Optimal. Leaf size=198 \[ \frac {2 \sqrt {f+g x} \left (2 e g (-4 a e g-b d g+5 b e f)-c \left (3 d^2 g^2-10 d e f g+15 e^2 f^2\right )\right )}{15 e^2 \sqrt {d+e x} (e f-d g)^3}-\frac {2 \sqrt {f+g x} \left (a+\frac {d (c d-b e)}{e^2}\right )}{5 (d+e x)^{5/2} (e f-d g)}+\frac {2 \sqrt {f+g x} (2 c d (5 e f-3 d g)-e (-4 a e g-b d g+5 b e f))}{15 e^2 (d+e x)^{3/2} (e f-d g)^2} \]
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Rubi [A] time = 0.21, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {949, 78, 37} \[ \frac {2 \sqrt {f+g x} \left (2 e g (-4 a e g-b d g+5 b e f)-c \left (3 d^2 g^2-10 d e f g+15 e^2 f^2\right )\right )}{15 e^2 \sqrt {d+e x} (e f-d g)^3}-\frac {2 \sqrt {f+g x} \left (a+\frac {d (c d-b e)}{e^2}\right )}{5 (d+e x)^{5/2} (e f-d g)}+\frac {2 \sqrt {f+g x} (2 c d (5 e f-3 d g)-e (-4 a e g-b d g+5 b e f))}{15 e^2 (d+e x)^{3/2} (e f-d g)^2} \]
Antiderivative was successfully verified.
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Rule 37
Rule 78
Rule 949
Rubi steps
\begin {align*} \int \frac {a+b x+c x^2}{(d+e x)^{7/2} \sqrt {f+g x}} \, dx &=-\frac {2 \left (a+\frac {d (c d-b e)}{e^2}\right ) \sqrt {f+g x}}{5 (e f-d g) (d+e x)^{5/2}}-\frac {2 \int \frac {\frac {c d (5 e f-d g)-e (5 b e f-b d g-4 a e g)}{2 e^2}-\frac {5}{2} c \left (f-\frac {d g}{e}\right ) x}{(d+e x)^{5/2} \sqrt {f+g x}} \, dx}{5 (e f-d g)}\\ &=-\frac {2 \left (a+\frac {d (c d-b e)}{e^2}\right ) \sqrt {f+g x}}{5 (e f-d g) (d+e x)^{5/2}}+\frac {2 (2 c d (5 e f-3 d g)-e (5 b e f-b d g-4 a e g)) \sqrt {f+g x}}{15 e^2 (e f-d g)^2 (d+e x)^{3/2}}-\frac {\left (2 e g (5 b e f-b d g-4 a e g)-c \left (15 e^2 f^2-10 d e f g+3 d^2 g^2\right )\right ) \int \frac {1}{(d+e x)^{3/2} \sqrt {f+g x}} \, dx}{15 e^2 (e f-d g)^2}\\ &=-\frac {2 \left (a+\frac {d (c d-b e)}{e^2}\right ) \sqrt {f+g x}}{5 (e f-d g) (d+e x)^{5/2}}+\frac {2 (2 c d (5 e f-3 d g)-e (5 b e f-b d g-4 a e g)) \sqrt {f+g x}}{15 e^2 (e f-d g)^2 (d+e x)^{3/2}}+\frac {2 \left (2 e g (5 b e f-b d g-4 a e g)-c \left (15 e^2 f^2-10 d e f g+3 d^2 g^2\right )\right ) \sqrt {f+g x}}{15 e^2 (e f-d g)^3 \sqrt {d+e x}}\\ \end {align*}
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Mathematica [A] time = 0.23, size = 178, normalized size = 0.90 \[ -\frac {2 \sqrt {f+g x} \left (a \left (15 d^2 g^2-10 d e g (f-2 g x)+e^2 \left (3 f^2-4 f g x+8 g^2 x^2\right )\right )+b \left (5 d^2 g (g x-2 f)+2 d e \left (f^2-13 f g x+g^2 x^2\right )+5 e^2 f x (f-2 g x)\right )+c \left (d^2 \left (8 f^2-4 f g x+3 g^2 x^2\right )+10 d e f x (2 f-g x)+15 e^2 f^2 x^2\right )\right )}{15 (d+e x)^{5/2} (e f-d g)^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 25.37, size = 353, normalized size = 1.78 \[ -\frac {2 \, {\left (15 \, a d^{2} g^{2} + {\left (8 \, c d^{2} + 2 \, b d e + 3 \, a e^{2}\right )} f^{2} - 10 \, {\left (b d^{2} + a d e\right )} f g + {\left (15 \, c e^{2} f^{2} - 10 \, {\left (c d e + b e^{2}\right )} f g + {\left (3 \, c d^{2} + 2 \, b d e + 8 \, a e^{2}\right )} g^{2}\right )} x^{2} + {\left (5 \, {\left (4 \, c d e + b e^{2}\right )} f^{2} - 2 \, {\left (2 \, c d^{2} + 13 \, b d e + 2 \, a e^{2}\right )} f g + 5 \, {\left (b d^{2} + 4 \, a d e\right )} g^{2}\right )} x\right )} \sqrt {e x + d} \sqrt {g x + f}}{15 \, {\left (d^{3} e^{3} f^{3} - 3 \, d^{4} e^{2} f^{2} g + 3 \, d^{5} e f g^{2} - d^{6} g^{3} + {\left (e^{6} f^{3} - 3 \, d e^{5} f^{2} g + 3 \, d^{2} e^{4} f g^{2} - d^{3} e^{3} g^{3}\right )} x^{3} + 3 \, {\left (d e^{5} f^{3} - 3 \, d^{2} e^{4} f^{2} g + 3 \, d^{3} e^{3} f g^{2} - d^{4} e^{2} g^{3}\right )} x^{2} + 3 \, {\left (d^{2} e^{4} f^{3} - 3 \, d^{3} e^{3} f^{2} g + 3 \, d^{4} e^{2} f g^{2} - d^{5} e g^{3}\right )} x\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.85, size = 1080, normalized size = 5.45 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.01, size = 238, normalized size = 1.20 \[ \frac {2 \sqrt {g x +f}\, \left (8 a \,e^{2} g^{2} x^{2}+2 b d e \,g^{2} x^{2}-10 b \,e^{2} f g \,x^{2}+3 c \,d^{2} g^{2} x^{2}-10 c d e f g \,x^{2}+15 c \,e^{2} f^{2} x^{2}+20 a d e \,g^{2} x -4 a \,e^{2} f g x +5 b \,d^{2} g^{2} x -26 b d e f g x +5 b \,e^{2} f^{2} x -4 c \,d^{2} f g x +20 c d e \,f^{2} x +15 a \,d^{2} g^{2}-10 a d e f g +3 a \,e^{2} f^{2}-10 b \,d^{2} f g +2 b d e \,f^{2}+8 c \,d^{2} f^{2}\right )}{15 \left (e x +d \right )^{\frac {5}{2}} \left (g^{3} d^{3}-3 d^{2} e f \,g^{2}+3 d \,e^{2} f^{2} g -e^{3} f^{3}\right )} \]
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 = 4.30, size = 260, normalized size = 1.31 \[ \frac {\sqrt {f+g\,x}\,\left (\frac {16\,c\,d^2\,f^2-20\,b\,d^2\,f\,g+30\,a\,d^2\,g^2+4\,b\,d\,e\,f^2-20\,a\,d\,e\,f\,g+6\,a\,e^2\,f^2}{15\,e^2\,{\left (d\,g-e\,f\right )}^3}+\frac {x\,\left (-8\,c\,d^2\,f\,g+10\,b\,d^2\,g^2+40\,c\,d\,e\,f^2-52\,b\,d\,e\,f\,g+40\,a\,d\,e\,g^2+10\,b\,e^2\,f^2-8\,a\,e^2\,f\,g\right )}{15\,e^2\,{\left (d\,g-e\,f\right )}^3}+\frac {x^2\,\left (6\,c\,d^2\,g^2-20\,c\,d\,e\,f\,g+4\,b\,d\,e\,g^2+30\,c\,e^2\,f^2-20\,b\,e^2\,f\,g+16\,a\,e^2\,g^2\right )}{15\,e^2\,{\left (d\,g-e\,f\right )}^3}\right )}{x^2\,\sqrt {d+e\,x}+\frac {d^2\,\sqrt {d+e\,x}}{e^2}+\frac {2\,d\,x\,\sqrt {d+e\,x}}{e}} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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