Optimal. Leaf size=156 \[ -\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a-b x) \sqrt {c+d x^2}}{x (a+b x)}+\frac {a \sqrt {d} \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{a+b x}-\frac {b \sqrt {c} \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a+b x} \]
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Rubi [A] time = 0.11, antiderivative size = 156, 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, 813, 844, 217, 206, 266, 63, 208} \[ -\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a-b x) \sqrt {c+d x^2}}{x (a+b x)}+\frac {a \sqrt {d} \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{a+b x}-\frac {b \sqrt {c} \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a+b x} \]
Antiderivative was successfully verified.
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Rule 63
Rule 206
Rule 208
Rule 217
Rule 266
Rule 813
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^2} \, 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^2} \, dx}{2 a b+2 b^2 x}\\ &=-\frac {(a-b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {-4 b^2 c-4 a b d x}{x \sqrt {c+d x^2}} \, dx}{2 \left (2 a b+2 b^2 x\right )}\\ &=-\frac {(a-b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x (a+b x)}+\frac {\left (2 b^2 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 (2 a b d \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 {(a-b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x (a+b x)}+\frac {\left (b^2 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 (2 a b d \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 {(a-b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x (a+b x)}+\frac {a \sqrt {d} \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{a+b x}+\frac {\left (2 b^2 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 {(a-b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x (a+b x)}+\frac {a \sqrt {d} \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{a+b x}-\frac {b \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 = 0.28, size = 118, normalized size = 0.76 \[ \frac {\sqrt {(a+b x)^2} \left (\frac {(b x-a) \sqrt {c+d x^2}}{x}+\frac {a \sqrt {c} \sqrt {d} \sqrt {\frac {d x^2}{c}+1} \sinh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{\sqrt {c+d x^2}}-b \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )\right )}{a+b x} \]
Antiderivative was successfully verified.
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fricas [A] time = 1.05, size = 333, normalized size = 2.13 \[ \left [\frac {a \sqrt {d} x \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + b \sqrt {c} x \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, \sqrt {d x^{2} + c} {\left (b x - a\right )}}{2 \, x}, -\frac {2 \, a \sqrt {-d} x \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - b \sqrt {c} x \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 2 \, \sqrt {d x^{2} + c} {\left (b x - a\right )}}{2 \, x}, \frac {2 \, b \sqrt {-c} x \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + a \sqrt {d} x \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, \sqrt {d x^{2} + c} {\left (b x - a\right )}}{2 \, x}, -\frac {a \sqrt {-d} x \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - b \sqrt {-c} x \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) - \sqrt {d x^{2} + c} {\left (b x - a\right )}}{x}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.26, size = 126, normalized size = 0.81 \[ \frac {2 \, b c \arctan \left (-\frac {\sqrt {d} x - \sqrt {d x^{2} + c}}{\sqrt {-c}}\right ) \mathrm {sgn}\left (b x + a\right )}{\sqrt {-c}} - a \sqrt {d} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right ) \mathrm {sgn}\left (b x + a\right ) + \sqrt {d x^{2} + c} b \mathrm {sgn}\left (b x + a\right ) + \frac {2 \, a c \sqrt {d} \mathrm {sgn}\left (b x + a\right )}{{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.01, size = 118, normalized size = 0.76 \[ \frac {\left (a c d x \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )-b \,c^{\frac {3}{2}} \sqrt {d}\, x \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )+\sqrt {d \,x^{2}+c}\, a \,d^{\frac {3}{2}} x^{2}+\sqrt {d \,x^{2}+c}\, b c \sqrt {d}\, x -\left (d \,x^{2}+c \right )^{\frac {3}{2}} a \sqrt {d}\right ) \mathrm {csgn}\left (b x +a \right )}{c \sqrt {d}\, x} \]
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 \[ \int \frac {\sqrt {d x^{2} + c} \sqrt {{\left (b x + a\right )}^{2}}}{x^{2}}\,{d x} \]
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 \[ \int \frac {\sqrt {{\left (a+b\,x\right )}^2}\,\sqrt {d\,x^2+c}}{x^2} \,d x \]
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 \[ \int \frac {\sqrt {c + d x^{2}} \sqrt {\left (a + b x\right )^{2}}}{x^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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