\(\int \frac {(c x^2)^{5/2} (a+b x)^n}{x^2} \, dx\) [940]

Optimal result

Integrand size = 20, antiderivative size = 143 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^n}{x^2} \, dx=-\frac {a^3 c^2 \sqrt {c x^2} (a+b x)^{1+n}}{b^4 (1+n) x}+\frac {3 a^2 c^2 \sqrt {c x^2} (a+b x)^{2+n}}{b^4 (2+n) x}-\frac {3 a c^2 \sqrt {c x^2} (a+b x)^{3+n}}{b^4 (3+n) x}+\frac {c^2 \sqrt {c x^2} (a+b x)^{4+n}}{b^4 (4+n) x} \]

[Out]

-a^3*c^2*(b*x+a)^(1+n)*(c*x^2)^(1/2)/b^4/(1+n)/x+3*a^2*c^2*(b*x+a)^(2+n)*(c*x^2)^(1/2)/b^4/(2+n)/x-3*a*c^2*(b*
x+a)^(3+n)*(c*x^2)^(1/2)/b^4/(3+n)/x+c^2*(b*x+a)^(4+n)*(c*x^2)^(1/2)/b^4/(4+n)/x

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 45} \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^n}{x^2} \, dx=-\frac {a^3 c^2 \sqrt {c x^2} (a+b x)^{n+1}}{b^4 (n+1) x}+\frac {3 a^2 c^2 \sqrt {c x^2} (a+b x)^{n+2}}{b^4 (n+2) x}-\frac {3 a c^2 \sqrt {c x^2} (a+b x)^{n+3}}{b^4 (n+3) x}+\frac {c^2 \sqrt {c x^2} (a+b x)^{n+4}}{b^4 (n+4) x} \]

[In]

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

[Out]

-((a^3*c^2*Sqrt[c*x^2]*(a + b*x)^(1 + n))/(b^4*(1 + n)*x)) + (3*a^2*c^2*Sqrt[c*x^2]*(a + b*x)^(2 + n))/(b^4*(2
+ n)*x) - (3*a*c^2*Sqrt[c*x^2]*(a + b*x)^(3 + n))/(b^4*(3 + n)*x) + (c^2*Sqrt[c*x^2]*(a + b*x)^(4 + n))/(b^4*
(4 + n)*x)

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 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {\left (c^2 \sqrt {c x^2}\right ) \int x^3 (a+b x)^n \, dx}{x} \\ & = \frac {\left (c^2 \sqrt {c x^2}\right ) \int \left (-\frac {a^3 (a+b x)^n}{b^3}+\frac {3 a^2 (a+b x)^{1+n}}{b^3}-\frac {3 a (a+b x)^{2+n}}{b^3}+\frac {(a+b x)^{3+n}}{b^3}\right ) \, dx}{x} \\ & = -\frac {a^3 c^2 \sqrt {c x^2} (a+b x)^{1+n}}{b^4 (1+n) x}+\frac {3 a^2 c^2 \sqrt {c x^2} (a+b x)^{2+n}}{b^4 (2+n) x}-\frac {3 a c^2 \sqrt {c x^2} (a+b x)^{3+n}}{b^4 (3+n) x}+\frac {c^2 \sqrt {c x^2} (a+b x)^{4+n}}{b^4 (4+n) x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.69 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^n}{x^2} \, dx=\frac {c \left (c x^2\right )^{3/2} (a+b x)^{1+n} \left (-6 a^3+6 a^2 b (1+n) x-3 a b^2 \left (2+3 n+n^2\right ) x^2+b^3 \left (6+11 n+6 n^2+n^3\right ) x^3\right )}{b^4 (1+n) (2+n) (3+n) (4+n) x^3} \]

[In]

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

[Out]

(c*(c*x^2)^(3/2)*(a + b*x)^(1 + n)*(-6*a^3 + 6*a^2*b*(1 + n)*x - 3*a*b^2*(2 + 3*n + n^2)*x^2 + b^3*(6 + 11*n +
6*n^2 + n^3)*x^3))/(b^4*(1 + n)*(2 + n)*(3 + n)*(4 + n)*x^3)

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.95

method	result	size
gosper	\(-\frac {\left (c \,x^{2}\right )^{\frac {5}{2}} \left (b x +a \right )^{1+n} \left (-b^{3} n^{3} x^{3}-6 b^{3} n^{2} x^{3}+3 a \,b^{2} n^{2} x^{2}-11 b^{3} n \,x^{3}+9 a \,b^{2} n \,x^{2}-6 b^{3} x^{3}-6 a^{2} b n x +6 a \,b^{2} x^{2}-6 a^{2} b x +6 a^{3}\right )}{b^{4} x^{5} \left (n^{4}+10 n^{3}+35 n^{2}+50 n +24\right )}\)	\(136\)
risch	\(-\frac {c^{2} \sqrt {c \,x^{2}}\, \left (-b^{4} n^{3} x^{4}-a \,b^{3} n^{3} x^{3}-6 b^{4} n^{2} x^{4}-3 a \,b^{3} n^{2} x^{3}-11 b^{4} n \,x^{4}+3 a^{2} b^{2} n^{2} x^{2}-2 x^{3} a n \,b^{3}-6 b^{4} x^{4}+3 a^{2} n \,x^{2} b^{2}-6 x \,a^{3} n b +6 a^{4}\right ) \left (b x +a \right )^{n}}{x \left (3+n \right ) \left (4+n \right ) \left (2+n \right ) \left (1+n \right ) b^{4}}\)	\(159\)

[In]

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

[Out]

-1/b^4/x^5*(c*x^2)^(5/2)*(b*x+a)^(1+n)/(n^4+10*n^3+35*n^2+50*n+24)*(-b^3*n^3*x^3-6*b^3*n^2*x^3+3*a*b^2*n^2*x^2
-11*b^3*n*x^3+9*a*b^2*n*x^2-6*b^3*x^3-6*a^2*b*n*x+6*a*b^2*x^2-6*a^2*b*x+6*a^3)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.30 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^n}{x^2} \, dx=\frac {{\left (6 \, a^{3} b c^{2} n x - 6 \, a^{4} c^{2} + {\left (b^{4} c^{2} n^{3} + 6 \, b^{4} c^{2} n^{2} + 11 \, b^{4} c^{2} n + 6 \, b^{4} c^{2}\right )} x^{4} + {\left (a b^{3} c^{2} n^{3} + 3 \, a b^{3} c^{2} n^{2} + 2 \, a b^{3} c^{2} n\right )} x^{3} - 3 \, {\left (a^{2} b^{2} c^{2} n^{2} + a^{2} b^{2} c^{2} n\right )} x^{2}\right )} \sqrt {c x^{2}} {\left (b x + a\right )}^{n}}{{\left (b^{4} n^{4} + 10 \, b^{4} n^{3} + 35 \, b^{4} n^{2} + 50 \, b^{4} n + 24 \, b^{4}\right )} x} \]

[In]

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

[Out]

(6*a^3*b*c^2*n*x - 6*a^4*c^2 + (b^4*c^2*n^3 + 6*b^4*c^2*n^2 + 11*b^4*c^2*n + 6*b^4*c^2)*x^4 + (a*b^3*c^2*n^3 +
3*a*b^3*c^2*n^2 + 2*a*b^3*c^2*n)*x^3 - 3*(a^2*b^2*c^2*n^2 + a^2*b^2*c^2*n)*x^2)*sqrt(c*x^2)*(b*x + a)^n/((b^4
*n^4 + 10*b^4*n^3 + 35*b^4*n^2 + 50*b^4*n + 24*b^4)*x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.81 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^n}{x^2} \, dx=\frac {{\left ({\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} b^{4} c^{\frac {5}{2}} x^{4} + {\left (n^{3} + 3 \, n^{2} + 2 \, n\right )} a b^{3} c^{\frac {5}{2}} x^{3} - 3 \, {\left (n^{2} + n\right )} a^{2} b^{2} c^{\frac {5}{2}} x^{2} + 6 \, a^{3} b c^{\frac {5}{2}} n x - 6 \, a^{4} c^{\frac {5}{2}}\right )} {\left (b x + a\right )}^{n}}{{\left (n^{4} + 10 \, n^{3} + 35 \, n^{2} + 50 \, n + 24\right )} b^{4}} \]

[In]

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

[Out]

((n^3 + 6*n^2 + 11*n + 6)*b^4*c^(5/2)*x^4 + (n^3 + 3*n^2 + 2*n)*a*b^3*c^(5/2)*x^3 - 3*(n^2 + n)*a^2*b^2*c^(5/2
)*x^2 + 6*a^3*b*c^(5/2)*n*x - 6*a^4*c^(5/2))*(b*x + a)^n/((n^4 + 10*n^3 + 35*n^2 + 50*n + 24)*b^4)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.60 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^n}{x^2} \, dx=\frac {{\left (a+b\,x\right )}^n\,\left (\frac {c^2\,x^4\,\sqrt {c\,x^2}\,\left (n^3+6\,n^2+11\,n+6\right )}{n^4+10\,n^3+35\,n^2+50\,n+24}-\frac {6\,a^4\,c^2\,\sqrt {c\,x^2}}{b^4\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}+\frac {6\,a^3\,c^2\,n\,x\,\sqrt {c\,x^2}}{b^3\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}+\frac {a\,c^2\,n\,x^3\,\sqrt {c\,x^2}\,\left (n^2+3\,n+2\right )}{b\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}-\frac {3\,a^2\,c^2\,n\,x^2\,\sqrt {c\,x^2}\,\left (n+1\right )}{b^2\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}\right )}{x} \]

[In]

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

[Out]

((a + b*x)^n*((c^2*x^4*(c*x^2)^(1/2)*(11*n + 6*n^2 + n^3 + 6))/(50*n + 35*n^2 + 10*n^3 + n^4 + 24) - (6*a^4*c^
2*(c*x^2)^(1/2))/(b^4*(50*n + 35*n^2 + 10*n^3 + n^4 + 24)) + (6*a^3*c^2*n*x*(c*x^2)^(1/2))/(b^3*(50*n + 35*n^2
+ 10*n^3 + n^4 + 24)) + (a*c^2*n*x^3*(c*x^2)^(1/2)*(3*n + n^2 + 2))/(b*(50*n + 35*n^2 + 10*n^3 + n^4 + 24)) -
(3*a^2*c^2*n*x^2*(c*x^2)^(1/2)*(n + 1))/(b^2*(50*n + 35*n^2 + 10*n^3 + n^4 + 24))))/x