sqrt {a^2-b^2 x^2} \left (315 a^5 \sin ^{-1}\left (\frac {b x}{a}\right )+\sqrt {1-\frac {b^2 x^2}{a^2}} \left (-448 a^5-75 a^4 b x+256 a^3 b^2 x^2+350 a^2 b^3 x^3+192 a b^4 x^4+40 b^5 x^5\right )\right )}{240 b \sqrt {1-\frac {b^2 x^2}{a^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a^2 - b^2*x^2]*(Sqrt[1 - (b^2*x^2)/a^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) + 315*a^5*ArcSin[(b*x)/a]))/(240*b*Sqrt[1 - (b^2*x^2)/a^2])

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fricas [A]  time = 0.89, size = 106, normalized size = 0.61 \[ -\frac {630 \, a^{6} \arctan \left (-\frac {a - \sqrt {-b^{2} x^{2} + a^{2}}}{b x}\right ) - {\left (40 \, b^{5} x^{5} + 192 \, a b^{4} x^{4} + 350 \, a^{2} b^{3} x^{3} + 256 \, a^{3} b^{2} x^{2} - 75 \, a^{4} b x - 448 \, a^{5}\right )} \sqrt {-b^{2} x^{2} + a^{2}}}{240 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.20, size = 91, normalized size = 0.53 \[ \frac {21 \, a^{6} \arcsin \left (\frac {b x}{a}\right ) \mathrm {sgn}\relax (a) \mathrm {sgn}\relax (b)}{16 \, {\left | b \right |}} - \frac {1}{240} \, {\left (\frac {448 \, a^{5}}{b} + {\left (75 \, a^{4} - 2 \, {\left (128 \, a^{3} b + {\left (175 \, a^{2} b^{2} + 4 \, {\left (5 \, b^{4} x + 24 \, a b^{3}\right )} x\right )} x\right )} x\right )} x\right )} \sqrt {-b^{2} x^{2} + a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.05, size = 139, normalized size = 0.80 \[ \frac {21 a^{6} \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-b^{2} x^{2}+a^{2}}}\right )}{16 \sqrt {b^{2}}}+\frac {21 \sqrt {-b^{2} x^{2}+a^{2}}\, a^{4} x}{16}-\frac {\left (-b^{2} x^{2}+a^{2}\right )^{\frac {3}{2}} b^{2} x^{3}}{6}-\frac {4 \left (-b^{2} x^{2}+a^{2}\right )^{\frac {3}{2}} a b \,x^{2}}{5}-\frac {13 \left (-b^{2} x^{2}+a^{2}\right )^{\frac {3}{2}} a^{2} x}{8}-\frac {28 \left (-b^{2} x^{2}+a^{2}\right )^{\frac {3}{2}} a^{3}}{15 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.98, size = 121, normalized size = 0.70 \[ -\frac {1}{6} \, {\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}} b^{2} x^{3} + \frac {21 \, a^{6} \arcsin \left (\frac {b x}{a}\right )}{16 \, b} + \frac {21}{16} \, \sqrt {-b^{2} x^{2} + a^{2}} a^{4} x - \frac {4}{5} \, {\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}} a b x^{2} - \frac {13}{8} \, {\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}} a^{2} x - \frac {28 \, {\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}} a^{3}}{15 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {a^2-b^2\,x^2}\,{\left (a+b\,x\right )}^4 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 14.61, size = 700, normalized size = 4.05 \[ a^{4} \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 ) + 4 a^{3} 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 ) + 6 a^{2} 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 ) + 4 a 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 ) + b^{4} \left (\begin {cases} - \frac {i a^{6} \operatorname {acosh}{\left (\frac {b x}{a} \right )}}{16 b^{5}} + \frac {i a^{5} x}{16 b^{4} \sqrt {-1 + \frac {b^{2} x^{2}}{a^{2}}}} - \frac {i a^{3} x^{3}}{48 b^{2} \sqrt {-1 + \frac {b^{2} x^{2}}{a^{2}}}} - \frac {5 i a x^{5}}{24 \sqrt {-1 + \frac {b^{2} x^{2}}{a^{2}}}} + \frac {i b^{2} x^{7}}{6 a \sqrt {-1 + \frac {b^{2} x^{2}}{a^{2}}}} & \text {for}\: \left |{\frac {b^{2} x^{2}}{a^{2}}}\right | > 1 \\\frac {a^{6} \operatorname {asin}{\left (\frac {b x}{a} \right )}}{16 b^{5}} - \frac {a^{5} x}{16 b^{4} \sqrt {1 - \frac {b^{2} x^{2}}{a^{2}}}} + \frac {a^{3} x^{3}}{48 b^{2} \sqrt {1 - \frac {b^{2} x^{2}}{a^{2}}}} + \frac {5 a x^{5}}{24 \sqrt {1 - \frac {b^{2} x^{2}}{a^{2}}}} - \frac {b^{2} x^{7}}{6 a \sqrt {1 - \frac {b^{2} x^{2}}{a^{2}}}} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**4*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)) +
4*a**3*b*Piecewise((x**2*sqrt(a**2)/2, Eq(b**2, 0)), (-(a**2 - b**2*x**2)**(3/2)/(3*b**2), True)) + 6*a**2*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)) + 4*a*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)) + b*
*4*Piecewise((-I*a**6*acosh(b*x/a)/(16*b**5) + I*a**5*x/(16*b**4*sqrt(-1 + b**2*x**2/a**2)) - I*a**3*x**3/(48*
b**2*sqrt(-1 + b**2*x**2/a**2)) - 5*I*a*x**5/(24*sqrt(-1 + b**2*x**2/a**2)) + I*b**2*x**7/(6*a*sqrt(-1 + b**2*
x**2/a**2)), Abs(b**2*x**2/a**2) > 1), (a**6*asin(b*x/a)/(16*b**5) - a**5*x/(16*b**4*sqrt(1 - b**2*x**2/a**2))
+ a**3*x**3/(48*b**2*sqrt(1 - b**2*x**2/a**2)) + 5*a*x**5/(24*sqrt(1 - b**2*x**2/a**2)) - b**2*x**7/(6*a*sqrt
(1 - b**2*x**2/a**2)), True))

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