\(\int \frac {(c x^2)^{3/2} (a+b x)^n}{x} \, dx\) [932]

Optimal result

Integrand size = 20, antiderivative size = 99 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^n}{x} \, dx=\frac {a^2 c \sqrt {c x^2} (a+b x)^{1+n}}{b^3 (1+n) x}-\frac {2 a c \sqrt {c x^2} (a+b x)^{2+n}}{b^3 (2+n) x}+\frac {c \sqrt {c x^2} (a+b x)^{3+n}}{b^3 (3+n) x} \]

[Out]

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

Rubi [A] (verified)

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

[In]

Int[((c*x^2)^(3/2)*(a + b*x)^n)/x,x]

[Out]

(a^2*c*Sqrt[c*x^2]*(a + b*x)^(1 + n))/(b^3*(1 + n)*x) - (2*a*c*Sqrt[c*x^2]*(a + b*x)^(2 + n))/(b^3*(2 + n)*x)
+ (c*Sqrt[c*x^2]*(a + b*x)^(3 + n))/(b^3*(3 + 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 \sqrt {c x^2}\right ) \int x^2 (a+b x)^n \, dx}{x} \\ & = \frac {\left (c \sqrt {c x^2}\right ) \int \left (\frac {a^2 (a+b x)^n}{b^2}-\frac {2 a (a+b x)^{1+n}}{b^2}+\frac {(a+b x)^{2+n}}{b^2}\right ) \, dx}{x} \\ & = \frac {a^2 c \sqrt {c x^2} (a+b x)^{1+n}}{b^3 (1+n) x}-\frac {2 a c \sqrt {c x^2} (a+b x)^{2+n}}{b^3 (2+n) x}+\frac {c \sqrt {c x^2} (a+b x)^{3+n}}{b^3 (3+n) x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.71 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^n}{x} \, dx=\frac {c^2 x (a+b x)^{1+n} \left (2 a^2-2 a b (1+n) x+b^2 \left (2+3 n+n^2\right ) x^2\right )}{b^3 (1+n) (2+n) (3+n) \sqrt {c x^2}} \]

[In]

Integrate[((c*x^2)^(3/2)*(a + b*x)^n)/x,x]

[Out]

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

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.84

method	result	size
gosper	\(\frac {\left (c \,x^{2}\right )^{\frac {3}{2}} \left (b x +a \right )^{1+n} \left (b^{2} n^{2} x^{2}+3 b^{2} n \,x^{2}-2 a b n x +2 b^{2} x^{2}-2 a b x +2 a^{2}\right )}{b^{3} x^{3} \left (n^{3}+6 n^{2}+11 n +6\right )}\)	\(83\)
risch	\(\frac {c \sqrt {c \,x^{2}}\, \left (b^{3} n^{2} x^{3}+a \,b^{2} n^{2} x^{2}+3 b^{3} n \,x^{3}+a \,b^{2} n \,x^{2}+2 b^{3} x^{3}-2 a^{2} b n x +2 a^{3}\right ) \left (b x +a \right )^{n}}{x \left (2+n \right ) \left (3+n \right ) \left (1+n \right ) b^{3}}\)	\(99\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.14 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^n}{x} \, dx=-\frac {{\left (2 \, a^{2} b c n x - 2 \, a^{3} c - {\left (b^{3} c n^{2} + 3 \, b^{3} c n + 2 \, b^{3} c\right )} x^{3} - {\left (a b^{2} c n^{2} + a b^{2} c n\right )} x^{2}\right )} \sqrt {c x^{2}} {\left (b x + a\right )}^{n}}{{\left (b^{3} n^{3} + 6 \, b^{3} n^{2} + 11 \, b^{3} n + 6 \, b^{3}\right )} x} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral((c*x**2)**(3/2)*(a + b*x)**n/x, x)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.81 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^n}{x} \, dx=\frac {{\left ({\left (n^{2} + 3 \, n + 2\right )} b^{3} c^{\frac {3}{2}} x^{3} + {\left (n^{2} + n\right )} a b^{2} c^{\frac {3}{2}} x^{2} - 2 \, a^{2} b c^{\frac {3}{2}} n x + 2 \, a^{3} c^{\frac {3}{2}}\right )} {\left (b x + a\right )}^{n}}{{\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} b^{3}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.47 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^n}{x} \, dx=\frac {{\left (a+b\,x\right )}^n\,\left (\frac {c\,x^3\,\sqrt {c\,x^2}\,\left (n^2+3\,n+2\right )}{n^3+6\,n^2+11\,n+6}+\frac {2\,a^3\,c\,\sqrt {c\,x^2}}{b^3\,\left (n^3+6\,n^2+11\,n+6\right )}-\frac {2\,a^2\,c\,n\,x\,\sqrt {c\,x^2}}{b^2\,\left (n^3+6\,n^2+11\,n+6\right )}+\frac {a\,c\,n\,x^2\,\sqrt {c\,x^2}\,\left (n+1\right )}{b\,\left (n^3+6\,n^2+11\,n+6\right )}\right )}{x} \]

[In]

int(((c*x^2)^(3/2)*(a + b*x)^n)/x,x)

[Out]

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