3.1137 \(\int (a+a x)^{5/2} (c-c x)^{5/2} \, dx\)

Optimal. Leaf size=126 \[ \frac{5}{16} a^2 c^2 x \sqrt{a x+a} \sqrt{c-c x}+\frac{5}{8} a^{5/2} c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a x+a}}{\sqrt{a} \sqrt{c-c x}}\right )+\frac{5}{24} a c x (a x+a)^{3/2} (c-c x)^{3/2}+\frac{1}{6} x (a x+a)^{5/2} (c-c x)^{5/2} \]

[Out]

(5*a^2*c^2*x*Sqrt[a + a*x]*Sqrt[c - c*x])/16 + (5*a*c*x*(a + a*x)^(3/2)*(c - c*x)^(3/2))/24 + (x*(a + a*x)^(5/
2)*(c - c*x)^(5/2))/6 + (5*a^(5/2)*c^(5/2)*ArcTan[(Sqrt[c]*Sqrt[a + a*x])/(Sqrt[a]*Sqrt[c - c*x])])/8

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Rubi [A]  time = 0.0548007, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {38, 63, 217, 203} \[ \frac{5}{16} a^2 c^2 x \sqrt{a x+a} \sqrt{c-c x}+\frac{5}{8} a^{5/2} c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a x+a}}{\sqrt{a} \sqrt{c-c x}}\right )+\frac{5}{24} a c x (a x+a)^{3/2} (c-c x)^{3/2}+\frac{1}{6} x (a x+a)^{5/2} (c-c x)^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*x)^(5/2)*(c - c*x)^(5/2),x]

[Out]

(5*a^2*c^2*x*Sqrt[a + a*x]*Sqrt[c - c*x])/16 + (5*a*c*x*(a + a*x)^(3/2)*(c - c*x)^(3/2))/24 + (x*(a + a*x)^(5/
2)*(c - c*x)^(5/2))/6 + (5*a^(5/2)*c^(5/2)*ArcTan[(Sqrt[c]*Sqrt[a + a*x])/(Sqrt[a]*Sqrt[c - c*x])])/8

Rule 38

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(x*(a + b*x)^m*(c + d*x)^m)/(2*m + 1)
, x] + Dist[(2*a*c*m)/(2*m + 1), Int[(a + b*x)^(m - 1)*(c + d*x)^(m - 1), x], x] /; FreeQ[{a, b, c, d}, x] &&
EqQ[b*c + a*d, 0] && IGtQ[m + 1/2, 0]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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+a x)^{5/2} (c-c x)^{5/2} \, dx &=\frac{1}{6} x (a+a x)^{5/2} (c-c x)^{5/2}+\frac{1}{6} (5 a c) \int (a+a x)^{3/2} (c-c x)^{3/2} \, dx\\ &=\frac{5}{24} a c x (a+a x)^{3/2} (c-c x)^{3/2}+\frac{1}{6} x (a+a x)^{5/2} (c-c x)^{5/2}+\frac{1}{8} \left (5 a^2 c^2\right ) \int \sqrt{a+a x} \sqrt{c-c x} \, dx\\ &=\frac{5}{16} a^2 c^2 x \sqrt{a+a x} \sqrt{c-c x}+\frac{5}{24} a c x (a+a x)^{3/2} (c-c x)^{3/2}+\frac{1}{6} x (a+a x)^{5/2} (c-c x)^{5/2}+\frac{1}{16} \left (5 a^3 c^3\right ) \int \frac{1}{\sqrt{a+a x} \sqrt{c-c x}} \, dx\\ &=\frac{5}{16} a^2 c^2 x \sqrt{a+a x} \sqrt{c-c x}+\frac{5}{24} a c x (a+a x)^{3/2} (c-c x)^{3/2}+\frac{1}{6} x (a+a x)^{5/2} (c-c x)^{5/2}+\frac{1}{8} \left (5 a^2 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2 c-\frac{c x^2}{a}}} \, dx,x,\sqrt{a+a x}\right )\\ &=\frac{5}{16} a^2 c^2 x \sqrt{a+a x} \sqrt{c-c x}+\frac{5}{24} a c x (a+a x)^{3/2} (c-c x)^{3/2}+\frac{1}{6} x (a+a x)^{5/2} (c-c x)^{5/2}+\frac{1}{8} \left (5 a^2 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{c x^2}{a}} \, dx,x,\frac{\sqrt{a+a x}}{\sqrt{c-c x}}\right )\\ &=\frac{5}{16} a^2 c^2 x \sqrt{a+a x} \sqrt{c-c x}+\frac{5}{24} a c x (a+a x)^{3/2} (c-c x)^{3/2}+\frac{1}{6} x (a+a x)^{5/2} (c-c x)^{5/2}+\frac{5}{8} a^{5/2} c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+a x}}{\sqrt{a} \sqrt{c-c x}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0980313, size = 114, normalized size = 0.9 \[ \frac{c^{3/2} (a (x+1))^{5/2} \sqrt{c-c x} \left (\sqrt{c} x \sqrt{x+1} \left (8 x^5-8 x^4-26 x^3+26 x^2+33 x-33\right )+30 \sqrt{c-c x} \sin ^{-1}\left (\frac{\sqrt{c-c x}}{\sqrt{2} \sqrt{c}}\right )\right )}{48 (x-1) (x+1)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*x)^(5/2)*(c - c*x)^(5/2),x]

[Out]

(c^(3/2)*(a*(1 + x))^(5/2)*Sqrt[c - c*x]*(Sqrt[c]*x*Sqrt[1 + x]*(-33 + 33*x + 26*x^2 - 26*x^3 - 8*x^4 + 8*x^5)
+ 30*Sqrt[c - c*x]*ArcSin[Sqrt[c - c*x]/(Sqrt[2]*Sqrt[c])]))/(48*(-1 + x)*(1 + x)^(5/2))

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Maple [B]  time = 0.013, size = 193, normalized size = 1.5 \begin{align*} -{\frac{1}{6\,c} \left ( ax+a \right ) ^{{\frac{5}{2}}} \left ( -cx+c \right ) ^{{\frac{7}{2}}}}-{\frac{a}{6\,c} \left ( ax+a \right ) ^{{\frac{3}{2}}} \left ( -cx+c \right ) ^{{\frac{7}{2}}}}-{\frac{{a}^{2}}{8\,c}\sqrt{ax+a} \left ( -cx+c \right ) ^{{\frac{7}{2}}}}+{\frac{{a}^{2}}{24} \left ( -cx+c \right ) ^{{\frac{5}{2}}}\sqrt{ax+a}}+{\frac{5\,{a}^{2}c}{48} \left ( -cx+c \right ) ^{{\frac{3}{2}}}\sqrt{ax+a}}+{\frac{5\,{a}^{2}{c}^{2}}{16}\sqrt{ax+a}\sqrt{-cx+c}}+{\frac{5\,{a}^{3}{c}^{3}}{16}\sqrt{ \left ( -cx+c \right ) \left ( ax+a \right ) }\arctan \left ({x\sqrt{ac}{\frac{1}{\sqrt{-ac{x}^{2}+ac}}}} \right ){\frac{1}{\sqrt{ax+a}}}{\frac{1}{\sqrt{-cx+c}}}{\frac{1}{\sqrt{ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+a)^(5/2)*(-c*x+c)^(5/2),x)

[Out]

-1/6/c*(a*x+a)^(5/2)*(-c*x+c)^(7/2)-1/6*a/c*(a*x+a)^(3/2)*(-c*x+c)^(7/2)-1/8*a^2/c*(a*x+a)^(1/2)*(-c*x+c)^(7/2
)+1/24*a^2*(-c*x+c)^(5/2)*(a*x+a)^(1/2)+5/48*a^2*c*(-c*x+c)^(3/2)*(a*x+a)^(1/2)+5/16*a^2*c^2*(-c*x+c)^(1/2)*(a
*x+a)^(1/2)+5/16*a^3*c^3*((-c*x+c)*(a*x+a))^(1/2)/(-c*x+c)^(1/2)/(a*x+a)^(1/2)/(a*c)^(1/2)*arctan((a*c)^(1/2)*
x/(-a*c*x^2+a*c)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+a)^(5/2)*(-c*x+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.66731, size = 478, normalized size = 3.79 \begin{align*} \left [\frac{5}{32} \, \sqrt{-a c} a^{2} c^{2} \log \left (2 \, a c x^{2} + 2 \, \sqrt{-a c} \sqrt{a x + a} \sqrt{-c x + c} x - a c\right ) + \frac{1}{48} \,{\left (8 \, a^{2} c^{2} x^{5} - 26 \, a^{2} c^{2} x^{3} + 33 \, a^{2} c^{2} x\right )} \sqrt{a x + a} \sqrt{-c x + c}, -\frac{5}{16} \, \sqrt{a c} a^{2} c^{2} \arctan \left (\frac{\sqrt{a c} \sqrt{a x + a} \sqrt{-c x + c} x}{a c x^{2} - a c}\right ) + \frac{1}{48} \,{\left (8 \, a^{2} c^{2} x^{5} - 26 \, a^{2} c^{2} x^{3} + 33 \, a^{2} c^{2} x\right )} \sqrt{a x + a} \sqrt{-c x + c}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+a)^(5/2)*(-c*x+c)^(5/2),x, algorithm="fricas")

[Out]

[5/32*sqrt(-a*c)*a^2*c^2*log(2*a*c*x^2 + 2*sqrt(-a*c)*sqrt(a*x + a)*sqrt(-c*x + c)*x - a*c) + 1/48*(8*a^2*c^2*
x^5 - 26*a^2*c^2*x^3 + 33*a^2*c^2*x)*sqrt(a*x + a)*sqrt(-c*x + c), -5/16*sqrt(a*c)*a^2*c^2*arctan(sqrt(a*c)*sq
rt(a*x + a)*sqrt(-c*x + c)*x/(a*c*x^2 - a*c)) + 1/48*(8*a^2*c^2*x^5 - 26*a^2*c^2*x^3 + 33*a^2*c^2*x)*sqrt(a*x
+ a)*sqrt(-c*x + c)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \left (x + 1\right )\right )^{\frac{5}{2}} \left (- c \left (x - 1\right )\right )^{\frac{5}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+a)**(5/2)*(-c*x+c)**(5/2),x)

[Out]

Integral((a*(x + 1))**(5/2)*(-c*(x - 1))**(5/2), x)

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Giac [B]  time = 1.35166, size = 478, normalized size = 3.79 \begin{align*} -\frac{{\left (\frac{6 \, a^{3} c \log \left ({\left | -\sqrt{-a c} \sqrt{a x + a} + \sqrt{-{\left (a x + a\right )} a c + 2 \, a^{2} c} \right |}\right )}{\sqrt{-a c}} - \sqrt{-{\left (a x + a\right )} a c + 2 \, a^{2} c}{\left ({\left (2 \,{\left ({\left (a x + a\right )}{\left (4 \,{\left (a x + a\right )}{\left (\frac{a x + a}{a^{4}} - \frac{5}{a^{3}}\right )} + \frac{39}{a^{2}}\right )} - \frac{37}{a}\right )}{\left (a x + a\right )} + 31\right )}{\left (a x + a\right )} - 3 \, a\right )} \sqrt{a x + a}\right )} c^{2}{\left | a \right |}}{48 \, a} - \frac{{\left (\frac{2 \, a^{3} c \log \left ({\left | -\sqrt{-a c} \sqrt{a x + a} + \sqrt{-{\left (a x + a\right )} a c + 2 \, a^{2} c} \right |}\right )}{\sqrt{-a c}} - \sqrt{-{\left (a x + a\right )} a c + 2 \, a^{2} c} \sqrt{a x + a} a x\right )} c^{2}{\left | a \right |}}{2 \, a} + \frac{{\left (\frac{2 \, a^{3} c \log \left ({\left | -\sqrt{-a c} \sqrt{a x + a} + \sqrt{-{\left (a x + a\right )} a c + 2 \, a^{2} c} \right |}\right )}{\sqrt{-a c}} - \sqrt{-{\left (a x + a\right )} a c + 2 \, a^{2} c}{\left ({\left (a x + a\right )}{\left (2 \,{\left (a x + a\right )}{\left (\frac{a x + a}{a^{2}} - \frac{3}{a}\right )} + 5\right )} - a\right )} \sqrt{a x + a}\right )} c^{2}{\left | a \right |}}{4 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+a)^(5/2)*(-c*x+c)^(5/2),x, algorithm="giac")

[Out]

-1/48*(6*a^3*c*log(abs(-sqrt(-a*c)*sqrt(a*x + a) + sqrt(-(a*x + a)*a*c + 2*a^2*c)))/sqrt(-a*c) - sqrt(-(a*x +
a)*a*c + 2*a^2*c)*((2*((a*x + a)*(4*(a*x + a)*((a*x + a)/a^4 - 5/a^3) + 39/a^2) - 37/a)*(a*x + a) + 31)*(a*x +
a) - 3*a)*sqrt(a*x + a))*c^2*abs(a)/a - 1/2*(2*a^3*c*log(abs(-sqrt(-a*c)*sqrt(a*x + a) + sqrt(-(a*x + a)*a*c
+ 2*a^2*c)))/sqrt(-a*c) - sqrt(-(a*x + a)*a*c + 2*a^2*c)*sqrt(a*x + a)*a*x)*c^2*abs(a)/a + 1/4*(2*a^3*c*log(ab
s(-sqrt(-a*c)*sqrt(a*x + a) + sqrt(-(a*x + a)*a*c + 2*a^2*c)))/sqrt(-a*c) - sqrt(-(a*x + a)*a*c + 2*a^2*c)*((a
*x + a)*(2*(a*x + a)*((a*x + a)/a^2 - 3/a) + 5) - a)*sqrt(a*x + a))*c^2*abs(a)/a