Optimal. Leaf size=223 \[ \frac {2 \sqrt {e (c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right )}{d e}-\frac {4 b \sqrt {(c+d x)^2+1} \sqrt {e (c+d x)}}{d e (c+d x+1)}-\frac {2 b (c+d x+1) \sqrt {\frac {(c+d x)^2+1}{(c+d x+1)^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{d \sqrt {e} \sqrt {(c+d x)^2+1}}+\frac {4 b (c+d x+1) \sqrt {\frac {(c+d x)^2+1}{(c+d x+1)^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{d \sqrt {e} \sqrt {(c+d x)^2+1}} \]
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Rubi [A] time = 0.21, antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {5865, 5661, 329, 305, 220, 1196} \[ \frac {2 \sqrt {e (c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right )}{d e}-\frac {4 b \sqrt {(c+d x)^2+1} \sqrt {e (c+d x)}}{d e (c+d x+1)}-\frac {2 b (c+d x+1) \sqrt {\frac {(c+d x)^2+1}{(c+d x+1)^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{d \sqrt {e} \sqrt {(c+d x)^2+1}}+\frac {4 b (c+d x+1) \sqrt {\frac {(c+d x)^2+1}{(c+d x+1)^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{d \sqrt {e} \sqrt {(c+d x)^2+1}} \]
Antiderivative was successfully verified.
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Rule 220
Rule 305
Rule 329
Rule 1196
Rule 5661
Rule 5865
Rubi steps
\begin {align*} \int \frac {a+b \sinh ^{-1}(c+d x)}{\sqrt {c e+d e x}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {a+b \sinh ^{-1}(x)}{\sqrt {e x}} \, dx,x,c+d x\right )}{d}\\ &=\frac {2 \sqrt {e (c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right )}{d e}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {\sqrt {e x}}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{d e}\\ &=\frac {2 \sqrt {e (c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right )}{d e}-\frac {(4 b) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+\frac {x^4}{e^2}}} \, dx,x,\sqrt {e (c+d x)}\right )}{d e^2}\\ &=\frac {2 \sqrt {e (c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right )}{d e}-\frac {(4 b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^4}{e^2}}} \, dx,x,\sqrt {e (c+d x)}\right )}{d e}+\frac {(4 b) \operatorname {Subst}\left (\int \frac {1-\frac {x^2}{e}}{\sqrt {1+\frac {x^4}{e^2}}} \, dx,x,\sqrt {e (c+d x)}\right )}{d e}\\ &=-\frac {4 b \sqrt {e (c+d x)} \sqrt {1+(c+d x)^2}}{d e (1+c+d x)}+\frac {2 \sqrt {e (c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right )}{d e}+\frac {4 b (1+c+d x) \sqrt {\frac {1+(c+d x)^2}{(1+c+d x)^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{d \sqrt {e} \sqrt {1+(c+d x)^2}}-\frac {2 b (1+c+d x) \sqrt {\frac {1+(c+d x)^2}{(1+c+d x)^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{d \sqrt {e} \sqrt {1+(c+d x)^2}}\\ \end {align*}
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Mathematica [C] time = 0.03, size = 61, normalized size = 0.27 \[ -\frac {2 \sqrt {e (c+d x)} \left (2 b (c+d x) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-(c+d x)^2\right )-3 \left (a+b \sinh ^{-1}(c+d x)\right )\right )}{3 d e} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \operatorname {arsinh}\left (d x + c\right ) + a}{\sqrt {d e x + c e}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arsinh}\left (d x + c\right ) + a}{\sqrt {d e x + c e}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.01, size = 161, normalized size = 0.72 \[ \frac {2 a \sqrt {d e x +c e}+2 b \left (\sqrt {d e x +c e}\, \arcsinh \left (\frac {d e x +c e}{e}\right )-\frac {2 i \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \left (\EllipticF \left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )-\EllipticE \left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )\right )}{\sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{d e} \]
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 \[ -\frac {1}{2} \, b {\left (\frac {\frac {i \, \sqrt {2} \sqrt {e} {\left (\log \left (\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {d x + c}\right )} + 1\right ) - \log \left (-\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {d x + c}\right )} + 1\right )\right )} - i \, \sqrt {2} \sqrt {e} {\left (\log \left (\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {d x + c}\right )} + 1\right ) - \log \left (-\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {d x + c}\right )} + 1\right )\right )} - \sqrt {2} \sqrt {e} \log \left (d x + \sqrt {2} \sqrt {d x + c} + c + 1\right ) + \sqrt {2} \sqrt {e} \log \left (d x - \sqrt {2} \sqrt {d x + c} + c + 1\right )}{e} + \frac {8 \, \sqrt {d x + c}}{\sqrt {e}}}{d} - \frac {4 \, {\left (d \sqrt {e} x + c \sqrt {e}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )}{\sqrt {d x + c} d e} + 2 \, \int \frac {2 \, {\left (d \sqrt {e} x + c \sqrt {e}\right )}}{{\left (d^{3} e x^{3} + 3 \, c d^{2} e x^{2} + c^{3} e + c e + {\left (3 \, c^{2} d e + d e\right )} x + {\left (d^{2} e x^{2} + 2 \, c d e x + c^{2} e + e\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} \sqrt {d x + c}}\,{d x}\right )} + \frac {2 \, \sqrt {d e x + c e} a}{d e} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\mathrm {asinh}\left (c+d\,x\right )}{\sqrt {c\,e+d\,e\,x}} \,d x \]
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 \[ \int \frac {a + b \operatorname {asinh}{\left (c + d x \right )}}{\sqrt {e \left (c + d x\right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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