Integrand size = 20, antiderivative size = 128 \[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)} \, dx=\frac {d (b c+2 a d) \sqrt {c+d x}}{a b}-\frac {c (c+d x)^{3/2}}{a x}+\frac {c^{3/2} (2 b c-5 a d) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a^2}-\frac {2 (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^2 b^{3/2}} \]
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Time = 0.11 (sec) , antiderivative size = 128, 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, 159, 162, 65, 214} \[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)} \, dx=-\frac {2 (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^2 b^{3/2}}+\frac {c^{3/2} (2 b c-5 a d) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a^2}+\frac {d \sqrt {c+d x} (2 a d+b c)}{a b}-\frac {c (c+d x)^{3/2}}{a x} \]
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Rule 65
Rule 100
Rule 159
Rule 162
Rule 214
Rubi steps \begin{align*} \text {integral}& = -\frac {c (c+d x)^{3/2}}{a x}-\frac {\int \frac {\sqrt {c+d x} \left (\frac {1}{2} c (2 b c-5 a d)-\frac {1}{2} d (b c+2 a d) x\right )}{x (a+b x)} \, dx}{a} \\ & = \frac {d (b c+2 a d) \sqrt {c+d x}}{a b}-\frac {c (c+d x)^{3/2}}{a x}-\frac {2 \int \frac {\frac {1}{4} b c^2 (2 b c-5 a d)+\frac {1}{4} d \left (b^2 c^2-6 a b c d+2 a^2 d^2\right ) x}{x (a+b x) \sqrt {c+d x}} \, dx}{a b} \\ & = \frac {d (b c+2 a d) \sqrt {c+d x}}{a b}-\frac {c (c+d x)^{3/2}}{a x}-\frac {\left (c^2 (2 b c-5 a d)\right ) \int \frac {1}{x \sqrt {c+d x}} \, dx}{2 a^2}+\frac {(b c-a d)^3 \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{a^2 b} \\ & = \frac {d (b c+2 a d) \sqrt {c+d x}}{a b}-\frac {c (c+d x)^{3/2}}{a x}-\frac {\left (c^2 (2 b c-5 a d)\right ) \text {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{a^2 d}+\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^2 b d} \\ & = \frac {d (b c+2 a d) \sqrt {c+d x}}{a b}-\frac {c (c+d x)^{3/2}}{a x}+\frac {c^{3/2} (2 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a^2}-\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^2 b^{3/2}} \\ \end{align*}
Time = 0.32 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.89 \[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)} \, dx=\frac {-\frac {a \sqrt {c+d x} \left (b c^2-2 a d^2 x\right )}{b x}-\frac {2 (-b c+a d)^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{b^{3/2}}+c^{3/2} (2 b c-5 a d) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a^2} \]
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Time = 0.60 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.97
method | result | size |
pseudoelliptic | \(\frac {-2 x \left (a d -b c \right )^{3} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )+2 \sqrt {\left (a d -b c \right ) b}\, \left (x b \left (c^{\frac {5}{2}} b -\frac {5 a d \,c^{\frac {3}{2}}}{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )+\sqrt {d x +c}\, a \left (x a \,d^{2}-\frac {b \,c^{2}}{2}\right )\right )}{b \,a^{2} \sqrt {\left (a d -b c \right ) b}\, x}\) | \(124\) |
derivativedivides | \(2 d^{2} \left (\frac {\sqrt {d x +c}}{b}-\frac {c^{2} \left (\frac {a \sqrt {d x +c}}{2 x}+\frac {\left (5 a d -2 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{2 \sqrt {c}}\right )}{a^{2} d^{2}}+\frac {\left (-a^{3} d^{3}+3 a^{2} b c \,d^{2}-3 a \,b^{2} c^{2} d +b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{b \,a^{2} d^{2} \sqrt {\left (a d -b c \right ) b}}\right )\) | \(149\) |
default | \(2 d^{2} \left (\frac {\sqrt {d x +c}}{b}-\frac {c^{2} \left (\frac {a \sqrt {d x +c}}{2 x}+\frac {\left (5 a d -2 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{2 \sqrt {c}}\right )}{a^{2} d^{2}}+\frac {\left (-a^{3} d^{3}+3 a^{2} b c \,d^{2}-3 a \,b^{2} c^{2} d +b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{b \,a^{2} d^{2} \sqrt {\left (a d -b c \right ) b}}\right )\) | \(149\) |
risch | \(-\frac {c^{2} \sqrt {d x +c}}{a x}+\frac {d \left (\frac {2 a d \sqrt {d x +c}}{b}+\frac {\left (-2 a^{3} d^{3}+6 a^{2} b c \,d^{2}-6 a \,b^{2} c^{2} d +2 b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{a b d \sqrt {\left (a d -b c \right ) b}}-\frac {c^{\frac {3}{2}} \left (5 a d -2 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a d}\right )}{a}\) | \(153\) |
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Time = 0.33 (sec) , antiderivative size = 638, normalized size of antiderivative = 4.98 \[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)} \, dx=\left [\frac {2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) - {\left (2 \, b^{2} c^{2} - 5 \, a b c d\right )} \sqrt {c} x \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (2 \, a^{2} d^{2} x - a b c^{2}\right )} \sqrt {d x + c}}{2 \, a^{2} b x}, -\frac {4 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) + {\left (2 \, b^{2} c^{2} - 5 \, a b c d\right )} \sqrt {c} x \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) - 2 \, {\left (2 \, a^{2} d^{2} x - a b c^{2}\right )} \sqrt {d x + c}}{2 \, a^{2} b x}, -\frac {{\left (2 \, b^{2} c^{2} - 5 \, a b c d\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) - {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) - {\left (2 \, a^{2} d^{2} x - a b c^{2}\right )} \sqrt {d x + c}}{a^{2} b x}, -\frac {2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) + {\left (2 \, b^{2} c^{2} - 5 \, a b c d\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) - {\left (2 \, a^{2} d^{2} x - a b c^{2}\right )} \sqrt {d x + c}}{a^{2} b x}\right ] \]
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\[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{2}}}{x^{2} \left (a + b x\right )}\, dx \]
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Exception generated. \[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)} \, dx=\text {Exception raised: ValueError} \]
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Time = 0.29 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.19 \[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)} \, dx=\frac {2 \, \sqrt {d x + c} d^{2}}{b} - \frac {\sqrt {d x + c} c^{2}}{a x} - \frac {{\left (2 \, b c^{3} - 5 \, a c^{2} d\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{a^{2} \sqrt {-c}} + \frac {2 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} a^{2} b} \]
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Time = 0.95 (sec) , antiderivative size = 1402, normalized size of antiderivative = 10.95 \[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)} \, dx=\frac {2\,d^2\,\sqrt {c+d\,x}}{b}+\frac {b\,c^2\,d\,\sqrt {c+d\,x}}{a\,\left (b\,c-b\,\left (c+d\,x\right )\right )}-\frac {\mathrm {atan}\left (\frac {c^3\,d^5\,\sqrt {c+d\,x}\,\sqrt {-a^5\,b^3\,d^5+5\,a^4\,b^4\,c\,d^4-10\,a^3\,b^5\,c^2\,d^3+10\,a^2\,b^6\,c^3\,d^2-5\,a\,b^7\,c^4\,d+b^8\,c^5}\,160{}\mathrm {i}}{740\,a\,b^3\,c^5\,d^6-340\,b^4\,c^6\,d^5-128\,a^4\,c^2\,d^9+448\,a^3\,b\,c^3\,d^8+\frac {16\,a^5\,c\,d^{10}}{b}-796\,a^2\,b^2\,c^4\,d^7+\frac {60\,b^5\,c^7\,d^4}{a}}-\frac {c^2\,d^6\,\sqrt {c+d\,x}\,\sqrt {-a^5\,b^3\,d^5+5\,a^4\,b^4\,c\,d^4-10\,a^3\,b^5\,c^2\,d^3+10\,a^2\,b^6\,c^3\,d^2-5\,a\,b^7\,c^4\,d+b^8\,c^5}\,80{}\mathrm {i}}{16\,a^4\,c\,d^{10}+740\,b^4\,c^5\,d^6-796\,a\,b^3\,c^4\,d^7-128\,a^3\,b\,c^2\,d^9+448\,a^2\,b^2\,c^3\,d^8-\frac {340\,b^5\,c^6\,d^5}{a}+\frac {60\,b^6\,c^7\,d^4}{a^2}}-\frac {c^4\,d^4\,\sqrt {c+d\,x}\,\sqrt {-a^5\,b^3\,d^5+5\,a^4\,b^4\,c\,d^4-10\,a^3\,b^5\,c^2\,d^3+10\,a^2\,b^6\,c^3\,d^2-5\,a\,b^7\,c^4\,d+b^8\,c^5}\,60{}\mathrm {i}}{448\,a^4\,c^3\,d^8+60\,b^4\,c^7\,d^4-340\,a\,b^3\,c^6\,d^5-796\,a^3\,b\,c^4\,d^7+\frac {16\,a^6\,c\,d^{10}}{b^2}+740\,a^2\,b^2\,c^5\,d^6-\frac {128\,a^5\,c^2\,d^9}{b}}+\frac {a\,c\,d^7\,\sqrt {c+d\,x}\,\sqrt {-a^5\,b^3\,d^5+5\,a^4\,b^4\,c\,d^4-10\,a^3\,b^5\,c^2\,d^3+10\,a^2\,b^6\,c^3\,d^2-5\,a\,b^7\,c^4\,d+b^8\,c^5}\,16{}\mathrm {i}}{740\,b^5\,c^5\,d^6-796\,a\,b^4\,c^4\,d^7+448\,a^2\,b^3\,c^3\,d^8-128\,a^3\,b^2\,c^2\,d^9-\frac {340\,b^6\,c^6\,d^5}{a}+\frac {60\,b^7\,c^7\,d^4}{a^2}+16\,a^4\,b\,c\,d^{10}}\right )\,\sqrt {-b^3\,{\left (a\,d-b\,c\right )}^5}\,2{}\mathrm {i}}{a^2\,b^3}+\frac {\mathrm {atan}\left (\frac {a^3\,d^9\,\sqrt {c^3}\,\sqrt {c+d\,x}\,40{}\mathrm {i}}{40\,a^3\,c^2\,d^9-790\,b^3\,c^5\,d^6+696\,a\,b^2\,c^4\,d^7-256\,a^2\,b\,c^3\,d^8+\frac {370\,b^4\,c^6\,d^5}{a}-\frac {60\,b^5\,c^7\,d^4}{a^2}}+\frac {b^2\,c^3\,d^6\,\sqrt {c^3}\,\sqrt {c+d\,x}\,790{}\mathrm {i}}{256\,a^2\,c^3\,d^8+790\,b^2\,c^5\,d^6-\frac {370\,b^3\,c^6\,d^5}{a}-\frac {40\,a^3\,c^2\,d^9}{b}+\frac {60\,b^4\,c^7\,d^4}{a^2}-696\,a\,b\,c^4\,d^7}-\frac {b^3\,c^4\,d^5\,\sqrt {c^3}\,\sqrt {c+d\,x}\,370{}\mathrm {i}}{256\,a^3\,c^3\,d^8-370\,b^3\,c^6\,d^5+790\,a\,b^2\,c^5\,d^6-696\,a^2\,b\,c^4\,d^7+\frac {60\,b^4\,c^7\,d^4}{a}-\frac {40\,a^4\,c^2\,d^9}{b}}+\frac {b^4\,c^5\,d^4\,\sqrt {c^3}\,\sqrt {c+d\,x}\,60{}\mathrm {i}}{256\,a^4\,c^3\,d^8+60\,b^4\,c^7\,d^4-370\,a\,b^3\,c^6\,d^5-696\,a^3\,b\,c^4\,d^7+790\,a^2\,b^2\,c^5\,d^6-\frac {40\,a^5\,c^2\,d^9}{b}}+\frac {a^2\,c\,d^8\,\sqrt {c^3}\,\sqrt {c+d\,x}\,256{}\mathrm {i}}{256\,a^2\,c^3\,d^8+790\,b^2\,c^5\,d^6-\frac {370\,b^3\,c^6\,d^5}{a}-\frac {40\,a^3\,c^2\,d^9}{b}+\frac {60\,b^4\,c^7\,d^4}{a^2}-696\,a\,b\,c^4\,d^7}-\frac {a\,b\,c^2\,d^7\,\sqrt {c^3}\,\sqrt {c+d\,x}\,696{}\mathrm {i}}{256\,a^2\,c^3\,d^8+790\,b^2\,c^5\,d^6-\frac {370\,b^3\,c^6\,d^5}{a}-\frac {40\,a^3\,c^2\,d^9}{b}+\frac {60\,b^4\,c^7\,d^4}{a^2}-696\,a\,b\,c^4\,d^7}\right )\,\left (5\,a\,d-2\,b\,c\right )\,\sqrt {c^3}\,1{}\mathrm {i}}{a^2} \]
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