### 3.789 $$\int (a+b x)^2 (a^2-b^2 x^2)^{3/2} \, dx$$

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

[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)

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Rubi [A]  time = 0.0437812, antiderivative size = 131, normalized size of antiderivative = 1., 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} $\frac{7}{16} a^4 x \sqrt{a^2-b^2 x^2}+\frac{7}{24} a^2 x \left (a^2-b^2 x^2\right )^{3/2}-\frac{7 a \left (a^2-b^2 x^2\right )^{5/2}}{30 b}-\frac{(a+b x) \left (a^2-b^2 x^2\right )^{5/2}}{6 b}+\frac{7 a^6 \tan ^{-1}\left (\frac{b x}{\sqrt{a^2-b^2 x^2}}\right )}{16 b}$

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 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
]

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 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 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 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])

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*}

Mathematica [A]  time = 0.183362, size = 123, normalized size = 0.94 $\frac{\sqrt{a^2-b^2 x^2} \left (\sqrt{1-\frac{b^2 x^2}{a^2}} \left (192 a^3 b^2 x^2+10 a^2 b^3 x^3+135 a^4 b x-96 a^5-96 a b^4 x^4-40 b^5 x^5\right )+105 a^5 \sin ^{-1}\left (\frac{b x}{a}\right )\right )}{240 b \sqrt{1-\frac{b^2 x^2}{a^2}}}$

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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Maple [A]  time = 0.051, size = 111, normalized size = 0.9 \begin{align*} -{\frac{x}{6} \left ( -{b}^{2}{x}^{2}+{a}^{2} \right ) ^{{\frac{5}{2}}}}+{\frac{7\,{a}^{2}x}{24} \left ( -{b}^{2}{x}^{2}+{a}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{7\,{a}^{4}x}{16}\sqrt{-{b}^{2}{x}^{2}+{a}^{2}}}+{\frac{7\,{a}^{6}}{16}\arctan \left ({x\sqrt{{b}^{2}}{\frac{1}{\sqrt{-{b}^{2}{x}^{2}+{a}^{2}}}}} \right ){\frac{1}{\sqrt{{b}^{2}}}}}-{\frac{2\,a}{5\,b} \left ( -{b}^{2}{x}^{2}+{a}^{2} \right ) ^{{\frac{5}{2}}}} \end{align*}

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

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Maxima [A]  time = 1.6062, size = 139, normalized size = 1.06 \begin{align*} \frac{7 \, a^{6} \arcsin \left (\frac{b^{2} x}{\sqrt{a^{2} b^{2}}}\right )}{16 \, \sqrt{b^{2}}} + \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{align*}

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^2*x/sqrt(a^2*b^2))/sqrt(b^2) + 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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Fricas [A]  time = 2.08246, size = 231, normalized size = 1.76 \begin{align*} -\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{align*}

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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Sympy [C]  time = 9.46096, size = 498, normalized size = 3.8 \begin{align*} 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}\: \frac{\left |{b^{2} x^{2}}\right |}{\left |{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}\: \frac{\left |{b^{2} x^{2}}\right |}{\left |{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{align*}

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)/Abs(a**2) > 1), (a**2*asin(b*x/a)/(2*b) + a*x*sqrt(1 - b**2*x**2/a**2)/2, Tru
e)) + 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*s
qrt(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)/Abs(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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Giac [A]  time = 1.24505, size = 124, normalized size = 0.95 \begin{align*} \frac{7 \, 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{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{align*}

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)