3.8.88 \(\int (a+b x)^3 (a^2-b^2 x^2)^{3/2} \, dx\) [788]

3.8.88.1 Optimal result
3.8.88.2 Mathematica [A] (verified)
3.8.88.3 Rubi [A] (verified)
3.8.88.4 Maple [A] (verified)
3.8.88.5 Fricas [A] (verification not implemented)
3.8.88.6 Sympy [A] (verification not implemented)
3.8.88.7 Maxima [A] (verification not implemented)
3.8.88.8 Giac [A] (verification not implemented)
3.8.88.9 Mupad [F(-1)]

3.8.88.1 Optimal result

Integrand size = 24, antiderivative size = 164 \[ \int (a+b x)^3 \left (a^2-b^2 x^2\right )^{3/2} \, dx=\frac {9}{16} a^5 x \sqrt {a^2-b^2 x^2}+\frac {3}{8} a^3 x \left (a^2-b^2 x^2\right )^{3/2}-\frac {3 a^2 \left (a^2-b^2 x^2\right )^{5/2}}{10 b}-\frac {3 a (a+b x) \left (a^2-b^2 x^2\right )^{5/2}}{14 b}-\frac {(a+b x)^2 \left (a^2-b^2 x^2\right )^{5/2}}{7 b}+\frac {9 a^7 \arctan \left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{16 b} \]

output
3/8*a^3*x*(-b^2*x^2+a^2)^(3/2)-3/10*a^2*(-b^2*x^2+a^2)^(5/2)/b-3/14*a*(b*x 
+a)*(-b^2*x^2+a^2)^(5/2)/b-1/7*(b*x+a)^2*(-b^2*x^2+a^2)^(5/2)/b+9/16*a^7*a 
rctan(b*x/(-b^2*x^2+a^2)^(1/2))/b+9/16*a^5*x*(-b^2*x^2+a^2)^(1/2)
 
3.8.88.2 Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.81 \[ \int (a+b x)^3 \left (a^2-b^2 x^2\right )^{3/2} \, dx=\frac {\sqrt {a^2-b^2 x^2} \left (-368 a^6+245 a^5 b x+656 a^4 b^2 x^2+350 a^3 b^3 x^3-208 a^2 b^4 x^4-280 a b^5 x^5-80 b^6 x^6\right )}{560 b}-\frac {9 a^7 \log \left (-\sqrt {-b^2} x+\sqrt {a^2-b^2 x^2}\right )}{16 \sqrt {-b^2}} \]

input
Integrate[(a + b*x)^3*(a^2 - b^2*x^2)^(3/2),x]
 
output
(Sqrt[a^2 - b^2*x^2]*(-368*a^6 + 245*a^5*b*x + 656*a^4*b^2*x^2 + 350*a^3*b 
^3*x^3 - 208*a^2*b^4*x^4 - 280*a*b^5*x^5 - 80*b^6*x^6))/(560*b) - (9*a^7*L 
og[-(Sqrt[-b^2]*x) + Sqrt[a^2 - b^2*x^2]])/(16*Sqrt[-b^2])
 
3.8.88.3 Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {469, 469, 455, 211, 211, 224, 216}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 469

\(\displaystyle \frac {9}{7} a \int (a+b x)^2 \left (a^2-b^2 x^2\right )^{3/2}dx-\frac {(a+b x)^2 \left (a^2-b^2 x^2\right )^{5/2}}{7 b}\)

\(\Big \downarrow \) 469

\(\displaystyle \frac {9}{7} a \left (\frac {7}{6} a \int (a+b x) \left (a^2-b^2 x^2\right )^{3/2}dx-\frac {(a+b x) \left (a^2-b^2 x^2\right )^{5/2}}{6 b}\right )-\frac {(a+b x)^2 \left (a^2-b^2 x^2\right )^{5/2}}{7 b}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {9}{7} a \left (\frac {7}{6} a \left (a \int \left (a^2-b^2 x^2\right )^{3/2}dx-\frac {\left (a^2-b^2 x^2\right )^{5/2}}{5 b}\right )-\frac {(a+b x) \left (a^2-b^2 x^2\right )^{5/2}}{6 b}\right )-\frac {(a+b x)^2 \left (a^2-b^2 x^2\right )^{5/2}}{7 b}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {9}{7} a \left (\frac {7}{6} a \left (a \left (\frac {3}{4} a^2 \int \sqrt {a^2-b^2 x^2}dx+\frac {1}{4} x \left (a^2-b^2 x^2\right )^{3/2}\right )-\frac {\left (a^2-b^2 x^2\right )^{5/2}}{5 b}\right )-\frac {(a+b x) \left (a^2-b^2 x^2\right )^{5/2}}{6 b}\right )-\frac {(a+b x)^2 \left (a^2-b^2 x^2\right )^{5/2}}{7 b}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {9}{7} a \left (\frac {7}{6} a \left (a \left (\frac {3}{4} a^2 \left (\frac {1}{2} a^2 \int \frac {1}{\sqrt {a^2-b^2 x^2}}dx+\frac {1}{2} x \sqrt {a^2-b^2 x^2}\right )+\frac {1}{4} x \left (a^2-b^2 x^2\right )^{3/2}\right )-\frac {\left (a^2-b^2 x^2\right )^{5/2}}{5 b}\right )-\frac {(a+b x) \left (a^2-b^2 x^2\right )^{5/2}}{6 b}\right )-\frac {(a+b x)^2 \left (a^2-b^2 x^2\right )^{5/2}}{7 b}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {9}{7} a \left (\frac {7}{6} a \left (a \left (\frac {3}{4} a^2 \left (\frac {1}{2} a^2 \int \frac {1}{\frac {b^2 x^2}{a^2-b^2 x^2}+1}d\frac {x}{\sqrt {a^2-b^2 x^2}}+\frac {1}{2} x \sqrt {a^2-b^2 x^2}\right )+\frac {1}{4} x \left (a^2-b^2 x^2\right )^{3/2}\right )-\frac {\left (a^2-b^2 x^2\right )^{5/2}}{5 b}\right )-\frac {(a+b x) \left (a^2-b^2 x^2\right )^{5/2}}{6 b}\right )-\frac {(a+b x)^2 \left (a^2-b^2 x^2\right )^{5/2}}{7 b}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {9}{7} a \left (\frac {7}{6} a \left (a \left (\frac {3}{4} a^2 \left (\frac {a^2 \arctan \left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{2 b}+\frac {1}{2} x \sqrt {a^2-b^2 x^2}\right )+\frac {1}{4} x \left (a^2-b^2 x^2\right )^{3/2}\right )-\frac {\left (a^2-b^2 x^2\right )^{5/2}}{5 b}\right )-\frac {(a+b x) \left (a^2-b^2 x^2\right )^{5/2}}{6 b}\right )-\frac {(a+b x)^2 \left (a^2-b^2 x^2\right )^{5/2}}{7 b}\)

input
Int[(a + b*x)^3*(a^2 - b^2*x^2)^(3/2),x]
 
output
-1/7*((a + b*x)^2*(a^2 - b^2*x^2)^(5/2))/b + (9*a*(-1/6*((a + b*x)*(a^2 - 
b^2*x^2)^(5/2))/b + (7*a*(-1/5*(a^2 - b^2*x^2)^(5/2)/b + a*((x*(a^2 - b^2* 
x^2)^(3/2))/4 + (3*a^2*((x*Sqrt[a^2 - b^2*x^2])/2 + (a^2*ArcTan[(b*x)/Sqrt 
[a^2 - b^2*x^2]])/(2*b)))/4)))/6))/7
 

3.8.88.3.1 Defintions of rubi rules used

rule 211
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

rule 216
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 224
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 455
Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( 
a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c   Int[(a + b*x^2)^p, x], x] 
/; FreeQ[{a, b, c, d, p}, x] &&  !LeQ[p, -1]
 

rule 469
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 1))), x] + Simp[2*c* 
((n + p)/(n + 2*p + 1))   Int[(c + d*x)^(n - 1)*(a + b*x^2)^p, x], x] /; Fr 
eeQ[{a, b, c, d, p}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[n, 0] && NeQ[n + 2* 
p + 1, 0] && IntegerQ[2*p]
 
3.8.88.4 Maple [A] (verified)

Time = 2.80 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.71

method result size
risch \(-\frac {\left (80 b^{6} x^{6}+280 a \,x^{5} b^{5}+208 a^{2} x^{4} b^{4}-350 a^{3} x^{3} b^{3}-656 a^{4} x^{2} b^{2}-245 a^{5} x b +368 a^{6}\right ) \sqrt {-b^{2} x^{2}+a^{2}}}{560 b}+\frac {9 a^{7} \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-b^{2} x^{2}+a^{2}}}\right )}{16 \sqrt {b^{2}}}\) \(116\)
default \(a^{3} \left (\frac {x \left (-b^{2} x^{2}+a^{2}\right )^{\frac {3}{2}}}{4}+\frac {3 a^{2} \left (\frac {x \sqrt {-b^{2} x^{2}+a^{2}}}{2}+\frac {a^{2} \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-b^{2} x^{2}+a^{2}}}\right )}{2 \sqrt {b^{2}}}\right )}{4}\right )+b^{3} \left (-\frac {x^{2} \left (-b^{2} x^{2}+a^{2}\right )^{\frac {5}{2}}}{7 b^{2}}-\frac {2 a^{2} \left (-b^{2} x^{2}+a^{2}\right )^{\frac {5}{2}}}{35 b^{4}}\right )+3 a \,b^{2} \left (-\frac {x \left (-b^{2} x^{2}+a^{2}\right )^{\frac {5}{2}}}{6 b^{2}}+\frac {a^{2} \left (\frac {x \left (-b^{2} x^{2}+a^{2}\right )^{\frac {3}{2}}}{4}+\frac {3 a^{2} \left (\frac {x \sqrt {-b^{2} x^{2}+a^{2}}}{2}+\frac {a^{2} \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-b^{2} x^{2}+a^{2}}}\right )}{2 \sqrt {b^{2}}}\right )}{4}\right )}{6 b^{2}}\right )-\frac {3 a^{2} \left (-b^{2} x^{2}+a^{2}\right )^{\frac {5}{2}}}{5 b}\) \(258\)

input
int((b*x+a)^3*(-b^2*x^2+a^2)^(3/2),x,method=_RETURNVERBOSE)
 
output
-1/560*(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)/b*(-b^2*x^2+a^2)^(1/2)+9/16*a^7/(b^2)^(1/2)*ar 
ctan((b^2)^(1/2)*x/(-b^2*x^2+a^2)^(1/2))
 
3.8.88.5 Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.71 \[ \int (a+b x)^3 \left (a^2-b^2 x^2\right )^{3/2} \, dx=-\frac {630 \, a^{7} \arctan \left (-\frac {a - \sqrt {-b^{2} x^{2} + a^{2}}}{b x}\right ) + {\left (80 \, b^{6} x^{6} + 280 \, a b^{5} x^{5} + 208 \, a^{2} b^{4} x^{4} - 350 \, a^{3} b^{3} x^{3} - 656 \, a^{4} b^{2} x^{2} - 245 \, a^{5} b x + 368 \, a^{6}\right )} \sqrt {-b^{2} x^{2} + a^{2}}}{560 \, b} \]

input
integrate((b*x+a)^3*(-b^2*x^2+a^2)^(3/2),x, algorithm="fricas")
 
output
-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 - 350*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
 
3.8.88.6 Sympy [A] (verification not implemented)

Time = 0.61 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.08 \[ \int (a+b x)^3 \left (a^2-b^2 x^2\right )^{3/2} \, dx=\begin {cases} \frac {9 a^{7} \left (\begin {cases} \frac {\log {\left (- 2 b^{2} x + 2 \sqrt {- b^{2}} \sqrt {a^{2} - b^{2} x^{2}} \right )}}{\sqrt {- b^{2}}} & \text {for}\: a^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- b^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{16} + \sqrt {a^{2} - b^{2} x^{2}} \left (- \frac {23 a^{6}}{35 b} + \frac {7 a^{5} x}{16} + \frac {41 a^{4} b x^{2}}{35} + \frac {5 a^{3} b^{2} x^{3}}{8} - \frac {13 a^{2} b^{3} x^{4}}{35} - \frac {a b^{4} x^{5}}{2} - \frac {b^{5} x^{6}}{7}\right ) & \text {for}\: b^{2} \neq 0 \\\left (a^{2}\right )^{\frac {3}{2}} \left (\begin {cases} a^{3} x & \text {for}\: b = 0 \\\frac {\left (a + b x\right )^{4}}{4 b} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \]

input
integrate((b*x+a)**3*(-b**2*x**2+a**2)**(3/2),x)
 
output
Piecewise((9*a**7*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)*(-23*a**6/(35*b) + 7*a**5*x/16 + 41*a**4*b*x**2/3 
5 + 5*a**3*b**2*x**3/8 - 13*a**2*b**3*x**4/35 - a*b**4*x**5/2 - b**5*x**6/ 
7), Ne(b**2, 0)), ((a**2)**(3/2)*Piecewise((a**3*x, Eq(b, 0)), ((a + b*x)* 
*4/(4*b), True)), True))
 
3.8.88.7 Maxima [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.71 \[ \int (a+b x)^3 \left (a^2-b^2 x^2\right )^{3/2} \, dx=\frac {9 \, a^{7} \arcsin \left (\frac {b x}{a}\right )}{16 \, b} + \frac {9}{16} \, \sqrt {-b^{2} x^{2} + a^{2}} a^{5} x + \frac {3}{8} \, {\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}} a^{3} x - \frac {1}{7} \, {\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {5}{2}} b x^{2} - \frac {1}{2} \, {\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {5}{2}} a x - \frac {23 \, {\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {5}{2}} a^{2}}{35 \, b} \]

input
integrate((b*x+a)^3*(-b^2*x^2+a^2)^(3/2),x, algorithm="maxima")
 
output
9/16*a^7*arcsin(b*x/a)/b + 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
 
3.8.88.8 Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.63 \[ \int (a+b x)^3 \left (a^2-b^2 x^2\right )^{3/2} \, dx=\frac {9 \, a^{7} \arcsin \left (\frac {b x}{a}\right ) \mathrm {sgn}\left (a\right ) \mathrm {sgn}\left (b\right )}{16 \, {\left | b \right |}} - \frac {1}{560} \, {\left (\frac {368 \, a^{6}}{b} - {\left (245 \, a^{5} + 2 \, {\left (328 \, a^{4} b + {\left (175 \, a^{3} b^{2} - 4 \, {\left (26 \, a^{2} b^{3} + 5 \, {\left (2 \, b^{5} x + 7 \, a b^{4}\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \sqrt {-b^{2} x^{2} + a^{2}} \]

input
integrate((b*x+a)^3*(-b^2*x^2+a^2)^(3/2),x, algorithm="giac")
 
output
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*(26*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)
 
3.8.88.9 Mupad [F(-1)]

Timed out. \[ \int (a+b x)^3 \left (a^2-b^2 x^2\right )^{3/2} \, dx=\int {\left (a^2-b^2\,x^2\right )}^{3/2}\,{\left (a+b\,x\right )}^3 \,d x \]

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