Optimal. Leaf size=118 \[ \frac {2 d (2 b c-a d) \sqrt {c+d x}}{b^2}+\frac {2 d (c+d x)^{3/2}}{3 b}-\frac {2 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a}+\frac {2 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a b^{5/2}} \]
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Rubi [A]
time = 0.09, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {86, 159, 162,
65, 214} \begin {gather*} \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 b^{5/2}}+\frac {2 d \sqrt {c+d x} (2 b c-a d)}{b^2}-\frac {2 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a}+\frac {2 d (c+d x)^{3/2}}{3 b} \end {gather*}
Antiderivative was successfully verified.
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Rule 65
Rule 86
Rule 159
Rule 162
Rule 214
Rubi steps
\begin {align*} \int \frac {(c+d x)^{5/2}}{x (a+b x)} \, dx &=\frac {2 d (c+d x)^{3/2}}{3 b}+\frac {\int \frac {\sqrt {c+d x} \left (b c^2+d (2 b c-a d) x\right )}{x (a+b x)} \, dx}{b}\\ &=\frac {2 d (2 b c-a d) \sqrt {c+d x}}{b^2}+\frac {2 d (c+d x)^{3/2}}{3 b}+\frac {2 \int \frac {\frac {b^2 c^3}{2}+\frac {1}{2} d \left (3 b^2 c^2-3 a b c d+a^2 d^2\right ) x}{x (a+b x) \sqrt {c+d x}} \, dx}{b^2}\\ &=\frac {2 d (2 b c-a d) \sqrt {c+d x}}{b^2}+\frac {2 d (c+d x)^{3/2}}{3 b}+\frac {c^3 \int \frac {1}{x \sqrt {c+d x}} \, dx}{a}-\frac {(b c-a d)^3 \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{a b^2}\\ &=\frac {2 d (2 b c-a d) \sqrt {c+d x}}{b^2}+\frac {2 d (c+d x)^{3/2}}{3 b}+\frac {\left (2 c^3\right ) \text {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{a 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 b^2 d}\\ &=\frac {2 d (2 b c-a d) \sqrt {c+d x}}{b^2}+\frac {2 d (c+d x)^{3/2}}{3 b}-\frac {2 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a}+\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 b^{5/2}}\\ \end {align*}
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Mathematica [A]
time = 0.17, size = 107, normalized size = 0.91 \begin {gather*} \frac {2 d \sqrt {c+d x} (7 b c-3 a d+b d x)}{3 b^2}+\frac {2 (-b c+a d)^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{a b^{5/2}}-\frac {2 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.07, size = 145, normalized size = 1.23
method | result | size |
derivativedivides | \(2 d \left (-\frac {-\frac {b \left (d x +c \right )^{\frac {3}{2}}}{3}+a d \sqrt {d x +c}-2 b c \sqrt {d x +c}}{b^{2}}+\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{b^{2} a d \sqrt {\left (a d -b c \right ) b}}-\frac {c^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a d}\right )\) | \(145\) |
default | \(2 d \left (-\frac {-\frac {b \left (d x +c \right )^{\frac {3}{2}}}{3}+a d \sqrt {d x +c}-2 b c \sqrt {d x +c}}{b^{2}}+\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{b^{2} a d \sqrt {\left (a d -b c \right ) b}}-\frac {c^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a d}\right )\) | \(145\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 1.22, size = 598, normalized size = 5.07 \begin {gather*} \left [\frac {3 \, b^{2} c^{\frac {5}{2}} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \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 ) + 2 \, {\left (a b d^{2} x + 7 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {d x + c}}{3 \, a b^{2}}, \frac {3 \, b^{2} c^{\frac {5}{2}} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 6 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \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 ) + 2 \, {\left (a b d^{2} x + 7 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {d x + c}}{3 \, a b^{2}}, \frac {6 \, b^{2} \sqrt {-c} c^{2} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \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 ) + 2 \, {\left (a b d^{2} x + 7 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {d x + c}}{3 \, a b^{2}}, \frac {2 \, {\left (3 \, b^{2} \sqrt {-c} c^{2} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \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 (a b d^{2} x + 7 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {d x + c}\right )}}{3 \, a b^{2}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 34.08, size = 119, normalized size = 1.01 \begin {gather*} \frac {2 d \left (c + d x\right )^{\frac {3}{2}}}{3 b} + \frac {\sqrt {c + d x} \left (- 2 a d^{2} + 4 b c d\right )}{b^{2}} + \frac {2 c^{3} \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {- c}} \right )}}{a \sqrt {- c}} + \frac {2 \left (a d - b c\right )^{3} \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{a b^{3} \sqrt {\frac {a d - b c}{b}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.90, size = 154, normalized size = 1.31 \begin {gather*} \frac {2 \, c^{3} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{a \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 b^{2}} + \frac {2 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} b^{2} d + 6 \, \sqrt {d x + c} b^{2} c d - 3 \, \sqrt {d x + c} a b d^{2}\right )}}{3 \, b^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.77, size = 2048, normalized size = 17.36 \begin {gather*} \frac {2\,d\,{\left (c+d\,x\right )}^{3/2}}{3\,b}-\frac {2\,d\,\left (a\,d-2\,b\,c\right )\,\sqrt {c+d\,x}}{b^2}+\frac {\mathrm {atan}\left (\frac {\frac {\left (\frac {8\,\sqrt {c+d\,x}\,\left (a^6\,d^8-6\,a^5\,b\,c\,d^7+15\,a^4\,b^2\,c^2\,d^6-20\,a^3\,b^3\,c^3\,d^5+15\,a^2\,b^4\,c^4\,d^4-6\,a\,b^5\,c^5\,d^3+2\,b^6\,c^6\,d^2\right )}{b^3}+\frac {\left (\frac {8\,\left (a^4\,b^3\,c\,d^5-3\,a^3\,b^4\,c^2\,d^4+2\,a^2\,b^5\,c^3\,d^3\right )}{b^3}+\frac {8\,\left (a^3\,b^5\,d^3-2\,a^2\,b^6\,c\,d^2\right )\,\sqrt {c^5}\,\sqrt {c+d\,x}}{a\,b^3}\right )\,\sqrt {c^5}}{a}\right )\,\sqrt {c^5}\,1{}\mathrm {i}}{a}+\frac {\left (\frac {8\,\sqrt {c+d\,x}\,\left (a^6\,d^8-6\,a^5\,b\,c\,d^7+15\,a^4\,b^2\,c^2\,d^6-20\,a^3\,b^3\,c^3\,d^5+15\,a^2\,b^4\,c^4\,d^4-6\,a\,b^5\,c^5\,d^3+2\,b^6\,c^6\,d^2\right )}{b^3}-\frac {\left (\frac {8\,\left (a^4\,b^3\,c\,d^5-3\,a^3\,b^4\,c^2\,d^4+2\,a^2\,b^5\,c^3\,d^3\right )}{b^3}-\frac {8\,\left (a^3\,b^5\,d^3-2\,a^2\,b^6\,c\,d^2\right )\,\sqrt {c^5}\,\sqrt {c+d\,x}}{a\,b^3}\right )\,\sqrt {c^5}}{a}\right )\,\sqrt {c^5}\,1{}\mathrm {i}}{a}}{\frac {16\,\left (a^5\,c^3\,d^8-6\,a^4\,b\,c^4\,d^7+15\,a^3\,b^2\,c^5\,d^6-19\,a^2\,b^3\,c^6\,d^5+12\,a\,b^4\,c^7\,d^4-3\,b^5\,c^8\,d^3\right )}{b^3}-\frac {\left (\frac {8\,\sqrt {c+d\,x}\,\left (a^6\,d^8-6\,a^5\,b\,c\,d^7+15\,a^4\,b^2\,c^2\,d^6-20\,a^3\,b^3\,c^3\,d^5+15\,a^2\,b^4\,c^4\,d^4-6\,a\,b^5\,c^5\,d^3+2\,b^6\,c^6\,d^2\right )}{b^3}+\frac {\left (\frac {8\,\left (a^4\,b^3\,c\,d^5-3\,a^3\,b^4\,c^2\,d^4+2\,a^2\,b^5\,c^3\,d^3\right )}{b^3}+\frac {8\,\left (a^3\,b^5\,d^3-2\,a^2\,b^6\,c\,d^2\right )\,\sqrt {c^5}\,\sqrt {c+d\,x}}{a\,b^3}\right )\,\sqrt {c^5}}{a}\right )\,\sqrt {c^5}}{a}+\frac {\left (\frac {8\,\sqrt {c+d\,x}\,\left (a^6\,d^8-6\,a^5\,b\,c\,d^7+15\,a^4\,b^2\,c^2\,d^6-20\,a^3\,b^3\,c^3\,d^5+15\,a^2\,b^4\,c^4\,d^4-6\,a\,b^5\,c^5\,d^3+2\,b^6\,c^6\,d^2\right )}{b^3}-\frac {\left (\frac {8\,\left (a^4\,b^3\,c\,d^5-3\,a^3\,b^4\,c^2\,d^4+2\,a^2\,b^5\,c^3\,d^3\right )}{b^3}-\frac {8\,\left (a^3\,b^5\,d^3-2\,a^2\,b^6\,c\,d^2\right )\,\sqrt {c^5}\,\sqrt {c+d\,x}}{a\,b^3}\right )\,\sqrt {c^5}}{a}\right )\,\sqrt {c^5}}{a}}\right )\,\sqrt {c^5}\,2{}\mathrm {i}}{a}+\frac {\mathrm {atan}\left (\frac {\frac {\sqrt {-b^5\,{\left (a\,d-b\,c\right )}^5}\,\left (\frac {8\,\sqrt {c+d\,x}\,\left (a^6\,d^8-6\,a^5\,b\,c\,d^7+15\,a^4\,b^2\,c^2\,d^6-20\,a^3\,b^3\,c^3\,d^5+15\,a^2\,b^4\,c^4\,d^4-6\,a\,b^5\,c^5\,d^3+2\,b^6\,c^6\,d^2\right )}{b^3}+\frac {\sqrt {-b^5\,{\left (a\,d-b\,c\right )}^5}\,\left (\frac {8\,\left (a^4\,b^3\,c\,d^5-3\,a^3\,b^4\,c^2\,d^4+2\,a^2\,b^5\,c^3\,d^3\right )}{b^3}+\frac {8\,\left (a^3\,b^5\,d^3-2\,a^2\,b^6\,c\,d^2\right )\,\sqrt {-b^5\,{\left (a\,d-b\,c\right )}^5}\,\sqrt {c+d\,x}}{a\,b^8}\right )}{a\,b^5}\right )\,1{}\mathrm {i}}{a\,b^5}+\frac {\sqrt {-b^5\,{\left (a\,d-b\,c\right )}^5}\,\left (\frac {8\,\sqrt {c+d\,x}\,\left (a^6\,d^8-6\,a^5\,b\,c\,d^7+15\,a^4\,b^2\,c^2\,d^6-20\,a^3\,b^3\,c^3\,d^5+15\,a^2\,b^4\,c^4\,d^4-6\,a\,b^5\,c^5\,d^3+2\,b^6\,c^6\,d^2\right )}{b^3}-\frac {\sqrt {-b^5\,{\left (a\,d-b\,c\right )}^5}\,\left (\frac {8\,\left (a^4\,b^3\,c\,d^5-3\,a^3\,b^4\,c^2\,d^4+2\,a^2\,b^5\,c^3\,d^3\right )}{b^3}-\frac {8\,\left (a^3\,b^5\,d^3-2\,a^2\,b^6\,c\,d^2\right )\,\sqrt {-b^5\,{\left (a\,d-b\,c\right )}^5}\,\sqrt {c+d\,x}}{a\,b^8}\right )}{a\,b^5}\right )\,1{}\mathrm {i}}{a\,b^5}}{\frac {16\,\left (a^5\,c^3\,d^8-6\,a^4\,b\,c^4\,d^7+15\,a^3\,b^2\,c^5\,d^6-19\,a^2\,b^3\,c^6\,d^5+12\,a\,b^4\,c^7\,d^4-3\,b^5\,c^8\,d^3\right )}{b^3}-\frac {\sqrt {-b^5\,{\left (a\,d-b\,c\right )}^5}\,\left (\frac {8\,\sqrt {c+d\,x}\,\left (a^6\,d^8-6\,a^5\,b\,c\,d^7+15\,a^4\,b^2\,c^2\,d^6-20\,a^3\,b^3\,c^3\,d^5+15\,a^2\,b^4\,c^4\,d^4-6\,a\,b^5\,c^5\,d^3+2\,b^6\,c^6\,d^2\right )}{b^3}+\frac {\sqrt {-b^5\,{\left (a\,d-b\,c\right )}^5}\,\left (\frac {8\,\left (a^4\,b^3\,c\,d^5-3\,a^3\,b^4\,c^2\,d^4+2\,a^2\,b^5\,c^3\,d^3\right )}{b^3}+\frac {8\,\left (a^3\,b^5\,d^3-2\,a^2\,b^6\,c\,d^2\right )\,\sqrt {-b^5\,{\left (a\,d-b\,c\right )}^5}\,\sqrt {c+d\,x}}{a\,b^8}\right )}{a\,b^5}\right )}{a\,b^5}+\frac {\sqrt {-b^5\,{\left (a\,d-b\,c\right )}^5}\,\left (\frac {8\,\sqrt {c+d\,x}\,\left (a^6\,d^8-6\,a^5\,b\,c\,d^7+15\,a^4\,b^2\,c^2\,d^6-20\,a^3\,b^3\,c^3\,d^5+15\,a^2\,b^4\,c^4\,d^4-6\,a\,b^5\,c^5\,d^3+2\,b^6\,c^6\,d^2\right )}{b^3}-\frac {\sqrt {-b^5\,{\left (a\,d-b\,c\right )}^5}\,\left (\frac {8\,\left (a^4\,b^3\,c\,d^5-3\,a^3\,b^4\,c^2\,d^4+2\,a^2\,b^5\,c^3\,d^3\right )}{b^3}-\frac {8\,\left (a^3\,b^5\,d^3-2\,a^2\,b^6\,c\,d^2\right )\,\sqrt {-b^5\,{\left (a\,d-b\,c\right )}^5}\,\sqrt {c+d\,x}}{a\,b^8}\right )}{a\,b^5}\right )}{a\,b^5}}\right )\,\sqrt {-b^5\,{\left (a\,d-b\,c\right )}^5}\,2{}\mathrm {i}}{a\,b^5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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