Optimal. Leaf size=308 \[ \frac {d g-e f}{2 e^2 (d+e x)^{3/2} (2 c d-b e) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {3 c \sqrt {d+e x} (-4 b e g+3 c d g+5 c e f)}{4 e^2 (2 c d-b e)^3 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {-4 b e g+3 c d g+5 c e f}{4 e^2 \sqrt {d+e x} (2 c d-b e)^2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {3 c (-4 b e g+3 c d g+5 c e f) \tanh ^{-1}\left (\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt {d+e x} \sqrt {2 c d-b e}}\right )}{4 e^2 (2 c d-b e)^{7/2}} \]
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Rubi [A] time = 0.47, antiderivative size = 308, 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 = {792, 672, 666, 660, 208} \begin {gather*} -\frac {e f-d g}{2 e^2 (d+e x)^{3/2} (2 c d-b e) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {3 c \sqrt {d+e x} (-4 b e g+3 c d g+5 c e f)}{4 e^2 (2 c d-b e)^3 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {-4 b e g+3 c d g+5 c e f}{4 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 {3 c (-4 b e g+3 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 )}{4 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 792
Rubi steps
\begin {align*} \int \frac {f+g x}{(d+e x)^{3/2} \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}} \, dx &=-\frac {e f-d g}{2 e^2 (2 c d-b e) (d+e x)^{3/2} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {(5 c e f+3 c d g-4 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}{4 e (2 c d-b e)}\\ &=-\frac {e f-d g}{2 e^2 (2 c d-b e) (d+e x)^{3/2} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {5 c e f+3 c d g-4 b e g}{4 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 {(3 c (5 c e f+3 c d g-4 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}{8 e (2 c d-b e)^2}\\ &=-\frac {e f-d g}{2 e^2 (2 c d-b e) (d+e x)^{3/2} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {5 c e f+3 c d g-4 b e g}{4 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 {3 c (5 c e f+3 c d g-4 b e g) \sqrt {d+e x}}{4 e^2 (2 c d-b e)^3 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {(3 c (5 c e f+3 c d g-4 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}{8 e (2 c d-b e)^3}\\ &=-\frac {e f-d g}{2 e^2 (2 c d-b e) (d+e x)^{3/2} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {5 c e f+3 c d g-4 b e g}{4 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 {3 c (5 c e f+3 c d g-4 b e g) \sqrt {d+e x}}{4 e^2 (2 c d-b e)^3 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {(3 c (5 c e f+3 c d g-4 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 )}{4 (2 c d-b e)^3}\\ &=-\frac {e f-d g}{2 e^2 (2 c d-b e) (d+e x)^{3/2} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {5 c e f+3 c d g-4 b e g}{4 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 {3 c (5 c e f+3 c d g-4 b e g) \sqrt {d+e x}}{4 e^2 (2 c d-b e)^3 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {3 c (5 c e f+3 c d g-4 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 )}{4 e^2 (2 c d-b e)^{7/2}}\\ \end {align*}
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Mathematica [C] time = 0.10, size = 129, normalized size = 0.42 \begin {gather*} \frac {\frac {c (d+e x)^2 (-4 b e g+3 c d g+5 c e f) \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};\frac {-c d+b e+c e x}{b e-2 c d}\right )}{e (b e-2 c d)^2}+\frac {d g}{e}-f}{2 e (d+e x)^{3/2} (2 c d-b e) \sqrt {(d+e x) (c (d-e x)-b e)}} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 6.52, size = 362, normalized size = 1.18 \begin {gather*} \frac {\sqrt {-b e (d+e x)-c (d+e x)^2+2 c d (d+e x)} \left (-4 b^2 e^2 g (d+e x)+2 b^2 d e^2 g-2 b^2 e^3 f-8 b c d^2 e g+5 b c e^2 f (d+e x)+8 b c d e^2 f+11 b c d e g (d+e x)-12 b c e g (d+e x)^2+8 c^2 d^3 g-8 c^2 d^2 e f-6 c^2 d^2 g (d+e x)-10 c^2 d e f (d+e x)+15 c^2 e f (d+e x)^2+9 c^2 d g (d+e x)^2\right )}{4 e^2 (d+e x)^{5/2} (b e-2 c d)^3 (b e+c (d+e x)-2 c d)}+\frac {3 \left (-4 b c e g+3 c^2 d g+5 c^2 e f\right ) \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 )}{4 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 = 0.48, size = 1976, normalized size = 6.42
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.09, size = 824, normalized size = 2.68 \begin {gather*} -\frac {\sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}\, \left (12 \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 )-9 \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 )+24 \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 )+12 \sqrt {b e -2 c d}\, b c \,e^{3} g \,x^{2}-18 \sqrt {-c e x -b e +c d}\, c^{2} d^{2} e g x \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-30 \sqrt {-c e x -b e +c d}\, c^{2} d \,e^{2} f x \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-9 \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}+4 \sqrt {b e -2 c d}\, b^{2} e^{3} g x +12 \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 )+13 \sqrt {b e -2 c d}\, b c d \,e^{2} g x -5 \sqrt {b e -2 c d}\, b c \,e^{3} f x -9 \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 )-12 \sqrt {b e -2 c d}\, c^{2} d^{2} e g x -20 \sqrt {b e -2 c d}\, c^{2} d \,e^{2} f x +2 \sqrt {b e -2 c d}\, b^{2} d \,e^{2} g +2 \sqrt {b e -2 c d}\, b^{2} e^{3} f +9 \sqrt {b e -2 c d}\, b c \,d^{2} e g -13 \sqrt {b e -2 c d}\, b c d \,e^{2} f -11 \sqrt {b e -2 c d}\, c^{2} d^{3} g +3 \sqrt {b e -2 c d}\, c^{2} d^{2} e f \right )}{4 \left (e x +d \right )^{\frac {5}{2}} \left (c e x +b e -c d \right ) \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 {g x + f}{{\left (-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e\right )}^{\frac {3}{2}} {\left (e x + d\right )}^{\frac {3}{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 {f+g\,x}{{\left (d+e\,x\right )}^{3/2}\,{\left (c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x\right )}^{3/2}} \,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 {f + g x}{\left (- \left (d + e x\right ) \left (b e - c d + c e x\right )\right )^{\frac {3}{2}} \left (d + e x\right )^{\frac {3}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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