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

Optimal. Leaf size=164 \[ \frac{9}{16} a^5 x \sqrt{a^2-b^2 x^2}+\frac{3}{8} a^3 x \left (a^2-b^2 x^2\right )^{3/2}-\frac{3 a^2 \left (a^2-b^2 x^2\right )^{5/2}}{10 b}-\frac{3 a (a+b x) \left (a^2-b^2 x^2\right )^{5/2}}{14 b}-\frac{(a+b x)^2 \left (a^2-b^2 x^2\right )^{5/2}}{7 b}+\frac{9 a^7 \tan ^{-1}\left (\frac{b x}{\sqrt{a^2-b^2 x^2}}\right )}{16 b} \]

[Out]

(9*a^5*x*Sqrt[a^2 - b^2*x^2])/16 + (3*a^3*x*(a^2 - b^2*x^2)^(3/2))/8 - (3*a^2*(a^2 - b^2*x^2)^(5/2))/(10*b) -
(3*a*(a + b*x)*(a^2 - b^2*x^2)^(5/2))/(14*b) - ((a + b*x)^2*(a^2 - b^2*x^2)^(5/2))/(7*b) + (9*a^7*ArcTan[(b*x)
/Sqrt[a^2 - b^2*x^2]])/(16*b)

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Rubi [A]  time = 0.0629918, antiderivative size = 164, normalized size of antiderivative = 1., 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} \[ \frac{9}{16} a^5 x \sqrt{a^2-b^2 x^2}+\frac{3}{8} a^3 x \left (a^2-b^2 x^2\right )^{3/2}-\frac{3 a^2 \left (a^2-b^2 x^2\right )^{5/2}}{10 b}-\frac{3 a (a+b x) \left (a^2-b^2 x^2\right )^{5/2}}{14 b}-\frac{(a+b x)^2 \left (a^2-b^2 x^2\right )^{5/2}}{7 b}+\frac{9 a^7 \tan ^{-1}\left (\frac{b x}{\sqrt{a^2-b^2 x^2}}\right )}{16 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(9*a^5*x*Sqrt[a^2 - b^2*x^2])/16 + (3*a^3*x*(a^2 - b^2*x^2)^(3/2))/8 - (3*a^2*(a^2 - b^2*x^2)^(5/2))/(10*b) -
(3*a*(a + b*x)*(a^2 - b^2*x^2)^(5/2))/(14*b) - ((a + b*x)^2*(a^2 - b^2*x^2)^(5/2))/(7*b) + (9*a^7*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)^3 \left (a^2-b^2 x^2\right )^{3/2} \, dx &=-\frac{(a+b x)^2 \left (a^2-b^2 x^2\right )^{5/2}}{7 b}+\frac{1}{7} (9 a) \int (a+b x)^2 \left (a^2-b^2 x^2\right )^{3/2} \, dx\\ &=-\frac{3 a (a+b x) \left (a^2-b^2 x^2\right )^{5/2}}{14 b}-\frac{(a+b x)^2 \left (a^2-b^2 x^2\right )^{5/2}}{7 b}+\frac{1}{2} \left (3 a^2\right ) \int (a+b x) \left (a^2-b^2 x^2\right )^{3/2} \, dx\\ &=-\frac{3 a^2 \left (a^2-b^2 x^2\right )^{5/2}}{10 b}-\frac{3 a (a+b x) \left (a^2-b^2 x^2\right )^{5/2}}{14 b}-\frac{(a+b x)^2 \left (a^2-b^2 x^2\right )^{5/2}}{7 b}+\frac{1}{2} \left (3 a^3\right ) \int \left (a^2-b^2 x^2\right )^{3/2} \, dx\\ &=\frac{3}{8} a^3 x \left (a^2-b^2 x^2\right )^{3/2}-\frac{3 a^2 \left (a^2-b^2 x^2\right )^{5/2}}{10 b}-\frac{3 a (a+b x) \left (a^2-b^2 x^2\right )^{5/2}}{14 b}-\frac{(a+b x)^2 \left (a^2-b^2 x^2\right )^{5/2}}{7 b}+\frac{1}{8} \left (9 a^5\right ) \int \sqrt{a^2-b^2 x^2} \, dx\\ &=\frac{9}{16} a^5 x \sqrt{a^2-b^2 x^2}+\frac{3}{8} a^3 x \left (a^2-b^2 x^2\right )^{3/2}-\frac{3 a^2 \left (a^2-b^2 x^2\right )^{5/2}}{10 b}-\frac{3 a (a+b x) \left (a^2-b^2 x^2\right )^{5/2}}{14 b}-\frac{(a+b x)^2 \left (a^2-b^2 x^2\right )^{5/2}}{7 b}+\frac{1}{16} \left (9 a^7\right ) \int \frac{1}{\sqrt{a^2-b^2 x^2}} \, dx\\ &=\frac{9}{16} a^5 x \sqrt{a^2-b^2 x^2}+\frac{3}{8} a^3 x \left (a^2-b^2 x^2\right )^{3/2}-\frac{3 a^2 \left (a^2-b^2 x^2\right )^{5/2}}{10 b}-\frac{3 a (a+b x) \left (a^2-b^2 x^2\right )^{5/2}}{14 b}-\frac{(a+b x)^2 \left (a^2-b^2 x^2\right )^{5/2}}{7 b}+\frac{1}{16} \left (9 a^7\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{9}{16} a^5 x \sqrt{a^2-b^2 x^2}+\frac{3}{8} a^3 x \left (a^2-b^2 x^2\right )^{3/2}-\frac{3 a^2 \left (a^2-b^2 x^2\right )^{5/2}}{10 b}-\frac{3 a (a+b x) \left (a^2-b^2 x^2\right )^{5/2}}{14 b}-\frac{(a+b x)^2 \left (a^2-b^2 x^2\right )^{5/2}}{7 b}+\frac{9 a^7 \tan ^{-1}\left (\frac{b x}{\sqrt{a^2-b^2 x^2}}\right )}{16 b}\\ \end{align*}

Mathematica [A]  time = 0.224929, size = 134, normalized size = 0.82 \[ \frac{\sqrt{a^2-b^2 x^2} \left (\sqrt{1-\frac{b^2 x^2}{a^2}} \left (656 a^4 b^2 x^2+350 a^3 b^3 x^3-208 a^2 b^4 x^4+245 a^5 b x-368 a^6-280 a b^5 x^5-80 b^6 x^6\right )+315 a^6 \sin ^{-1}\left (\frac{b x}{a}\right )\right )}{560 b \sqrt{1-\frac{b^2 x^2}{a^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^3*(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]*(-368*a^6 + 245*a^5*b*x + 656*a^4*b^2*x^2 + 350*a^3*b^3*x^3 - 20
8*a^2*b^4*x^4 - 280*a*b^5*x^5 - 80*b^6*x^6) + 315*a^6*ArcSin[(b*x)/a]))/(560*b*Sqrt[1 - (b^2*x^2)/a^2])

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Maple [A]  time = 0.055, size = 134, normalized size = 0.8 \begin{align*} -{\frac{b{x}^{2}}{7} \left ( -{b}^{2}{x}^{2}+{a}^{2} \right ) ^{{\frac{5}{2}}}}-{\frac{23\,{a}^{2}}{35\,b} \left ( -{b}^{2}{x}^{2}+{a}^{2} \right ) ^{{\frac{5}{2}}}}-{\frac{ax}{2} \left ( -{b}^{2}{x}^{2}+{a}^{2} \right ) ^{{\frac{5}{2}}}}+{\frac{3\,x{a}^{3}}{8} \left ( -{b}^{2}{x}^{2}+{a}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{9\,{a}^{5}x}{16}\sqrt{-{b}^{2}{x}^{2}+{a}^{2}}}+{\frac{9\,{a}^{7}}{16}\arctan \left ({x\sqrt{{b}^{2}}{\frac{1}{\sqrt{-{b}^{2}{x}^{2}+{a}^{2}}}}} \right ){\frac{1}{\sqrt{{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7*b*x^2*(-b^2*x^2+a^2)^(5/2)-23/35*a^2*(-b^2*x^2+a^2)^(5/2)/b-1/2*a*x*(-b^2*x^2+a^2)^(5/2)+3/8*a^3*x*(-b^2*
x^2+a^2)^(3/2)+9/16*a^5*x*(-b^2*x^2+a^2)^(1/2)+9/16*a^7/(b^2)^(1/2)*arctan((b^2)^(1/2)*x/(-b^2*x^2+a^2)^(1/2))

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Maxima [A]  time = 1.56827, size = 170, normalized size = 1.04 \begin{align*} \frac{9 \, a^{7} \arcsin \left (\frac{b^{2} x}{\sqrt{a^{2} b^{2}}}\right )}{16 \, \sqrt{b^{2}}} + \frac{9}{16} \, \sqrt{-b^{2} x^{2} + a^{2}} a^{5} x + \frac{3}{8} \,{\left (-b^{2} x^{2} + a^{2}\right )}^{\frac{3}{2}} a^{3} x - \frac{1}{7} \,{\left (-b^{2} x^{2} + a^{2}\right )}^{\frac{5}{2}} b x^{2} - \frac{1}{2} \,{\left (-b^{2} x^{2} + a^{2}\right )}^{\frac{5}{2}} a x - \frac{23 \,{\left (-b^{2} x^{2} + a^{2}\right )}^{\frac{5}{2}} a^{2}}{35 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9/16*a^7*arcsin(b^2*x/sqrt(a^2*b^2))/sqrt(b^2) + 9/16*sqrt(-b^2*x^2 + a^2)*a^5*x + 3/8*(-b^2*x^2 + a^2)^(3/2)*
a^3*x - 1/7*(-b^2*x^2 + a^2)^(5/2)*b*x^2 - 1/2*(-b^2*x^2 + a^2)^(5/2)*a*x - 23/35*(-b^2*x^2 + a^2)^(5/2)*a^2/b

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Fricas [A]  time = 2.05022, size = 259, normalized size = 1.58 \begin{align*} -\frac{630 \, a^{7} \arctan \left (-\frac{a - \sqrt{-b^{2} x^{2} + a^{2}}}{b x}\right ) +{\left (80 \, b^{6} x^{6} + 280 \, a b^{5} x^{5} + 208 \, a^{2} b^{4} x^{4} - 350 \, a^{3} b^{3} x^{3} - 656 \, a^{4} b^{2} x^{2} - 245 \, a^{5} b x + 368 \, a^{6}\right )} \sqrt{-b^{2} x^{2} + a^{2}}}{560 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/560*(630*a^7*arctan(-(a - sqrt(-b^2*x^2 + a^2))/(b*x)) + (80*b^6*x^6 + 280*a*b^5*x^5 + 208*a^2*b^4*x^4 - 35
0*a^3*b^3*x^3 - 656*a^4*b^2*x^2 - 245*a^5*b*x + 368*a^6)*sqrt(-b^2*x^2 + a^2))/b

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Sympy [C]  time = 13.6904, size = 821, normalized size = 5.01 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**5*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)) + 3*a**4*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*
*3*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)/Abs(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)) - 2*a**2*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)) - 3*a*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*sqr
t(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)) - b**5*Piecewise((-8*a**6*sqrt(a**2 - b**2*x**2)/(105*b**6
) - 4*a**4*x**2*sqrt(a**2 - b**2*x**2)/(105*b**4) - a**2*x**4*sqrt(a**2 - b**2*x**2)/(35*b**2) + x**6*sqrt(a**
2 - b**2*x**2)/7, Ne(b, 0)), (x**6*sqrt(a**2)/6, True))

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Giac [A]  time = 1.26588, size = 140, normalized size = 0.85 \begin{align*} \frac{9 \, a^{7} \arcsin \left (\frac{b x}{a}\right ) \mathrm{sgn}\left (a\right ) \mathrm{sgn}\left (b\right )}{16 \,{\left | b \right |}} - \frac{1}{560} \,{\left (\frac{368 \, a^{6}}{b} -{\left (245 \, a^{5} + 2 \,{\left (328 \, a^{4} b +{\left (175 \, a^{3} b^{2} - 4 \,{\left (26 \, a^{2} b^{3} + 5 \,{\left (2 \, b^{5} x + 7 \, a b^{4}\right )} x\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)^3*(-b^2*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

9/16*a^7*arcsin(b*x/a)*sgn(a)*sgn(b)/abs(b) - 1/560*(368*a^6/b - (245*a^5 + 2*(328*a^4*b + (175*a^3*b^2 - 4*(2
6*a^2*b^3 + 5*(2*b^5*x + 7*a*b^4)*x)*x)*x)*x)*x)*sqrt(-b^2*x^2 + a^2)