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

Optimal. Leaf size=179 \[ \frac {a^4 c^2 \sqrt {c x^2} (a+b x)^{n+1}}{b^5 (n+1) x}-\frac {4 a^3 c^2 \sqrt {c x^2} (a+b x)^{n+2}}{b^5 (n+2) x}+\frac {6 a^2 c^2 \sqrt {c x^2} (a+b x)^{n+3}}{b^5 (n+3) x}-\frac {4 a c^2 \sqrt {c x^2} (a+b x)^{n+4}}{b^5 (n+4) x}+\frac {c^2 \sqrt {c x^2} (a+b x)^{n+5}}{b^5 (n+5) x} \]

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Rubi [A]  time = 0.05, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 43} \begin {gather*} \frac {a^4 c^2 \sqrt {c x^2} (a+b x)^{n+1}}{b^5 (n+1) x}-\frac {4 a^3 c^2 \sqrt {c x^2} (a+b x)^{n+2}}{b^5 (n+2) x}+\frac {6 a^2 c^2 \sqrt {c x^2} (a+b x)^{n+3}}{b^5 (n+3) x}-\frac {4 a c^2 \sqrt {c x^2} (a+b x)^{n+4}}{b^5 (n+4) x}+\frac {c^2 \sqrt {c x^2} (a+b x)^{n+5}}{b^5 (n+5) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((c*x^2)^(5/2)*(a + b*x)^n)/x,x]

[Out]

(a^4*c^2*Sqrt[c*x^2]*(a + b*x)^(1 + n))/(b^5*(1 + n)*x) - (4*a^3*c^2*Sqrt[c*x^2]*(a + b*x)^(2 + n))/(b^5*(2 +
n)*x) + (6*a^2*c^2*Sqrt[c*x^2]*(a + b*x)^(3 + n))/(b^5*(3 + n)*x) - (4*a*c^2*Sqrt[c*x^2]*(a + b*x)^(4 + n))/(b
^5*(4 + n)*x) + (c^2*Sqrt[c*x^2]*(a + b*x)^(5 + n))/(b^5*(5 + 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 43

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*} \int \frac {\left (c x^2\right )^{5/2} (a+b x)^n}{x} \, dx &=\frac {\left (c^2 \sqrt {c x^2}\right ) \int x^4 (a+b x)^n \, dx}{x}\\ &=\frac {\left (c^2 \sqrt {c x^2}\right ) \int \left (\frac {a^4 (a+b x)^n}{b^4}-\frac {4 a^3 (a+b x)^{1+n}}{b^4}+\frac {6 a^2 (a+b x)^{2+n}}{b^4}-\frac {4 a (a+b x)^{3+n}}{b^4}+\frac {(a+b x)^{4+n}}{b^4}\right ) \, dx}{x}\\ &=\frac {a^4 c^2 \sqrt {c x^2} (a+b x)^{1+n}}{b^5 (1+n) x}-\frac {4 a^3 c^2 \sqrt {c x^2} (a+b x)^{2+n}}{b^5 (2+n) x}+\frac {6 a^2 c^2 \sqrt {c x^2} (a+b x)^{3+n}}{b^5 (3+n) x}-\frac {4 a c^2 \sqrt {c x^2} (a+b x)^{4+n}}{b^5 (4+n) x}+\frac {c^2 \sqrt {c x^2} (a+b x)^{5+n}}{b^5 (5+n) x}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 133, normalized size = 0.74 \begin {gather*} \frac {c \left (c x^2\right )^{3/2} (a+b x)^{n+1} \left (24 a^4-24 a^3 b (n+1) x+12 a^2 b^2 \left (n^2+3 n+2\right ) x^2-4 a b^3 \left (n^3+6 n^2+11 n+6\right ) x^3+b^4 \left (n^4+10 n^3+35 n^2+50 n+24\right ) x^4\right )}{b^5 (n+1) (n+2) (n+3) (n+4) (n+5) x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(c*x^2)^(3/2)*(a + b*x)^(1 + n)*(24*a^4 - 24*a^3*b*(1 + n)*x + 12*a^2*b^2*(2 + 3*n + n^2)*x^2 - 4*a*b^3*(6
+ 11*n + 6*n^2 + n^3)*x^3 + b^4*(24 + 50*n + 35*n^2 + 10*n^3 + n^4)*x^4))/(b^5*(1 + n)*(2 + n)*(3 + n)*(4 + n)
*(5 + n)*x^3)

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IntegrateAlgebraic [F]  time = 0.24, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c x^2\right )^{5/2} (a+b x)^n}{x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.15, size = 265, normalized size = 1.48 \begin {gather*} -\frac {{\left (24 \, a^{4} b c^{2} n x - 24 \, a^{5} c^{2} - {\left (b^{5} c^{2} n^{4} + 10 \, b^{5} c^{2} n^{3} + 35 \, b^{5} c^{2} n^{2} + 50 \, b^{5} c^{2} n + 24 \, b^{5} c^{2}\right )} x^{5} - {\left (a b^{4} c^{2} n^{4} + 6 \, a b^{4} c^{2} n^{3} + 11 \, a b^{4} c^{2} n^{2} + 6 \, a b^{4} c^{2} n\right )} x^{4} + 4 \, {\left (a^{2} b^{3} c^{2} n^{3} + 3 \, a^{2} b^{3} c^{2} n^{2} + 2 \, a^{2} b^{3} c^{2} n\right )} x^{3} - 12 \, {\left (a^{3} b^{2} c^{2} n^{2} + a^{3} b^{2} c^{2} n\right )} x^{2}\right )} \sqrt {c x^{2}} {\left (b x + a\right )}^{n}}{{\left (b^{5} n^{5} + 15 \, b^{5} n^{4} + 85 \, b^{5} n^{3} + 225 \, b^{5} n^{2} + 274 \, b^{5} n + 120 \, b^{5}\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(24*a^4*b*c^2*n*x - 24*a^5*c^2 - (b^5*c^2*n^4 + 10*b^5*c^2*n^3 + 35*b^5*c^2*n^2 + 50*b^5*c^2*n + 24*b^5*c^2)*
x^5 - (a*b^4*c^2*n^4 + 6*a*b^4*c^2*n^3 + 11*a*b^4*c^2*n^2 + 6*a*b^4*c^2*n)*x^4 + 4*(a^2*b^3*c^2*n^3 + 3*a^2*b^
3*c^2*n^2 + 2*a^2*b^3*c^2*n)*x^3 - 12*(a^3*b^2*c^2*n^2 + a^3*b^2*c^2*n)*x^2)*sqrt(c*x^2)*(b*x + a)^n/((b^5*n^5
+ 15*b^5*n^4 + 85*b^5*n^3 + 225*b^5*n^2 + 274*b^5*n + 120*b^5)*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 199, normalized size = 1.11 \begin {gather*} \frac {\left (b^{4} n^{4} x^{4}+10 b^{4} n^{3} x^{4}-4 a \,b^{3} n^{3} x^{3}+35 b^{4} n^{2} x^{4}-24 a \,b^{3} n^{2} x^{3}+50 b^{4} n \,x^{4}+12 a^{2} b^{2} n^{2} x^{2}-44 a \,b^{3} n \,x^{3}+24 b^{4} x^{4}+36 a^{2} b^{2} n \,x^{2}-24 a \,b^{3} x^{3}-24 a^{3} b n x +24 a^{2} b^{2} x^{2}-24 a^{3} b x +24 a^{4}\right ) \left (c \,x^{2}\right )^{\frac {5}{2}} \left (b x +a \right )^{n +1}}{\left (n^{5}+15 n^{4}+85 n^{3}+225 n^{2}+274 n +120\right ) b^{5} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x+a)^(n+1)*(b^4*n^4*x^4+10*b^4*n^3*x^4-4*a*b^3*n^3*x^3+35*b^4*n^2*x^4-24*a*b^3*n^2*x^3+50*b^4*n*x^4+12*a^2*
b^2*n^2*x^2-44*a*b^3*n*x^3+24*b^4*x^4+36*a^2*b^2*n*x^2-24*a*b^3*x^3-24*a^3*b*n*x+24*a^2*b^2*x^2-24*a^3*b*x+24*
a^4)*(c*x^2)^(5/2)/x^5/b^5/(n^5+15*n^4+85*n^3+225*n^2+274*n+120)

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maxima [A]  time = 1.45, size = 157, normalized size = 0.88 \begin {gather*} \frac {{\left ({\left (n^{4} + 10 \, n^{3} + 35 \, n^{2} + 50 \, n + 24\right )} b^{5} c^{\frac {5}{2}} x^{5} + {\left (n^{4} + 6 \, n^{3} + 11 \, n^{2} + 6 \, n\right )} a b^{4} c^{\frac {5}{2}} x^{4} - 4 \, {\left (n^{3} + 3 \, n^{2} + 2 \, n\right )} a^{2} b^{3} c^{\frac {5}{2}} x^{3} + 12 \, {\left (n^{2} + n\right )} a^{3} b^{2} c^{\frac {5}{2}} x^{2} - 24 \, a^{4} b c^{\frac {5}{2}} n x + 24 \, a^{5} c^{\frac {5}{2}}\right )} {\left (b x + a\right )}^{n}}{{\left (n^{5} + 15 \, n^{4} + 85 \, n^{3} + 225 \, n^{2} + 274 \, n + 120\right )} b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((n^4 + 10*n^3 + 35*n^2 + 50*n + 24)*b^5*c^(5/2)*x^5 + (n^4 + 6*n^3 + 11*n^2 + 6*n)*a*b^4*c^(5/2)*x^4 - 4*(n^3
+ 3*n^2 + 2*n)*a^2*b^3*c^(5/2)*x^3 + 12*(n^2 + n)*a^3*b^2*c^(5/2)*x^2 - 24*a^4*b*c^(5/2)*n*x + 24*a^5*c^(5/2)
)*(b*x + a)^n/((n^5 + 15*n^4 + 85*n^3 + 225*n^2 + 274*n + 120)*b^5)

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mupad [B]  time = 0.38, size = 319, normalized size = 1.78 \begin {gather*} \frac {{\left (a+b\,x\right )}^n\,\left (\frac {c^2\,x^5\,\sqrt {c\,x^2}\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}{n^5+15\,n^4+85\,n^3+225\,n^2+274\,n+120}+\frac {24\,a^5\,c^2\,\sqrt {c\,x^2}}{b^5\,\left (n^5+15\,n^4+85\,n^3+225\,n^2+274\,n+120\right )}-\frac {24\,a^4\,c^2\,n\,x\,\sqrt {c\,x^2}}{b^4\,\left (n^5+15\,n^4+85\,n^3+225\,n^2+274\,n+120\right )}+\frac {a\,c^2\,n\,x^4\,\sqrt {c\,x^2}\,\left (n^3+6\,n^2+11\,n+6\right )}{b\,\left (n^5+15\,n^4+85\,n^3+225\,n^2+274\,n+120\right )}+\frac {12\,a^3\,c^2\,n\,x^2\,\sqrt {c\,x^2}\,\left (n+1\right )}{b^3\,\left (n^5+15\,n^4+85\,n^3+225\,n^2+274\,n+120\right )}-\frac {4\,a^2\,c^2\,n\,x^3\,\sqrt {c\,x^2}\,\left (n^2+3\,n+2\right )}{b^2\,\left (n^5+15\,n^4+85\,n^3+225\,n^2+274\,n+120\right )}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((c*x^2)^(5/2)*(a + b*x)^n)/x,x)

[Out]

((a + b*x)^n*((c^2*x^5*(c*x^2)^(1/2)*(50*n + 35*n^2 + 10*n^3 + n^4 + 24))/(274*n + 225*n^2 + 85*n^3 + 15*n^4 +
n^5 + 120) + (24*a^5*c^2*(c*x^2)^(1/2))/(b^5*(274*n + 225*n^2 + 85*n^3 + 15*n^4 + n^5 + 120)) - (24*a^4*c^2*n
*x*(c*x^2)^(1/2))/(b^4*(274*n + 225*n^2 + 85*n^3 + 15*n^4 + n^5 + 120)) + (a*c^2*n*x^4*(c*x^2)^(1/2)*(11*n + 6
*n^2 + n^3 + 6))/(b*(274*n + 225*n^2 + 85*n^3 + 15*n^4 + n^5 + 120)) + (12*a^3*c^2*n*x^2*(c*x^2)^(1/2)*(n + 1)
)/(b^3*(274*n + 225*n^2 + 85*n^3 + 15*n^4 + n^5 + 120)) - (4*a^2*c^2*n*x^3*(c*x^2)^(1/2)*(3*n + n^2 + 2))/(b^2
*(274*n + 225*n^2 + 85*n^3 + 15*n^4 + n^5 + 120))))/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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