Optimal. Leaf size=160 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (2 a+b x) \sqrt {c+d x^2}}{2 (a+b x)}+\frac {b c \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d} (a+b x)}-\frac {a \sqrt {c} \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a+b x} \]
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Rubi [A] time = 0.12, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {1001, 815, 844, 217, 206, 266, 63, 208} \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (2 a+b x) \sqrt {c+d x^2}}{2 (a+b x)}+\frac {b c \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d} (a+b x)}-\frac {a \sqrt {c} \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a+b x} \]
Antiderivative was successfully verified.
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Rule 63
Rule 206
Rule 208
Rule 217
Rule 266
Rule 815
Rule 844
Rule 1001
Rubi steps
\begin {align*} \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (2 a b+2 b^2 x\right ) \sqrt {c+d x^2}}{x} \, dx}{2 a b+2 b^2 x}\\ &=\frac {(2 a+b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{2 (a+b x)}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {4 a b c d+2 b^2 c d x}{x \sqrt {c+d x^2}} \, dx}{2 d \left (2 a b+2 b^2 x\right )}\\ &=\frac {(2 a+b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{2 (a+b x)}+\frac {\left (2 a b c \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \frac {1}{x \sqrt {c+d x^2}} \, dx}{2 a b+2 b^2 x}+\frac {\left (b^2 c \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{2 a b+2 b^2 x}\\ &=\frac {(2 a+b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{2 (a+b x)}+\frac {\left (a b c \sqrt {a^2+2 a b x+b^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right )}{2 a b+2 b^2 x}+\frac {\left (b^2 c \sqrt {a^2+2 a b x+b^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{2 a b+2 b^2 x}\\ &=\frac {(2 a+b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{2 (a+b x)}+\frac {b c \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d} (a+b x)}+\frac {\left (2 a b c \sqrt {a^2+2 a b x+b^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{d \left (2 a b+2 b^2 x\right )}\\ &=\frac {(2 a+b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{2 (a+b x)}+\frac {b c \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d} (a+b x)}-\frac {a \sqrt {c} \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a+b x}\\ \end {align*}
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Mathematica [A] time = 0.14, size = 139, normalized size = 0.87 \[ \frac {\sqrt {(a+b x)^2} \left (\sqrt {d} \sqrt {\frac {d x^2}{c}+1} \left ((2 a+b x) \sqrt {c+d x^2}-2 a \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )\right )+b \sqrt {c} \sqrt {c+d x^2} \sinh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )\right )}{2 \sqrt {d} (a+b x) \sqrt {\frac {d x^2}{c}+1}} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.99, size = 341, normalized size = 2.13 \[ \left [\frac {b c \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, a \sqrt {c} d \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (b d x + 2 \, a d\right )} \sqrt {d x^{2} + c}}{4 \, d}, -\frac {b c \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - a \sqrt {c} d \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - {\left (b d x + 2 \, a d\right )} \sqrt {d x^{2} + c}}{2 \, d}, \frac {4 \, a \sqrt {-c} d \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + b c \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (b d x + 2 \, a d\right )} \sqrt {d x^{2} + c}}{4 \, d}, -\frac {b c \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - 2 \, a \sqrt {-c} d \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) - {\left (b d x + 2 \, a d\right )} \sqrt {d x^{2} + c}}{2 \, d}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.25, size = 102, normalized size = 0.64 \[ \frac {2 \, a c \arctan \left (-\frac {\sqrt {d} x - \sqrt {d x^{2} + c}}{\sqrt {-c}}\right ) \mathrm {sgn}\left (b x + a\right )}{\sqrt {-c}} - \frac {b c \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right ) \mathrm {sgn}\left (b x + a\right )}{2 \, \sqrt {d}} + \frac {1}{2} \, \sqrt {d x^{2} + c} {\left (b x \mathrm {sgn}\left (b x + a\right ) + 2 \, a \mathrm {sgn}\left (b x + a\right )\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.01, size = 94, normalized size = 0.59 \[ -\frac {\left (2 a \sqrt {c}\, \sqrt {d}\, \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )-b c \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )-\sqrt {d \,x^{2}+c}\, b \sqrt {d}\, x -2 \sqrt {d \,x^{2}+c}\, a \sqrt {d}\right ) \mathrm {csgn}\left (b x +a \right )}{2 \sqrt {d}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {d x^{2} + c} \sqrt {{\left (b x + a\right )}^{2}}}{x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {{\left (a+b\,x\right )}^2}\,\sqrt {d\,x^2+c}}{x} \,d x \]
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 \[ \int \frac {\sqrt {c + d x^{2}} \sqrt {\left (a + b x\right )^{2}}}{x}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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