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

Optimal. Leaf size=135 \[ \frac {5 a^6 c^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {c (a-b x)}}\right )}{8 b}+\frac {5}{16} a^4 c^2 x \sqrt {a+b x} \sqrt {a c-b c x}+\frac {5}{24} a^2 c x (a+b x)^{3/2} (a c-b c x)^{3/2}+\frac {1}{6} x (a+b x)^{5/2} (a c-b c x)^{5/2} \]

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Rubi [A]  time = 0.05, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {38, 63, 217, 203} \begin {gather*} \frac {5}{16} a^4 c^2 x \sqrt {a+b x} \sqrt {a c-b c x}+\frac {5 a^6 c^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {c (a-b x)}}\right )}{8 b}+\frac {5}{24} a^2 c x (a+b x)^{3/2} (a c-b c x)^{3/2}+\frac {1}{6} x (a+b x)^{5/2} (a c-b c x)^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a^4*c^2*x*Sqrt[a + b*x]*Sqrt[a*c - b*c*x])/16 + (5*a^2*c*x*(a + b*x)^(3/2)*(a*c - b*c*x)^(3/2))/24 + (x*(a
+ b*x)^(5/2)*(a*c - b*c*x)^(5/2))/6 + (5*a^6*c^(5/2)*ArcTan[(Sqrt[c]*Sqrt[a + b*x])/Sqrt[c*(a - b*x)]])/(8*b)

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 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]

Rubi steps

\begin {align*} \int (a+b x)^{5/2} (a c-b c x)^{5/2} \, dx &=\frac {1}{6} x (a+b x)^{5/2} (a c-b c x)^{5/2}+\frac {1}{6} \left (5 a^2 c\right ) \int (a+b x)^{3/2} (a c-b c x)^{3/2} \, dx\\ &=\frac {5}{24} a^2 c x (a+b x)^{3/2} (a c-b c x)^{3/2}+\frac {1}{6} x (a+b x)^{5/2} (a c-b c x)^{5/2}+\frac {1}{8} \left (5 a^4 c^2\right ) \int \sqrt {a+b x} \sqrt {a c-b c x} \, dx\\ &=\frac {5}{16} a^4 c^2 x \sqrt {a+b x} \sqrt {a c-b c x}+\frac {5}{24} a^2 c x (a+b x)^{3/2} (a c-b c x)^{3/2}+\frac {1}{6} x (a+b x)^{5/2} (a c-b c x)^{5/2}+\frac {1}{16} \left (5 a^6 c^3\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {a c-b c x}} \, dx\\ &=\frac {5}{16} a^4 c^2 x \sqrt {a+b x} \sqrt {a c-b c x}+\frac {5}{24} a^2 c x (a+b x)^{3/2} (a c-b c x)^{3/2}+\frac {1}{6} x (a+b x)^{5/2} (a c-b c x)^{5/2}+\frac {\left (5 a^6 c^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2 a c-c x^2}} \, dx,x,\sqrt {a+b x}\right )}{8 b}\\ &=\frac {5}{16} a^4 c^2 x \sqrt {a+b x} \sqrt {a c-b c x}+\frac {5}{24} a^2 c x (a+b x)^{3/2} (a c-b c x)^{3/2}+\frac {1}{6} x (a+b x)^{5/2} (a c-b c x)^{5/2}+\frac {\left (5 a^6 c^3\right ) \operatorname {Subst}\left (\int \frac {1}{1+c x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c (a-b x)}}\right )}{8 b}\\ &=\frac {5}{16} a^4 c^2 x \sqrt {a+b x} \sqrt {a c-b c x}+\frac {5}{24} a^2 c x (a+b x)^{3/2} (a c-b c x)^{3/2}+\frac {1}{6} x (a+b x)^{5/2} (a c-b c x)^{5/2}+\frac {5 a^6 c^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {c (a-b x)}}\right )}{8 b}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 120, normalized size = 0.89 \begin {gather*} \frac {c^3 \left (-30 a^{13/2} \sqrt {a-b x} \sqrt {\frac {b x}{a}+1} \sin ^{-1}\left (\frac {\sqrt {a-b x}}{\sqrt {2} \sqrt {a}}\right )+33 a^6 b x-59 a^4 b^3 x^3+34 a^2 b^5 x^5-8 b^7 x^7\right )}{48 b \sqrt {a+b x} \sqrt {c (a-b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c^3*(33*a^6*b*x - 59*a^4*b^3*x^3 + 34*a^2*b^5*x^5 - 8*b^7*x^7 - 30*a^(13/2)*Sqrt[a - b*x]*Sqrt[1 + (b*x)/a]*A
rcSin[Sqrt[a - b*x]/(Sqrt[2]*Sqrt[a])]))/(48*b*Sqrt[c*(a - b*x)]*Sqrt[a + b*x])

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IntegrateAlgebraic [A]  time = 0.33, size = 215, normalized size = 1.59 \begin {gather*} \frac {a^6 c^3 \sqrt {a c-b c x} \left (\frac {85 c^4 (a c-b c x)}{a+b x}+\frac {198 c^3 (a c-b c x)^2}{(a+b x)^2}-\frac {198 c^2 (a c-b c x)^3}{(a+b x)^3}-\frac {85 c (a c-b c x)^4}{(a+b x)^4}-\frac {15 (a c-b c x)^5}{(a+b x)^5}+15 c^5\right )}{24 b \sqrt {a+b x} \left (\frac {a c-b c x}{a+b x}+c\right )^6}-\frac {5 a^6 c^{5/2} \tan ^{-1}\left (\frac {\sqrt {a c-b c x}}{\sqrt {c} \sqrt {a+b x}}\right )}{8 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^6*c^3*Sqrt[a*c - b*c*x]*(15*c^5 + (85*c^4*(a*c - b*c*x))/(a + b*x) + (198*c^3*(a*c - b*c*x)^2)/(a + b*x)^2
- (198*c^2*(a*c - b*c*x)^3)/(a + b*x)^3 - (85*c*(a*c - b*c*x)^4)/(a + b*x)^4 - (15*(a*c - b*c*x)^5)/(a + b*x)^
5))/(24*b*Sqrt[a + b*x]*(c + (a*c - b*c*x)/(a + b*x))^6) - (5*a^6*c^(5/2)*ArcTan[Sqrt[a*c - b*c*x]/(Sqrt[c]*Sq
rt[a + b*x])])/(8*b)

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fricas [A]  time = 1.16, size = 232, normalized size = 1.72 \begin {gather*} \left [\frac {15 \, a^{6} \sqrt {-c} c^{2} \log \left (2 \, b^{2} c x^{2} + 2 \, \sqrt {-b c x + a c} \sqrt {b x + a} b \sqrt {-c} x - a^{2} c\right ) + 2 \, {\left (8 \, b^{5} c^{2} x^{5} - 26 \, a^{2} b^{3} c^{2} x^{3} + 33 \, a^{4} b c^{2} x\right )} \sqrt {-b c x + a c} \sqrt {b x + a}}{96 \, b}, -\frac {15 \, a^{6} c^{\frac {5}{2}} \arctan \left (\frac {\sqrt {-b c x + a c} \sqrt {b x + a} b \sqrt {c} x}{b^{2} c x^{2} - a^{2} c}\right ) - {\left (8 \, b^{5} c^{2} x^{5} - 26 \, a^{2} b^{3} c^{2} x^{3} + 33 \, a^{4} b c^{2} x\right )} \sqrt {-b c x + a c} \sqrt {b x + a}}{48 \, b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(15*a^6*sqrt(-c)*c^2*log(2*b^2*c*x^2 + 2*sqrt(-b*c*x + a*c)*sqrt(b*x + a)*b*sqrt(-c)*x - a^2*c) + 2*(8*b
^5*c^2*x^5 - 26*a^2*b^3*c^2*x^3 + 33*a^4*b*c^2*x)*sqrt(-b*c*x + a*c)*sqrt(b*x + a))/b, -1/48*(15*a^6*c^(5/2)*a
rctan(sqrt(-b*c*x + a*c)*sqrt(b*x + a)*b*sqrt(c)*x/(b^2*c*x^2 - a^2*c)) - (8*b^5*c^2*x^5 - 26*a^2*b^3*c^2*x^3
+ 33*a^4*b*c^2*x)*sqrt(-b*c*x + a*c)*sqrt(b*x + a))/b]

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [B]  time = 0.01, size = 243, normalized size = 1.80 \begin {gather*} \frac {5 \sqrt {\left (b x +a \right ) \left (-b c x +a c \right )}\, a^{6} c^{3} \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {-b^{2} c \,x^{2}+a^{2} c}}\right )}{16 \sqrt {-b c x +a c}\, \sqrt {b x +a}\, \sqrt {b^{2} c}}+\frac {5 \sqrt {-b c x +a c}\, \sqrt {b x +a}\, a^{5} c^{2}}{16 b}+\frac {5 \left (-b c x +a c \right )^{\frac {3}{2}} \sqrt {b x +a}\, a^{4} c}{48 b}+\frac {\left (-b c x +a c \right )^{\frac {5}{2}} \sqrt {b x +a}\, a^{3}}{24 b}-\frac {\sqrt {b x +a}\, \left (-b c x +a c \right )^{\frac {7}{2}} a^{2}}{8 b c}-\frac {\left (b x +a \right )^{\frac {3}{2}} \left (-b c x +a c \right )^{\frac {7}{2}} a}{6 b c}-\frac {\left (b x +a \right )^{\frac {5}{2}} \left (-b c x +a c \right )^{\frac {7}{2}}}{6 b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/b/c*(b*x+a)^(5/2)*(-b*c*x+a*c)^(7/2)-1/6*a/b/c*(b*x+a)^(3/2)*(-b*c*x+a*c)^(7/2)-1/8*a^2/b/c*(b*x+a)^(1/2)
*(-b*c*x+a*c)^(7/2)+1/24*a^3/b*(-b*c*x+a*c)^(5/2)*(b*x+a)^(1/2)+5/48*a^4*c/b*(-b*c*x+a*c)^(3/2)*(b*x+a)^(1/2)+
5/16*a^5*c^2/b*(-b*c*x+a*c)^(1/2)*(b*x+a)^(1/2)+5/16*a^6*c^3*((b*x+a)*(-b*c*x+a*c))^(1/2)/(-b*c*x+a*c)^(1/2)/(
b*x+a)^(1/2)/(b^2*c)^(1/2)*arctan((b^2*c)^(1/2)*x/(-b^2*c*x^2+a^2*c)^(1/2))

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maxima [A]  time = 3.12, size = 89, normalized size = 0.66 \begin {gather*} \frac {5 \, a^{6} c^{\frac {5}{2}} \arcsin \left (\frac {b x}{a}\right )}{16 \, b} + \frac {5}{16} \, \sqrt {-b^{2} c x^{2} + a^{2} c} a^{4} c^{2} x + \frac {5}{24} \, {\left (-b^{2} c x^{2} + a^{2} c\right )}^{\frac {3}{2}} a^{2} c x + \frac {1}{6} \, {\left (-b^{2} c x^{2} + a^{2} c\right )}^{\frac {5}{2}} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (a\,c-b\,c\,x\right )}^{5/2}\,{\left (a+b\,x\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (- c \left (- a + b x\right )\right )^{\frac {5}{2}} \left (a + b x\right )^{\frac {5}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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