Optimal. Leaf size=179 \[ \frac {a^4 c^2 \sqrt {c x^2} (a+b x)^{1+n}}{b^5 (1+n) x}-\frac {4 a^3 c^2 \sqrt {c x^2} (a+b x)^{2+n}}{b^5 (2+n) x}+\frac {6 a^2 c^2 \sqrt {c x^2} (a+b x)^{3+n}}{b^5 (3+n) x}-\frac {4 a c^2 \sqrt {c x^2} (a+b x)^{4+n}}{b^5 (4+n) x}+\frac {c^2 \sqrt {c x^2} (a+b x)^{5+n}}{b^5 (5+n) x} \]
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Rubi [A]
time = 0.04, 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, 45}
\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.
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Rule 15
Rule 45
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)^{1+n} \left (24 a^4-24 a^3 b (1+n) x+12 a^2 b^2 \left (2+3 n+n^2\right ) x^2-4 a b^3 \left (6+11 n+6 n^2+n^3\right ) x^3+b^4 \left (24+50 n+35 n^2+10 n^3+n^4\right ) x^4\right )}{b^5 (1+n) (2+n) (3+n) (4+n) (5+n) x^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.14, size = 199, normalized size = 1.11
method | result | size |
gosper | \(\frac {\left (b x +a \right )^{1+n} \left (b^{4} n^{4} x^{4}+10 b^{4} n^{3} x^{4}-4 a \,b^{3} n^{3} x^{3}+35 b^{4} n^{2} x^{4}-24 a \,b^{3} n^{2} x^{3}+50 b^{4} n \,x^{4}+12 a^{2} b^{2} n^{2} x^{2}-44 x^{3} a n \,b^{3}+24 b^{4} x^{4}+36 a^{2} n \,x^{2} b^{2}-24 a \,b^{3} x^{3}-24 a^{3} b n x +24 a^{2} b^{2} x^{2}-24 a^{3} b x +24 a^{4}\right ) \left (c \,x^{2}\right )^{\frac {5}{2}}}{x^{5} b^{5} \left (n^{5}+15 n^{4}+85 n^{3}+225 n^{2}+274 n +120\right )}\) | \(199\) |
risch | \(\frac {c^{2} \sqrt {c \,x^{2}}\, \left (b^{5} n^{4} x^{5}+a \,b^{4} n^{4} x^{4}+10 b^{5} n^{3} x^{5}+6 a \,b^{4} n^{3} x^{4}+35 b^{5} n^{2} x^{5}-4 a^{2} b^{3} n^{3} x^{3}+11 a \,b^{4} n^{2} x^{4}+50 b^{5} n \,x^{5}-12 a^{2} b^{3} n^{2} x^{3}+6 x^{4} a n \,b^{4}+24 b^{5} x^{5}+12 a^{3} b^{2} n^{2} x^{2}-8 a^{2} n \,x^{3} b^{3}+12 a^{3} n \,x^{2} b^{2}-24 a^{4} b n x +24 a^{5}\right ) \left (b x +a \right )^{n}}{x \left (4+n \right ) \left (5+n \right ) \left (3+n \right ) \left (2+n \right ) \left (1+n \right ) b^{5}}\) | \(224\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.30, 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.
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Fricas [A]
time = 0.57, 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.
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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.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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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.
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