Integrand size = 20, antiderivative size = 151 \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)} \, dx=\frac {c (4 b c-7 a d) \sqrt {c+d x}}{4 a^2 x}-\frac {c (c+d x)^{3/2}}{2 a x^2}-\frac {\sqrt {c} \left (8 b^2 c^2-20 a b c d+15 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{4 a^3}+\frac {2 (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^3 \sqrt {b}} \]
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Time = 0.10 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {100, 154, 162, 65, 214} \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)} \, dx=\frac {2 (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^3 \sqrt {b}}+\frac {c \sqrt {c+d x} (4 b c-7 a d)}{4 a^2 x}-\frac {\sqrt {c} \left (15 a^2 d^2-20 a b c d+8 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{4 a^3}-\frac {c (c+d x)^{3/2}}{2 a x^2} \]
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Rule 65
Rule 100
Rule 154
Rule 162
Rule 214
Rubi steps \begin{align*} \text {integral}& = -\frac {c (c+d x)^{3/2}}{2 a x^2}-\frac {\int \frac {\sqrt {c+d x} \left (\frac {1}{2} c (4 b c-7 a d)+\frac {1}{2} d (b c-4 a d) x\right )}{x^2 (a+b x)} \, dx}{2 a} \\ & = \frac {c (4 b c-7 a d) \sqrt {c+d x}}{4 a^2 x}-\frac {c (c+d x)^{3/2}}{2 a x^2}-\frac {\int \frac {-\frac {1}{4} c \left (8 b^2 c^2-20 a b c d+15 a^2 d^2\right )-\frac {1}{4} d \left (4 b^2 c^2-9 a b c d+8 a^2 d^2\right ) x}{x (a+b x) \sqrt {c+d x}} \, dx}{2 a^2} \\ & = \frac {c (4 b c-7 a d) \sqrt {c+d x}}{4 a^2 x}-\frac {c (c+d x)^{3/2}}{2 a x^2}-\frac {(b c-a d)^3 \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{a^3}+\frac {\left (c \left (8 b^2 c^2-20 a b c d+15 a^2 d^2\right )\right ) \int \frac {1}{x \sqrt {c+d x}} \, dx}{8 a^3} \\ & = \frac {c (4 b c-7 a d) \sqrt {c+d x}}{4 a^2 x}-\frac {c (c+d x)^{3/2}}{2 a x^2}-\frac {\left (2 (b c-a d)^3\right ) \text {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{a^3 d}+\frac {\left (c \left (8 b^2 c^2-20 a b c d+15 a^2 d^2\right )\right ) \text {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{4 a^3 d} \\ & = \frac {c (4 b c-7 a d) \sqrt {c+d x}}{4 a^2 x}-\frac {c (c+d x)^{3/2}}{2 a x^2}-\frac {\sqrt {c} \left (8 b^2 c^2-20 a b c d+15 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{4 a^3}+\frac {2 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^3 \sqrt {b}} \\ \end{align*}
Time = 0.36 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.87 \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)} \, dx=\frac {\frac {a c \sqrt {c+d x} (-2 a c+4 b c x-9 a d x)}{x^2}+\frac {8 (-b c+a d)^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{\sqrt {b}}-\sqrt {c} \left (8 b^2 c^2-20 a b c d+15 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{4 a^3} \]
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Time = 0.62 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.82
method | result | size |
pseudoelliptic | \(\frac {\frac {2 \left (a d -b c \right )^{3} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}+\frac {c \left (-\frac {\sqrt {d x +c}\, a \left (9 a d x -4 b c x +2 a c \right )}{x^{2}}-\frac {\left (15 a^{2} d^{2}-20 a b c d +8 b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{\sqrt {c}}\right )}{4}}{a^{3}}\) | \(124\) |
risch | \(-\frac {c \sqrt {d x +c}\, \left (9 a d x -4 b c x +2 a c \right )}{4 a^{2} x^{2}}+\frac {d \left (\frac {\left (8 a^{3} d^{3}-24 a^{2} b c \,d^{2}+24 a \,b^{2} c^{2} d -8 b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{a d \sqrt {\left (a d -b c \right ) b}}-\frac {\sqrt {c}\, \left (15 a^{2} d^{2}-20 a b c d +8 b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a d}\right )}{4 a^{2}}\) | \(164\) |
derivativedivides | \(2 d^{3} \left (-\frac {c \left (\frac {\left (\frac {9}{8} a^{2} d^{2}-\frac {1}{2} a b c d \right ) \left (d x +c \right )^{\frac {3}{2}}+\left (-\frac {7}{8} c \,a^{2} d^{2}+\frac {1}{2} b \,c^{2} d a \right ) \sqrt {d x +c}}{d^{2} x^{2}}+\frac {\left (15 a^{2} d^{2}-20 a b c d +8 b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{8 \sqrt {c}}\right )}{a^{3} d^{3}}+\frac {\left (a d -b c \right )^{3} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{a^{3} d^{3} \sqrt {\left (a d -b c \right ) b}}\right )\) | \(165\) |
default | \(2 d^{3} \left (-\frac {c \left (\frac {\left (\frac {9}{8} a^{2} d^{2}-\frac {1}{2} a b c d \right ) \left (d x +c \right )^{\frac {3}{2}}+\left (-\frac {7}{8} c \,a^{2} d^{2}+\frac {1}{2} b \,c^{2} d a \right ) \sqrt {d x +c}}{d^{2} x^{2}}+\frac {\left (15 a^{2} d^{2}-20 a b c d +8 b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{8 \sqrt {c}}\right )}{a^{3} d^{3}}+\frac {\left (a d -b c \right )^{3} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{a^{3} d^{3} \sqrt {\left (a d -b c \right ) b}}\right )\) | \(165\) |
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Time = 0.35 (sec) , antiderivative size = 713, normalized size of antiderivative = 4.72 \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)} \, dx=\left [\frac {8 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d + 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) + {\left (8 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt {c} x^{2} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) - 2 \, {\left (2 \, a^{2} c^{2} - {\left (4 \, a b c^{2} - 9 \, a^{2} c d\right )} x\right )} \sqrt {d x + c}}{8 \, a^{3} x^{2}}, \frac {16 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) + {\left (8 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt {c} x^{2} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) - 2 \, {\left (2 \, a^{2} c^{2} - {\left (4 \, a b c^{2} - 9 \, a^{2} c d\right )} x\right )} \sqrt {d x + c}}{8 \, a^{3} x^{2}}, \frac {{\left (8 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) + 4 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d + 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) - {\left (2 \, a^{2} c^{2} - {\left (4 \, a b c^{2} - 9 \, a^{2} c d\right )} x\right )} \sqrt {d x + c}}{4 \, a^{3} x^{2}}, \frac {8 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) + {\left (8 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) - {\left (2 \, a^{2} c^{2} - {\left (4 \, a b c^{2} - 9 \, a^{2} c d\right )} x\right )} \sqrt {d x + c}}{4 \, a^{3} x^{2}}\right ] \]
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\[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{2}}}{x^{3} \left (a + b x\right )}\, dx \]
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Exception generated. \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)} \, dx=\text {Exception raised: ValueError} \]
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Time = 0.30 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.31 \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)} \, dx=-\frac {2 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} a^{3}} + \frac {{\left (8 \, b^{2} c^{3} - 20 \, a b c^{2} d + 15 \, a^{2} c d^{2}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{4 \, a^{3} \sqrt {-c}} + \frac {4 \, {\left (d x + c\right )}^{\frac {3}{2}} b c^{2} d - 4 \, \sqrt {d x + c} b c^{3} d - 9 \, {\left (d x + c\right )}^{\frac {3}{2}} a c d^{2} + 7 \, \sqrt {d x + c} a c^{2} d^{2}}{4 \, a^{2} d^{2} x^{2}} \]
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Time = 0.75 (sec) , antiderivative size = 1204, normalized size of antiderivative = 7.97 \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)} \, dx=\frac {\frac {\left (7\,a\,c^2\,d^2-4\,b\,c^3\,d\right )\,\sqrt {c+d\,x}}{4\,a^2}-\frac {\left (9\,a\,c\,d^2-4\,b\,c^2\,d\right )\,{\left (c+d\,x\right )}^{3/2}}{4\,a^2}}{{\left (c+d\,x\right )}^2-2\,c\,\left (c+d\,x\right )+c^2}+\frac {2\,\mathrm {atanh}\left (\frac {95\,b^2\,c^2\,d^6\,\sqrt {c+d\,x}\,\sqrt {-a^5\,b\,d^5+5\,a^4\,b^2\,c\,d^4-10\,a^3\,b^3\,c^2\,d^3+10\,a^2\,b^4\,c^3\,d^2-5\,a\,b^5\,c^4\,d+b^6\,c^5}}{4\,\left (\frac {215\,b^5\,c^5\,d^6}{4}-\frac {469\,a\,b^4\,c^4\,d^7}{4}+\frac {517\,a^2\,b^3\,c^3\,d^8}{4}-\frac {287\,a^3\,b^2\,c^2\,d^9}{4}-\frac {10\,b^6\,c^6\,d^5}{a}+16\,a^4\,b\,c\,d^{10}\right )}+\frac {10\,b^3\,c^3\,d^5\,\sqrt {c+d\,x}\,\sqrt {-a^5\,b\,d^5+5\,a^4\,b^2\,c\,d^4-10\,a^3\,b^3\,c^2\,d^3+10\,a^2\,b^4\,c^3\,d^2-5\,a\,b^5\,c^4\,d+b^6\,c^5}}{-16\,a^5\,b\,c\,d^{10}+\frac {287\,a^4\,b^2\,c^2\,d^9}{4}-\frac {517\,a^3\,b^3\,c^3\,d^8}{4}+\frac {469\,a^2\,b^4\,c^4\,d^7}{4}-\frac {215\,a\,b^5\,c^5\,d^6}{4}+10\,b^6\,c^6\,d^5}+\frac {16\,b\,c\,d^7\,\sqrt {c+d\,x}\,\sqrt {-a^5\,b\,d^5+5\,a^4\,b^2\,c\,d^4-10\,a^3\,b^3\,c^2\,d^3+10\,a^2\,b^4\,c^3\,d^2-5\,a\,b^5\,c^4\,d+b^6\,c^5}}{\frac {469\,b^4\,c^4\,d^7}{4}-\frac {517\,a\,b^3\,c^3\,d^8}{4}+\frac {287\,a^2\,b^2\,c^2\,d^9}{4}-\frac {215\,b^5\,c^5\,d^6}{4\,a}+\frac {10\,b^6\,c^6\,d^5}{a^2}-16\,a^3\,b\,c\,d^{10}}\right )\,\sqrt {-b\,{\left (a\,d-b\,c\right )}^5}}{a^3\,b}-\frac {\sqrt {c}\,\mathrm {atanh}\left (\frac {3665\,b^2\,c^{3/2}\,d^9\,\sqrt {c+d\,x}}{32\,\left (\frac {3665\,b^2\,c^2\,d^9}{32}-30\,a\,b\,c\,d^{10}-\frac {5717\,b^3\,c^3\,d^8}{32\,a}+\frac {1143\,b^4\,c^4\,d^7}{8\,a^2}-\frac {235\,b^5\,c^5\,d^6}{4\,a^3}+\frac {10\,b^6\,c^6\,d^5}{a^4}\right )}+\frac {5717\,b^3\,c^{5/2}\,d^8\,\sqrt {c+d\,x}}{32\,\left (\frac {5717\,b^3\,c^3\,d^8}{32}-\frac {3665\,a\,b^2\,c^2\,d^9}{32}-\frac {1143\,b^4\,c^4\,d^7}{8\,a}+\frac {235\,b^5\,c^5\,d^6}{4\,a^2}-\frac {10\,b^6\,c^6\,d^5}{a^3}+30\,a^2\,b\,c\,d^{10}\right )}+\frac {1143\,b^4\,c^{7/2}\,d^7\,\sqrt {c+d\,x}}{8\,\left (\frac {1143\,b^4\,c^4\,d^7}{8}-\frac {5717\,a\,b^3\,c^3\,d^8}{32}+\frac {3665\,a^2\,b^2\,c^2\,d^9}{32}-\frac {235\,b^5\,c^5\,d^6}{4\,a}+\frac {10\,b^6\,c^6\,d^5}{a^2}-30\,a^3\,b\,c\,d^{10}\right )}+\frac {235\,b^5\,c^{9/2}\,d^6\,\sqrt {c+d\,x}}{4\,\left (\frac {235\,b^5\,c^5\,d^6}{4}-\frac {1143\,a\,b^4\,c^4\,d^7}{8}+\frac {5717\,a^2\,b^3\,c^3\,d^8}{32}-\frac {3665\,a^3\,b^2\,c^2\,d^9}{32}-\frac {10\,b^6\,c^6\,d^5}{a}+30\,a^4\,b\,c\,d^{10}\right )}+\frac {10\,b^6\,c^{11/2}\,d^5\,\sqrt {c+d\,x}}{-30\,a^5\,b\,c\,d^{10}+\frac {3665\,a^4\,b^2\,c^2\,d^9}{32}-\frac {5717\,a^3\,b^3\,c^3\,d^8}{32}+\frac {1143\,a^2\,b^4\,c^4\,d^7}{8}-\frac {235\,a\,b^5\,c^5\,d^6}{4}+10\,b^6\,c^6\,d^5}-\frac {30\,a\,b\,\sqrt {c}\,d^{10}\,\sqrt {c+d\,x}}{\frac {3665\,b^2\,c^2\,d^9}{32}-30\,a\,b\,c\,d^{10}-\frac {5717\,b^3\,c^3\,d^8}{32\,a}+\frac {1143\,b^4\,c^4\,d^7}{8\,a^2}-\frac {235\,b^5\,c^5\,d^6}{4\,a^3}+\frac {10\,b^6\,c^6\,d^5}{a^4}}\right )\,\left (15\,a^2\,d^2-20\,a\,b\,c\,d+8\,b^2\,c^2\right )}{4\,a^3} \]
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