Integrand size = 20, antiderivative size = 50 \[ \int \frac {\sqrt {c x^2} (a+b x)^n}{x^4} \, dx=-\frac {b^2 \sqrt {c x^2} (a+b x)^{1+n} \operatorname {Hypergeometric2F1}\left (3,1+n,2+n,1+\frac {b x}{a}\right )}{a^3 (1+n) x} \]
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Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 67} \[ \int \frac {\sqrt {c x^2} (a+b x)^n}{x^4} \, dx=-\frac {b^2 \sqrt {c x^2} (a+b x)^{n+1} \operatorname {Hypergeometric2F1}\left (3,n+1,n+2,\frac {b x}{a}+1\right )}{a^3 (n+1) x} \]
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Rule 15
Rule 67
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {c x^2} \int \frac {(a+b x)^n}{x^3} \, dx}{x} \\ & = -\frac {b^2 \sqrt {c x^2} (a+b x)^{1+n} \, _2F_1\left (3,1+n;2+n;1+\frac {b x}{a}\right )}{a^3 (1+n) x} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {c x^2} (a+b x)^n}{x^4} \, dx=-\frac {b^2 \sqrt {c x^2} (a+b x)^{1+n} \operatorname {Hypergeometric2F1}\left (3,1+n,2+n,1+\frac {b x}{a}\right )}{a^3 (1+n) x} \]
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\[\int \frac {\left (b x +a \right )^{n} \sqrt {c \,x^{2}}}{x^{4}}d x\]
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\[ \int \frac {\sqrt {c x^2} (a+b x)^n}{x^4} \, dx=\int { \frac {\sqrt {c x^{2}} {\left (b x + a\right )}^{n}}{x^{4}} \,d x } \]
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\[ \int \frac {\sqrt {c x^2} (a+b x)^n}{x^4} \, dx=\int \frac {\sqrt {c x^{2}} \left (a + b x\right )^{n}}{x^{4}}\, dx \]
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\[ \int \frac {\sqrt {c x^2} (a+b x)^n}{x^4} \, dx=\int { \frac {\sqrt {c x^{2}} {\left (b x + a\right )}^{n}}{x^{4}} \,d x } \]
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\[ \int \frac {\sqrt {c x^2} (a+b x)^n}{x^4} \, dx=\int { \frac {\sqrt {c x^{2}} {\left (b x + a\right )}^{n}}{x^{4}} \,d x } \]
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Timed out. \[ \int \frac {\sqrt {c x^2} (a+b x)^n}{x^4} \, dx=\int \frac {\sqrt {c\,x^2}\,{\left (a+b\,x\right )}^n}{x^4} \,d x \]
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