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

Optimal. Leaf size=131 \[ \frac {7}{24} a^2 x \left (a^2-b^2 x^2\right )^{3/2}-\frac {7 a \left (a^2-b^2 x^2\right )^{5/2}}{30 b}-\frac {(a+b x) \left (a^2-b^2 x^2\right )^{5/2}}{6 b}+\frac {7 a^6 \tan ^{-1}\left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{16 b}+\frac {7}{16} a^4 x \sqrt {a^2-b^2 x^2} \]

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Rubi [A]  time = 0.04, antiderivative size = 131, 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} \begin {gather*} \frac {7}{16} a^4 x \sqrt {a^2-b^2 x^2}+\frac {7}{24} a^2 x \left (a^2-b^2 x^2\right )^{3/2}-\frac {7 a \left (a^2-b^2 x^2\right )^{5/2}}{30 b}-\frac {(a+b x) \left (a^2-b^2 x^2\right )^{5/2}}{6 b}+\frac {7 a^6 \tan ^{-1}\left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{16 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7*a^4*x*Sqrt[a^2 - b^2*x^2])/16 + (7*a^2*x*(a^2 - b^2*x^2)^(3/2))/24 - (7*a*(a^2 - b^2*x^2)^(5/2))/(30*b) - (
(a + b*x)*(a^2 - b^2*x^2)^(5/2))/(6*b) + (7*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)^2 \left (a^2-b^2 x^2\right )^{3/2} \, dx &=-\frac {(a+b x) \left (a^2-b^2 x^2\right )^{5/2}}{6 b}+\frac {1}{6} (7 a) \int (a+b x) \left (a^2-b^2 x^2\right )^{3/2} \, dx\\ &=-\frac {7 a \left (a^2-b^2 x^2\right )^{5/2}}{30 b}-\frac {(a+b x) \left (a^2-b^2 x^2\right )^{5/2}}{6 b}+\frac {1}{6} \left (7 a^2\right ) \int \left (a^2-b^2 x^2\right )^{3/2} \, dx\\ &=\frac {7}{24} a^2 x \left (a^2-b^2 x^2\right )^{3/2}-\frac {7 a \left (a^2-b^2 x^2\right )^{5/2}}{30 b}-\frac {(a+b x) \left (a^2-b^2 x^2\right )^{5/2}}{6 b}+\frac {1}{8} \left (7 a^4\right ) \int \sqrt {a^2-b^2 x^2} \, dx\\ &=\frac {7}{16} a^4 x \sqrt {a^2-b^2 x^2}+\frac {7}{24} a^2 x \left (a^2-b^2 x^2\right )^{3/2}-\frac {7 a \left (a^2-b^2 x^2\right )^{5/2}}{30 b}-\frac {(a+b x) \left (a^2-b^2 x^2\right )^{5/2}}{6 b}+\frac {1}{16} \left (7 a^6\right ) \int \frac {1}{\sqrt {a^2-b^2 x^2}} \, dx\\ &=\frac {7}{16} a^4 x \sqrt {a^2-b^2 x^2}+\frac {7}{24} a^2 x \left (a^2-b^2 x^2\right )^{3/2}-\frac {7 a \left (a^2-b^2 x^2\right )^{5/2}}{30 b}-\frac {(a+b x) \left (a^2-b^2 x^2\right )^{5/2}}{6 b}+\frac {1}{16} \left (7 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 {7}{16} a^4 x \sqrt {a^2-b^2 x^2}+\frac {7}{24} a^2 x \left (a^2-b^2 x^2\right )^{3/2}-\frac {7 a \left (a^2-b^2 x^2\right )^{5/2}}{30 b}-\frac {(a+b x) \left (a^2-b^2 x^2\right )^{5/2}}{6 b}+\frac {7 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 = 0.18, size = 123, normalized size = 0.94 \begin {gather*} \frac {\sqrt {a^2-b^2 x^2} \left (105 a^5 \sin ^{-1}\left (\frac {b x}{a}\right )+\sqrt {1-\frac {b^2 x^2}{a^2}} \left (-96 a^5+135 a^4 b x+192 a^3 b^2 x^2+10 a^2 b^3 x^3-96 a b^4 x^4-40 b^5 x^5\right )\right )}{240 b \sqrt {1-\frac {b^2 x^2}{a^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a^2 - b^2*x^2]*(Sqrt[1 - (b^2*x^2)/a^2]*(-96*a^5 + 135*a^4*b*x + 192*a^3*b^2*x^2 + 10*a^2*b^3*x^3 - 96*a
*b^4*x^4 - 40*b^5*x^5) + 105*a^5*ArcSin[(b*x)/a]))/(240*b*Sqrt[1 - (b^2*x^2)/a^2])

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IntegrateAlgebraic [A]  time = 0.40, size = 125, normalized size = 0.95 \begin {gather*} \frac {7 a^6 \sqrt {-b^2} \log \left (\sqrt {a^2-b^2 x^2}-\sqrt {-b^2} x\right )}{16 b^2}+\frac {\sqrt {a^2-b^2 x^2} \left (-96 a^5+135 a^4 b x+192 a^3 b^2 x^2+10 a^2 b^3 x^3-96 a b^4 x^4-40 b^5 x^5\right )}{240 b} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)^2*(a^2 - b^2*x^2)^(3/2),x]

[Out]

(Sqrt[a^2 - b^2*x^2]*(-96*a^5 + 135*a^4*b*x + 192*a^3*b^2*x^2 + 10*a^2*b^3*x^3 - 96*a*b^4*x^4 - 40*b^5*x^5))/(
240*b) + (7*a^6*Sqrt[-b^2]*Log[-(Sqrt[-b^2]*x) + Sqrt[a^2 - b^2*x^2]])/(16*b^2)

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fricas [A]  time = 0.41, size = 105, normalized size = 0.80 \begin {gather*} -\frac {210 \, a^{6} \arctan \left (-\frac {a - \sqrt {-b^{2} x^{2} + a^{2}}}{b x}\right ) + {\left (40 \, b^{5} x^{5} + 96 \, a b^{4} x^{4} - 10 \, a^{2} b^{3} x^{3} - 192 \, a^{3} b^{2} x^{2} - 135 \, a^{4} b x + 96 \, a^{5}\right )} \sqrt {-b^{2} x^{2} + a^{2}}}{240 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*(210*a^6*arctan(-(a - sqrt(-b^2*x^2 + a^2))/(b*x)) + (40*b^5*x^5 + 96*a*b^4*x^4 - 10*a^2*b^3*x^3 - 192*
a^3*b^2*x^2 - 135*a^4*b*x + 96*a^5)*sqrt(-b^2*x^2 + a^2))/b

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giac [A]  time = 0.22, size = 92, normalized size = 0.70 \begin {gather*} \frac {7 \, 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 {96 \, a^{5}}{b} - {\left (135 \, a^{4} + 2 \, {\left (96 \, a^{3} b + {\left (5 \, a^{2} b^{2} - 4 \, {\left (5 \, b^{4} x + 12 \, a b^{3}\right )} x\right )} x\right )} x\right )} x\right )} \sqrt {-b^{2} x^{2} + a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7/16*a^6*arcsin(b*x/a)*sgn(a)*sgn(b)/abs(b) - 1/240*(96*a^5/b - (135*a^4 + 2*(96*a^3*b + (5*a^2*b^2 - 4*(5*b^4
*x + 12*a*b^3)*x)*x)*x)*x)*sqrt(-b^2*x^2 + a^2)

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maple [A]  time = 0.05, size = 111, normalized size = 0.85 \begin {gather*} \frac {7 a^{6} \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-b^{2} x^{2}+a^{2}}}\right )}{16 \sqrt {b^{2}}}+\frac {7 \sqrt {-b^{2} x^{2}+a^{2}}\, a^{4} x}{16}+\frac {7 \left (-b^{2} x^{2}+a^{2}\right )^{\frac {3}{2}} a^{2} x}{24}-\frac {\left (-b^{2} x^{2}+a^{2}\right )^{\frac {5}{2}} x}{6}-\frac {2 \left (-b^{2} x^{2}+a^{2}\right )^{\frac {5}{2}} a}{5 b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*x*(-b^2*x^2+a^2)^(5/2)+7/24*(-b^2*x^2+a^2)^(3/2)*a^2*x+7/16*(-b^2*x^2+a^2)^(1/2)*a^4*x+7/16/(b^2)^(1/2)*a
^6*arctan((b^2)^(1/2)/(-b^2*x^2+a^2)^(1/2)*x)-2/5*a*(-b^2*x^2+a^2)^(5/2)/b

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maxima [A]  time = 3.01, size = 93, normalized size = 0.71 \begin {gather*} \frac {7 \, a^{6} \arcsin \left (\frac {b x}{a}\right )}{16 \, b} + \frac {7}{16} \, \sqrt {-b^{2} x^{2} + a^{2}} a^{4} x + \frac {7}{24} \, {\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}} a^{2} x - \frac {1}{6} \, {\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {5}{2}} x - \frac {2 \, {\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {5}{2}} a}{5 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7/16*a^6*arcsin(b*x/a)/b + 7/16*sqrt(-b^2*x^2 + a^2)*a^4*x + 7/24*(-b^2*x^2 + a^2)^(3/2)*a^2*x - 1/6*(-b^2*x^2
+ a^2)^(5/2)*x - 2/5*(-b^2*x^2 + a^2)^(5/2)*a/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 11.12, size = 495, normalized size = 3.78 \begin {gather*} 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 ) + 2 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 ) - 2 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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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