Optimal. Leaf size=313 \[ \frac {\sqrt {d+e x} (-2 b e g-c d g+5 c e f)}{e^2 (2 c d-b e)^3 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {-2 b e g-c d g+5 c e f}{3 c e^2 \sqrt {d+e x} (2 c d-b e)^2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {2 \sqrt {d+e x} (-b e g+c d g+c e f)}{3 c e^2 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}-\frac {(-2 b e g-c d g+5 c e f) \tanh ^{-1}\left (\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt {d+e x} \sqrt {2 c d-b e}}\right )}{e^2 (2 c d-b e)^{7/2}} \]
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Rubi [A] time = 0.44, antiderivative size = 313, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.109, Rules used = {788, 672, 666, 660, 208} \begin {gather*} \frac {\sqrt {d+e x} (-2 b e g-c d g+5 c e f)}{e^2 (2 c d-b e)^3 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {-2 b e g-c d g+5 c e f}{3 c e^2 \sqrt {d+e x} (2 c d-b e)^2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {2 \sqrt {d+e x} (-b e g+c d g+c e f)}{3 c e^2 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}-\frac {(-2 b e g-c d g+5 c e f) \tanh ^{-1}\left (\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt {d+e x} \sqrt {2 c d-b e}}\right )}{e^2 (2 c d-b e)^{7/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 208
Rule 660
Rule 666
Rule 672
Rule 788
Rubi steps
\begin {align*} \int \frac {\sqrt {d+e x} (f+g x)}{\left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}} \, dx &=\frac {2 (c e f+c d g-b e g) \sqrt {d+e x}}{3 c e^2 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}+\frac {(5 c e f-c d g-2 b e g) \int \frac {1}{\sqrt {d+e x} \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}} \, dx}{3 c e (2 c d-b e)}\\ &=\frac {2 (c e f+c d g-b e g) \sqrt {d+e x}}{3 c e^2 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}-\frac {5 c e f-c d g-2 b e g}{3 c e^2 (2 c d-b e)^2 \sqrt {d+e x} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {(5 c e f-c d g-2 b e g) \int \frac {\sqrt {d+e x}}{\left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}} \, dx}{2 e (2 c d-b e)^2}\\ &=\frac {2 (c e f+c d g-b e g) \sqrt {d+e x}}{3 c e^2 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}-\frac {5 c e f-c d g-2 b e g}{3 c e^2 (2 c d-b e)^2 \sqrt {d+e x} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {(5 c e f-c d g-2 b e g) \sqrt {d+e x}}{e^2 (2 c d-b e)^3 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {(5 c e f-c d g-2 b e g) \int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx}{2 e (2 c d-b e)^3}\\ &=\frac {2 (c e f+c d g-b e g) \sqrt {d+e x}}{3 c e^2 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}-\frac {5 c e f-c d g-2 b e g}{3 c e^2 (2 c d-b e)^2 \sqrt {d+e x} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {(5 c e f-c d g-2 b e g) \sqrt {d+e x}}{e^2 (2 c d-b e)^3 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {(5 c e f-c d g-2 b e g) \operatorname {Subst}\left (\int \frac {1}{-2 c d e^2+b e^3+e^2 x^2} \, dx,x,\frac {\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}}{\sqrt {d+e x}}\right )}{(2 c d-b e)^3}\\ &=\frac {2 (c e f+c d g-b e g) \sqrt {d+e x}}{3 c e^2 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}-\frac {5 c e f-c d g-2 b e g}{3 c e^2 (2 c d-b e)^2 \sqrt {d+e x} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {(5 c e f-c d g-2 b e g) \sqrt {d+e x}}{e^2 (2 c d-b e)^3 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {(5 c e f-c d g-2 b e g) \tanh ^{-1}\left (\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt {2 c d-b e} \sqrt {d+e x}}\right )}{e^2 (2 c d-b e)^{7/2}}\\ \end {align*}
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Mathematica [C] time = 0.08, size = 135, normalized size = 0.43 \begin {gather*} \frac {(d+e x) (2 b e g+c d g-5 c e f) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {-c d+b e+c e x}{b e-2 c d}\right )-3 (2 c d-b e) (d g-e f)}{3 e^2 \sqrt {d+e x} (b e-2 c d)^2 (b e-c d+c e x) \sqrt {(d+e x) (c (d-e x)-b e)}} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 9.31, size = 354, normalized size = 1.13 \begin {gather*} \frac {\left (-b e (d+e x)-c (d+e x)^2+2 c d (d+e x)\right )^{3/2} \left (8 b^2 e^2 g (d+e x)+3 b^2 d e^2 g-3 b^2 e^3 f-12 b c d^2 e g-20 b c e^2 f (d+e x)+12 b c d e^2 f-12 b c d e g (d+e x)+6 b c e g (d+e x)^2+12 c^2 d^3 g-12 c^2 d^2 e f-8 c^2 d^2 g (d+e x)+40 c^2 d e f (d+e x)-15 c^2 e f (d+e x)^2+3 c^2 d g (d+e x)^2\right )}{3 e^2 (d+e x)^{5/2} (b e-2 c d)^3 (b e+c (d+e x)-2 c d)^3}+\frac {(-2 b e g-c d g+5 c e f) \tan ^{-1}\left (\frac {\sqrt {b e-2 c d} \sqrt {(d+e x) (2 c d-b e)-c (d+e x)^2}}{\sqrt {d+e x} (b e+c (d+e x)-2 c d)}\right )}{e^2 (2 c d-b e)^3 \sqrt {b e-2 c d}} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.46, size = 2106, normalized size = 6.73
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.08, size = 904, normalized size = 2.89 \begin {gather*} -\frac {\sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}\, \left (6 \sqrt {-c e x -b e +c d}\, b c \,e^{3} g \,x^{2} \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )+3 \sqrt {-c e x -b e +c d}\, c^{2} d \,e^{2} g \,x^{2} \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-15 \sqrt {-c e x -b e +c d}\, c^{2} e^{3} f \,x^{2} \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )+6 \sqrt {-c e x -b e +c d}\, b^{2} e^{3} g x \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )+3 \sqrt {-c e x -b e +c d}\, b c d \,e^{2} g x \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-15 \sqrt {-c e x -b e +c d}\, b c \,e^{3} f x \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )+6 \sqrt {b e -2 c d}\, b c \,e^{3} g \,x^{2}+3 \sqrt {b e -2 c d}\, c^{2} d \,e^{2} g \,x^{2}-15 \sqrt {b e -2 c d}\, c^{2} e^{3} f \,x^{2}+6 \sqrt {-c e x -b e +c d}\, b^{2} d \,e^{2} g \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )+8 \sqrt {b e -2 c d}\, b^{2} e^{3} g x -3 \sqrt {-c e x -b e +c d}\, b c \,d^{2} e g \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-15 \sqrt {-c e x -b e +c d}\, b c d \,e^{2} f \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-20 \sqrt {b e -2 c d}\, b c \,e^{3} f x -3 \sqrt {-c e x -b e +c d}\, c^{2} d^{3} g \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )+15 \sqrt {-c e x -b e +c d}\, c^{2} d^{2} e f \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-2 \sqrt {b e -2 c d}\, c^{2} d^{2} e g x +10 \sqrt {b e -2 c d}\, c^{2} d \,e^{2} f x +11 \sqrt {b e -2 c d}\, b^{2} d \,e^{2} g -3 \sqrt {b e -2 c d}\, b^{2} e^{3} f -18 \sqrt {b e -2 c d}\, b c \,d^{2} e g -8 \sqrt {b e -2 c d}\, b c d \,e^{2} f +7 \sqrt {b e -2 c d}\, c^{2} d^{3} g +13 \sqrt {b e -2 c d}\, c^{2} d^{2} e f \right )}{3 \left (e x +d \right )^{\frac {3}{2}} \left (c e x +b e -c d \right )^{2} \left (b e -2 c d \right )^{\frac {7}{2}} e^{2}} \end {gather*}
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 \begin {gather*} \int \frac {\sqrt {e x + d} {\left (g x + f\right )}}{{\left (-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e\right )}^{\frac {5}{2}}}\,{d x} \end {gather*}
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 \begin {gather*} \int \frac {\left (f+g\,x\right )\,\sqrt {d+e\,x}}{{\left (c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x\right )}^{5/2}} \,d x \end {gather*}
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 \begin {gather*} \int \frac {\sqrt {d + e x} \left (f + g x\right )}{\left (- \left (d + e x\right ) \left (b e - c d + c e x\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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