\(\int (c (a+b x)^3)^{5/2} \, dx\) [2813]

Optimal result

Integrand size = 13, antiderivative size = 30 \[ \int \left (c (a+b x)^3\right )^{5/2} \, dx=\frac {2 c^2 (a+b x)^7 \sqrt {c (a+b x)^3}}{17 b} \]

[Out]

2/17*c^2*(b*x+a)^7*(c*(b*x+a)^3)^(1/2)/b

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {253, 15, 30} \[ \int \left (c (a+b x)^3\right )^{5/2} \, dx=\frac {2 c^2 (a+b x)^7 \sqrt {c (a+b x)^3}}{17 b} \]

[In]

Int[(c*(a + b*x)^3)^(5/2),x]

[Out]

(2*c^2*(a + b*x)^7*Sqrt[c*(a + b*x)^3])/(17*b)

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 253

Int[((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[1/Coefficient[v, x, 1], Subst[Int[(a + b*x^n)^p, x], x,
v], x] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && NeQ[v, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \left (c x^3\right )^{5/2} \, dx,x,a+b x\right )}{b} \\ & = \frac {\left (c^2 \sqrt {c (a+b x)^3}\right ) \text {Subst}\left (\int x^{15/2} \, dx,x,a+b x\right )}{b (a+b x)^{3/2}} \\ & = \frac {2 c^2 (a+b x)^7 \sqrt {c (a+b x)^3}}{17 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \left (c (a+b x)^3\right )^{5/2} \, dx=\frac {2 (a+b x) \left (c (a+b x)^3\right )^{5/2}}{17 b} \]

[In]

Integrate[(c*(a + b*x)^3)^(5/2),x]

[Out]

(2*(a + b*x)*(c*(a + b*x)^3)^(5/2))/(17*b)

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73

method	result	size
gosper	\(\frac {2 \left (b x +a \right ) \left (c \left (b x +a \right )^{3}\right )^{\frac {5}{2}}}{17 b}\)	\(22\)
default	\(\frac {2 \left (c \left (b x +a \right )^{3}\right )^{\frac {5}{2}} \left (b c x +a c \right )^{\frac {17}{2}}}{17 \left (b x +a \right )^{5} \left (c \left (b x +a \right )\right )^{\frac {5}{2}} c^{6} b}\)	\(46\)
risch	\(\frac {2 c^{2} \sqrt {c \left (b x +a \right )^{3}}\, \left (b^{8} x^{8}+8 a \,x^{7} b^{7}+28 a^{2} x^{6} b^{6}+56 a^{3} x^{5} b^{5}+70 a^{4} x^{4} b^{4}+56 a^{5} b^{3} x^{3}+28 a^{6} x^{2} b^{2}+8 a^{7} x b +a^{8}\right )}{17 \left (b x +a \right ) b}\)	\(109\)
trager	\(\frac {2 c^{2} \left (b^{7} x^{7}+7 a \,b^{6} x^{6}+21 a^{2} b^{5} x^{5}+35 a^{3} b^{4} x^{4}+35 a^{4} b^{3} x^{3}+21 a^{5} b^{2} x^{2}+7 a^{6} b x +a^{7}\right ) \sqrt {b^{3} c \,x^{3}+3 a \,b^{2} c \,x^{2}+3 a^{2} b c x +c \,a^{3}}}{17 b}\)	\(114\)

[In]

int((c*(b*x+a)^3)^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/17*(b*x+a)*(c*(b*x+a)^3)^(5/2)/b

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 135 vs. \(2 (26) = 52\).

Time = 0.26 (sec) , antiderivative size = 135, normalized size of antiderivative = 4.50 \[ \int \left (c (a+b x)^3\right )^{5/2} \, dx=\frac {2 \, {\left (b^{7} c^{2} x^{7} + 7 \, a b^{6} c^{2} x^{6} + 21 \, a^{2} b^{5} c^{2} x^{5} + 35 \, a^{3} b^{4} c^{2} x^{4} + 35 \, a^{4} b^{3} c^{2} x^{3} + 21 \, a^{5} b^{2} c^{2} x^{2} + 7 \, a^{6} b c^{2} x + a^{7} c^{2}\right )} \sqrt {b^{3} c x^{3} + 3 \, a b^{2} c x^{2} + 3 \, a^{2} b c x + a^{3} c}}{17 \, b} \]

[In]

integrate((c*(b*x+a)^3)^(5/2),x, algorithm="fricas")

[Out]

2/17*(b^7*c^2*x^7 + 7*a*b^6*c^2*x^6 + 21*a^2*b^5*c^2*x^5 + 35*a^3*b^4*c^2*x^4 + 35*a^4*b^3*c^2*x^3 + 21*a^5*b^
2*c^2*x^2 + 7*a^6*b*c^2*x + a^7*c^2)*sqrt(b^3*c*x^3 + 3*a*b^2*c*x^2 + 3*a^2*b*c*x + a^3*c)/b

Sympy [F]

\[ \int \left (c (a+b x)^3\right )^{5/2} \, dx=\int \left (c \left (a + b x\right )^{3}\right )^{\frac {5}{2}}\, dx \]

[In]

integrate((c*(b*x+a)**3)**(5/2),x)

[Out]

Integral((c*(a + b*x)**3)**(5/2), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 108 vs. \(2 (26) = 52\).

Time = 0.22 (sec) , antiderivative size = 108, normalized size of antiderivative = 3.60 \[ \int \left (c (a+b x)^3\right )^{5/2} \, dx=\frac {2 \, {\left (b^{7} c^{\frac {5}{2}} x^{7} + 7 \, a b^{6} c^{\frac {5}{2}} x^{6} + 21 \, a^{2} b^{5} c^{\frac {5}{2}} x^{5} + 35 \, a^{3} b^{4} c^{\frac {5}{2}} x^{4} + 35 \, a^{4} b^{3} c^{\frac {5}{2}} x^{3} + 21 \, a^{5} b^{2} c^{\frac {5}{2}} x^{2} + 7 \, a^{6} b c^{\frac {5}{2}} x + a^{7} c^{\frac {5}{2}}\right )} {\left (b x + a\right )}^{\frac {3}{2}}}{17 \, b} \]

[In]

integrate((c*(b*x+a)^3)^(5/2),x, algorithm="maxima")

[Out]

2/17*(b^7*c^(5/2)*x^7 + 7*a*b^6*c^(5/2)*x^6 + 21*a^2*b^5*c^(5/2)*x^5 + 35*a^3*b^4*c^(5/2)*x^4 + 35*a^4*b^3*c^(
5/2)*x^3 + 21*a^5*b^2*c^(5/2)*x^2 + 7*a^6*b*c^(5/2)*x + a^7*c^(5/2))*(b*x + a)^(3/2)/b

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 851 vs. \(2 (26) = 52\).

Time = 0.30 (sec) , antiderivative size = 851, normalized size of antiderivative = 28.37 \[ \int \left (c (a+b x)^3\right )^{5/2} \, dx=\text {Too large to display} \]

[In]

integrate((c*(b*x+a)^3)^(5/2),x, algorithm="giac")

[Out]

2/109395*(109395*sqrt(b*c*x + a*c)*a^8*c^2*sgn(b*x + a) - 291720*(3*sqrt(b*c*x + a*c)*a*c - (b*c*x + a*c)^(3/2
))*a^7*c*sgn(b*x + a) + 204204*(15*sqrt(b*c*x + a*c)*a^2*c^2 - 10*(b*c*x + a*c)^(3/2)*a*c + 3*(b*c*x + a*c)^(5
/2))*a^6*sgn(b*x + a) - 175032*(35*sqrt(b*c*x + a*c)*a^3*c^3 - 35*(b*c*x + a*c)^(3/2)*a^2*c^2 + 21*(b*c*x + a*
c)^(5/2)*a*c - 5*(b*c*x + a*c)^(7/2))*a^5*sgn(b*x + a)/c + 24310*(315*sqrt(b*c*x + a*c)*a^4*c^4 - 420*(b*c*x +
a*c)^(3/2)*a^3*c^3 + 378*(b*c*x + a*c)^(5/2)*a^2*c^2 - 180*(b*c*x + a*c)^(7/2)*a*c + 35*(b*c*x + a*c)^(9/2))*
a^4*sgn(b*x + a)/c^2 - 8840*(693*sqrt(b*c*x + a*c)*a^5*c^5 - 1155*(b*c*x + a*c)^(3/2)*a^4*c^4 + 1386*(b*c*x +
a*c)^(5/2)*a^3*c^3 - 990*(b*c*x + a*c)^(7/2)*a^2*c^2 + 385*(b*c*x + a*c)^(9/2)*a*c - 63*(b*c*x + a*c)^(11/2))*
a^3*sgn(b*x + a)/c^3 + 1020*(3003*sqrt(b*c*x + a*c)*a^6*c^6 - 6006*(b*c*x + a*c)^(3/2)*a^5*c^5 + 9009*(b*c*x +
a*c)^(5/2)*a^4*c^4 - 8580*(b*c*x + a*c)^(7/2)*a^3*c^3 + 5005*(b*c*x + a*c)^(9/2)*a^2*c^2 - 1638*(b*c*x + a*c)
^(11/2)*a*c + 231*(b*c*x + a*c)^(13/2))*a^2*sgn(b*x + a)/c^4 - 136*(6435*sqrt(b*c*x + a*c)*a^7*c^7 - 15015*(b*
c*x + a*c)^(3/2)*a^6*c^6 + 27027*(b*c*x + a*c)^(5/2)*a^5*c^5 - 32175*(b*c*x + a*c)^(7/2)*a^4*c^4 + 25025*(b*c*
x + a*c)^(9/2)*a^3*c^3 - 12285*(b*c*x + a*c)^(11/2)*a^2*c^2 + 3465*(b*c*x + a*c)^(13/2)*a*c - 429*(b*c*x + a*c
)^(15/2))*a*sgn(b*x + a)/c^5 + (109395*sqrt(b*c*x + a*c)*a^8*c^8 - 291720*(b*c*x + a*c)^(3/2)*a^7*c^7 + 612612
*(b*c*x + a*c)^(5/2)*a^6*c^6 - 875160*(b*c*x + a*c)^(7/2)*a^5*c^5 + 850850*(b*c*x + a*c)^(9/2)*a^4*c^4 - 55692
0*(b*c*x + a*c)^(11/2)*a^3*c^3 + 235620*(b*c*x + a*c)^(13/2)*a^2*c^2 - 58344*(b*c*x + a*c)^(15/2)*a*c + 6435*(
b*c*x + a*c)^(17/2))*sgn(b*x + a)/c^6)/b

Mupad [B] (verification not implemented)

Time = 6.94 (sec) , antiderivative size = 110, normalized size of antiderivative = 3.67 \[ \int \left (c (a+b x)^3\right )^{5/2} \, dx=\sqrt {c\,{\left (a+b\,x\right )}^3}\,\left (\frac {14\,a^6\,c^2\,x}{17}+\frac {2\,a^7\,c^2}{17\,b}+\frac {2\,b^6\,c^2\,x^7}{17}+\frac {42\,a^5\,b\,c^2\,x^2}{17}+\frac {14\,a\,b^5\,c^2\,x^6}{17}+\frac {70\,a^4\,b^2\,c^2\,x^3}{17}+\frac {70\,a^3\,b^3\,c^2\,x^4}{17}+\frac {42\,a^2\,b^4\,c^2\,x^5}{17}\right ) \]

[In]

int((c*(a + b*x)^3)^(5/2),x)

[Out]

(c*(a + b*x)^3)^(1/2)*((14*a^6*c^2*x)/17 + (2*a^7*c^2)/(17*b) + (2*b^6*c^2*x^7)/17 + (42*a^5*b*c^2*x^2)/17 + (
14*a*b^5*c^2*x^6)/17 + (70*a^4*b^2*c^2*x^3)/17 + (70*a^3*b^3*c^2*x^4)/17 + (42*a^2*b^4*c^2*x^5)/17)