Optimal. Leaf size=164 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {d+e x}}{\sqrt {e} \sqrt {f+g x}}\right ) \left (4 e g (2 a e g-b (d g+e f))+c \left (3 d^2 g^2+2 d e f g+3 e^2 f^2\right )\right )}{4 e^{5/2} g^{5/2}}-\frac {\sqrt {d+e x} \sqrt {f+g x} (-4 b e g+5 c d g+3 c e f)}{4 e^2 g^2}+\frac {c (d+e x)^{3/2} \sqrt {f+g x}}{2 e^2 g} \]
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Rubi [A] time = 0.18, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {951, 80, 63, 217, 206} \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {d+e x}}{\sqrt {e} \sqrt {f+g x}}\right ) \left (4 e g (2 a e g-b (d g+e f))+c \left (3 d^2 g^2+2 d e f g+3 e^2 f^2\right )\right )}{4 e^{5/2} g^{5/2}}-\frac {\sqrt {d+e x} \sqrt {f+g x} (-4 b e g+5 c d g+3 c e f)}{4 e^2 g^2}+\frac {c (d+e x)^{3/2} \sqrt {f+g x}}{2 e^2 g} \]
Antiderivative was successfully verified.
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Rule 63
Rule 80
Rule 206
Rule 217
Rule 951
Rubi steps
\begin {align*} \int \frac {a+b x+c x^2}{\sqrt {d+e x} \sqrt {f+g x}} \, dx &=\frac {c (d+e x)^{3/2} \sqrt {f+g x}}{2 e^2 g}+\frac {\int \frac {\frac {1}{2} \left (4 a e^2 g-c d (3 e f+d g)\right )-\frac {1}{2} e (3 c e f+5 c d g-4 b e g) x}{\sqrt {d+e x} \sqrt {f+g x}} \, dx}{2 e^2 g}\\ &=-\frac {(3 c e f+5 c d g-4 b e g) \sqrt {d+e x} \sqrt {f+g x}}{4 e^2 g^2}+\frac {c (d+e x)^{3/2} \sqrt {f+g x}}{2 e^2 g}+\frac {\left (c \left (3 e^2 f^2+2 d e f g+3 d^2 g^2\right )+4 e g (2 a e g-b (e f+d g))\right ) \int \frac {1}{\sqrt {d+e x} \sqrt {f+g x}} \, dx}{8 e^2 g^2}\\ &=-\frac {(3 c e f+5 c d g-4 b e g) \sqrt {d+e x} \sqrt {f+g x}}{4 e^2 g^2}+\frac {c (d+e x)^{3/2} \sqrt {f+g x}}{2 e^2 g}+\frac {\left (c \left (3 e^2 f^2+2 d e f g+3 d^2 g^2\right )+4 e g (2 a e g-b (e f+d g))\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {f-\frac {d g}{e}+\frac {g x^2}{e}}} \, dx,x,\sqrt {d+e x}\right )}{4 e^3 g^2}\\ &=-\frac {(3 c e f+5 c d g-4 b e g) \sqrt {d+e x} \sqrt {f+g x}}{4 e^2 g^2}+\frac {c (d+e x)^{3/2} \sqrt {f+g x}}{2 e^2 g}+\frac {\left (c \left (3 e^2 f^2+2 d e f g+3 d^2 g^2\right )+4 e g (2 a e g-b (e f+d g))\right ) \operatorname {Subst}\left (\int \frac {1}{1-\frac {g x^2}{e}} \, dx,x,\frac {\sqrt {d+e x}}{\sqrt {f+g x}}\right )}{4 e^3 g^2}\\ &=-\frac {(3 c e f+5 c d g-4 b e g) \sqrt {d+e x} \sqrt {f+g x}}{4 e^2 g^2}+\frac {c (d+e x)^{3/2} \sqrt {f+g x}}{2 e^2 g}+\frac {\left (c \left (3 e^2 f^2+2 d e f g+3 d^2 g^2\right )+4 e g (2 a e g-b (e f+d g))\right ) \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {d+e x}}{\sqrt {e} \sqrt {f+g x}}\right )}{4 e^{5/2} g^{5/2}}\\ \end {align*}
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Mathematica [A] time = 0.77, size = 173, normalized size = 1.05 \[ \frac {\sqrt {e f-d g} \sqrt {\frac {e (f+g x)}{e f-d g}} \sinh ^{-1}\left (\frac {\sqrt {g} \sqrt {d+e x}}{\sqrt {e f-d g}}\right ) \left (4 e g (2 a e g-b (d g+e f))+c \left (3 d^2 g^2+2 d e f g+3 e^2 f^2\right )\right )+e \sqrt {g} \sqrt {d+e x} (f+g x) (4 b e g+c (-3 d g-3 e f+2 e g x))}{4 e^3 g^{5/2} \sqrt {f+g x}} \]
Antiderivative was successfully verified.
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fricas [A] time = 1.11, size = 380, normalized size = 2.32 \[ \left [\frac {{\left (3 \, c e^{2} f^{2} + 2 \, {\left (c d e - 2 \, b e^{2}\right )} f g + {\left (3 \, c d^{2} - 4 \, b d e + 8 \, a e^{2}\right )} g^{2}\right )} \sqrt {e g} \log \left (8 \, e^{2} g^{2} x^{2} + e^{2} f^{2} + 6 \, d e f g + d^{2} g^{2} + 4 \, {\left (2 \, e g x + e f + d g\right )} \sqrt {e g} \sqrt {e x + d} \sqrt {g x + f} + 8 \, {\left (e^{2} f g + d e g^{2}\right )} x\right ) + 4 \, {\left (2 \, c e^{2} g^{2} x - 3 \, c e^{2} f g - {\left (3 \, c d e - 4 \, b e^{2}\right )} g^{2}\right )} \sqrt {e x + d} \sqrt {g x + f}}{16 \, e^{3} g^{3}}, -\frac {{\left (3 \, c e^{2} f^{2} + 2 \, {\left (c d e - 2 \, b e^{2}\right )} f g + {\left (3 \, c d^{2} - 4 \, b d e + 8 \, a e^{2}\right )} g^{2}\right )} \sqrt {-e g} \arctan \left (\frac {{\left (2 \, e g x + e f + d g\right )} \sqrt {-e g} \sqrt {e x + d} \sqrt {g x + f}}{2 \, {\left (e^{2} g^{2} x^{2} + d e f g + {\left (e^{2} f g + d e g^{2}\right )} x\right )}}\right ) - 2 \, {\left (2 \, c e^{2} g^{2} x - 3 \, c e^{2} f g - {\left (3 \, c d e - 4 \, b e^{2}\right )} g^{2}\right )} \sqrt {e x + d} \sqrt {g x + f}}{8 \, e^{3} g^{3}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.26, size = 179, normalized size = 1.09 \[ \frac {1}{4} \, \sqrt {{\left (x e + d\right )} g e - d g e + f e^{2}} \sqrt {x e + d} {\left (\frac {2 \, {\left (x e + d\right )} c e^{\left (-3\right )}}{g} - \frac {{\left (5 \, c d g^{2} e^{5} + 3 \, c f g e^{6} - 4 \, b g^{2} e^{6}\right )} e^{\left (-8\right )}}{g^{3}}\right )} - \frac {{\left (3 \, c d^{2} g^{2} + 2 \, c d f g e - 4 \, b d g^{2} e + 3 \, c f^{2} e^{2} - 4 \, b f g e^{2} + 8 \, a g^{2} e^{2}\right )} e^{\left (-\frac {5}{2}\right )} \log \left ({\left | -\sqrt {x e + d} \sqrt {g} e^{\frac {1}{2}} + \sqrt {{\left (x e + d\right )} g e - d g e + f e^{2}} \right |}\right )}{4 \, g^{\frac {5}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.03, size = 425, normalized size = 2.59 \[ \frac {\left (8 a \,e^{2} g^{2} \ln \left (\frac {2 e g x +d g +e f +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}}{2 \sqrt {e g}}\right )-4 b d e \,g^{2} \ln \left (\frac {2 e g x +d g +e f +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}}{2 \sqrt {e g}}\right )-4 b \,e^{2} f g \ln \left (\frac {2 e g x +d g +e f +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}}{2 \sqrt {e g}}\right )+3 c \,d^{2} g^{2} \ln \left (\frac {2 e g x +d g +e f +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}}{2 \sqrt {e g}}\right )+2 c d e f g \ln \left (\frac {2 e g x +d g +e f +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}}{2 \sqrt {e g}}\right )+3 c \,e^{2} f^{2} \ln \left (\frac {2 e g x +d g +e f +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}}{2 \sqrt {e g}}\right )+4 \sqrt {e g}\, \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, c e g x +8 \sqrt {e g}\, \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, b e g -6 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}\, c d g -6 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}\, c e f \right ) \sqrt {e x +d}\, \sqrt {g x +f}}{8 \sqrt {e g}\, \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, e^{2} g^{2}} \]
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 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 22.38, size = 833, normalized size = 5.08 \[ \frac {\frac {\left (2\,b\,d\,g+2\,b\,e\,f\right )\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}{g^3\,\left (\sqrt {f+g\,x}-\sqrt {f}\right )}+\frac {\left (2\,b\,d\,g+2\,b\,e\,f\right )\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^3}{e\,g^2\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^3}-\frac {8\,b\,\sqrt {d}\,\sqrt {f}\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}{g^2\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^2}}{\frac {{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^4}{{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^4}+\frac {e^2}{g^2}-\frac {2\,e\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}{g\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^2}}-\frac {\frac {\left (\sqrt {d+e\,x}-\sqrt {d}\right )\,\left (\frac {3\,c\,d^2\,e\,g^2}{2}+c\,d\,e^2\,f\,g+\frac {3\,c\,e^3\,f^2}{2}\right )}{g^6\,\left (\sqrt {f+g\,x}-\sqrt {f}\right )}-\frac {{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^3\,\left (\frac {11\,c\,d^2\,g^2}{2}+25\,c\,d\,e\,f\,g+\frac {11\,c\,e^2\,f^2}{2}\right )}{g^5\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^3}+\frac {{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^7\,\left (\frac {3\,c\,d^2\,g^2}{2}+c\,d\,e\,f\,g+\frac {3\,c\,e^2\,f^2}{2}\right )}{e^2\,g^3\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^7}-\frac {{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^5\,\left (\frac {11\,c\,d^2\,g^2}{2}+25\,c\,d\,e\,f\,g+\frac {11\,c\,e^2\,f^2}{2}\right )}{e\,g^4\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^5}+\frac {\sqrt {d}\,\sqrt {f}\,\left (32\,c\,d\,g+32\,c\,e\,f\right )\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^4}{g^4\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^4}}{\frac {{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^8}{{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^8}+\frac {e^4}{g^4}-\frac {4\,e\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^6}{g\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^6}-\frac {4\,e^3\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}{g^3\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^2}+\frac {6\,e^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^4}{g^2\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^4}}-\frac {4\,a\,\mathrm {atan}\left (\frac {e\,\left (\sqrt {f+g\,x}-\sqrt {f}\right )}{\sqrt {-e\,g}\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}\right )}{\sqrt {-e\,g}}-\frac {2\,b\,\mathrm {atanh}\left (\frac {\sqrt {g}\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}{\sqrt {e}\,\left (\sqrt {f+g\,x}-\sqrt {f}\right )}\right )\,\left (d\,g+e\,f\right )}{e^{3/2}\,g^{3/2}}+\frac {c\,\mathrm {atanh}\left (\frac {\sqrt {g}\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}{\sqrt {e}\,\left (\sqrt {f+g\,x}-\sqrt {f}\right )}\right )\,\left (3\,d^2\,g^2+2\,d\,e\,f\,g+3\,e^2\,f^2\right )}{2\,e^{5/2}\,g^{5/2}} \]
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 \[ \int \frac {a + b x + c x^{2}}{\sqrt {d + e x} \sqrt {f + g x}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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