Optimal. Leaf size=277 \[ \frac {2 a^2 (b+3 c) (a+b x)^{3/2}}{3 b^3 (b-c)^3}-\frac {8 a^2 (a+b x)^{3/2}}{3 b^2 (b-c)^3}-\frac {2 a^2 (3 b+c) (a+c x)^{3/2}}{3 c^3 (b-c)^3}+\frac {8 a^2 (a+c x)^{3/2}}{3 c^2 (b-c)^3}+\frac {2 (b+3 c) (a+b x)^{7/2}}{7 b^3 (b-c)^3}-\frac {4 a (b+3 c) (a+b x)^{5/2}}{5 b^3 (b-c)^3}+\frac {8 a (a+b x)^{5/2}}{5 b^2 (b-c)^3}-\frac {2 (3 b+c) (a+c x)^{7/2}}{7 c^3 (b-c)^3}+\frac {4 a (3 b+c) (a+c x)^{5/2}}{5 c^3 (b-c)^3}-\frac {8 a (a+c x)^{5/2}}{5 c^2 (b-c)^3} \]
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Rubi [A] time = 0.32, antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {6690, 43} \[ \frac {2 a^2 (b+3 c) (a+b x)^{3/2}}{3 b^3 (b-c)^3}-\frac {8 a^2 (a+b x)^{3/2}}{3 b^2 (b-c)^3}-\frac {2 a^2 (3 b+c) (a+c x)^{3/2}}{3 c^3 (b-c)^3}+\frac {8 a^2 (a+c x)^{3/2}}{3 c^2 (b-c)^3}+\frac {2 (b+3 c) (a+b x)^{7/2}}{7 b^3 (b-c)^3}-\frac {4 a (b+3 c) (a+b x)^{5/2}}{5 b^3 (b-c)^3}+\frac {8 a (a+b x)^{5/2}}{5 b^2 (b-c)^3}-\frac {2 (3 b+c) (a+c x)^{7/2}}{7 c^3 (b-c)^3}+\frac {4 a (3 b+c) (a+c x)^{5/2}}{5 c^3 (b-c)^3}-\frac {8 a (a+c x)^{5/2}}{5 c^2 (b-c)^3} \]
Antiderivative was successfully verified.
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Rule 43
Rule 6690
Rubi steps
\begin {align*} \int \frac {x^4}{\left (\sqrt {a+b x}+\sqrt {a+c x}\right )^3} \, dx &=\frac {\int \left (4 a x \sqrt {a+b x}+b \left (1+\frac {3 c}{b}\right ) x^2 \sqrt {a+b x}-4 a x \sqrt {a+c x}-3 b \left (1+\frac {c}{3 b}\right ) x^2 \sqrt {a+c x}\right ) \, dx}{(b-c)^3}\\ &=\frac {(4 a) \int x \sqrt {a+b x} \, dx}{(b-c)^3}-\frac {(4 a) \int x \sqrt {a+c x} \, dx}{(b-c)^3}-\frac {(3 b+c) \int x^2 \sqrt {a+c x} \, dx}{(b-c)^3}+\frac {(b+3 c) \int x^2 \sqrt {a+b x} \, dx}{(b-c)^3}\\ &=\frac {(4 a) \int \left (-\frac {a \sqrt {a+b x}}{b}+\frac {(a+b x)^{3/2}}{b}\right ) \, dx}{(b-c)^3}-\frac {(4 a) \int \left (-\frac {a \sqrt {a+c x}}{c}+\frac {(a+c x)^{3/2}}{c}\right ) \, dx}{(b-c)^3}-\frac {(3 b+c) \int \left (\frac {a^2 \sqrt {a+c x}}{c^2}-\frac {2 a (a+c x)^{3/2}}{c^2}+\frac {(a+c x)^{5/2}}{c^2}\right ) \, dx}{(b-c)^3}+\frac {(b+3 c) \int \left (\frac {a^2 \sqrt {a+b x}}{b^2}-\frac {2 a (a+b x)^{3/2}}{b^2}+\frac {(a+b x)^{5/2}}{b^2}\right ) \, dx}{(b-c)^3}\\ &=-\frac {8 a^2 (a+b x)^{3/2}}{3 b^2 (b-c)^3}+\frac {2 a^2 (b+3 c) (a+b x)^{3/2}}{3 b^3 (b-c)^3}+\frac {8 a (a+b x)^{5/2}}{5 b^2 (b-c)^3}-\frac {4 a (b+3 c) (a+b x)^{5/2}}{5 b^3 (b-c)^3}+\frac {2 (b+3 c) (a+b x)^{7/2}}{7 b^3 (b-c)^3}+\frac {8 a^2 (a+c x)^{3/2}}{3 (b-c)^3 c^2}-\frac {2 a^2 (3 b+c) (a+c x)^{3/2}}{3 (b-c)^3 c^3}-\frac {8 a (a+c x)^{5/2}}{5 (b-c)^3 c^2}+\frac {4 a (3 b+c) (a+c x)^{5/2}}{5 (b-c)^3 c^3}-\frac {2 (3 b+c) (a+c x)^{7/2}}{7 (b-c)^3 c^3}\\ \end {align*}
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Mathematica [A] time = 0.39, size = 271, normalized size = 0.98 \[ \frac {2 \left (8 a^3 \left (b^4 \left (-\sqrt {a+c x}\right )+2 b^3 c \sqrt {a+c x}+c^4 \sqrt {a+b x}-2 b c^3 \sqrt {a+b x}\right )+4 a^2 b c x \left (b^3 \sqrt {a+c x}-2 b^2 c \sqrt {a+c x}-c^3 \sqrt {a+b x}+2 b c^2 \sqrt {a+b x}\right )+5 b^3 c^3 x^3 \left (-3 b \sqrt {a+c x}+3 c \sqrt {a+b x}+b \sqrt {a+b x}-c \sqrt {a+c x}\right )+a b^2 c^2 x^2 \left (-3 b^2 \sqrt {a+c x}+3 c^2 \sqrt {a+b x}+29 b c \left (\sqrt {a+b x}-\sqrt {a+c x}\right )\right )\right )}{35 b^3 c^3 (b-c)^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.48, size = 225, normalized size = 0.81 \[ -\frac {2 \, {\left ({\left (16 \, a^{3} b c^{3} - 8 \, a^{3} c^{4} - 5 \, {\left (b^{4} c^{3} + 3 \, b^{3} c^{4}\right )} x^{3} - {\left (29 \, a b^{3} c^{3} + 3 \, a b^{2} c^{4}\right )} x^{2} - 4 \, {\left (2 \, a^{2} b^{2} c^{3} - a^{2} b c^{4}\right )} x\right )} \sqrt {b x + a} + {\left (8 \, a^{3} b^{4} - 16 \, a^{3} b^{3} c + 5 \, {\left (3 \, b^{4} c^{3} + b^{3} c^{4}\right )} x^{3} + {\left (3 \, a b^{4} c^{2} + 29 \, a b^{3} c^{3}\right )} x^{2} - 4 \, {\left (a^{2} b^{4} c - 2 \, a^{2} b^{3} c^{2}\right )} x\right )} \sqrt {c x + a}\right )}}{35 \, {\left (b^{6} c^{3} - 3 \, b^{5} c^{4} + 3 \, b^{4} c^{5} - b^{3} c^{6}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 5.58, size = 932, normalized size = 3.36 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.01, size = 246, normalized size = 0.89 \[ \frac {8 \left (-\frac {\left (b x +a \right )^{\frac {3}{2}} a}{3}+\frac {\left (b x +a \right )^{\frac {5}{2}}}{5}\right ) a}{\left (b -c \right )^{3} b^{2}}-\frac {8 \left (-\frac {\left (c x +a \right )^{\frac {3}{2}} a}{3}+\frac {\left (c x +a \right )^{\frac {5}{2}}}{5}\right ) a}{\left (b -c \right )^{3} c^{2}}-\frac {6 \left (\frac {\left (c x +a \right )^{\frac {3}{2}} a^{2}}{3}-\frac {2 \left (c x +a \right )^{\frac {5}{2}} a}{5}+\frac {\left (c x +a \right )^{\frac {7}{2}}}{7}\right ) b}{\left (b -c \right )^{3} c^{3}}+\frac {\frac {2 \left (b x +a \right )^{\frac {3}{2}} a^{2}}{3}-\frac {4 \left (b x +a \right )^{\frac {5}{2}} a}{5}+\frac {2 \left (b x +a \right )^{\frac {7}{2}}}{7}}{\left (b -c \right )^{3} b^{2}}+\frac {6 \left (\frac {\left (b x +a \right )^{\frac {3}{2}} a^{2}}{3}-\frac {2 \left (b x +a \right )^{\frac {5}{2}} a}{5}+\frac {\left (b x +a \right )^{\frac {7}{2}}}{7}\right ) c}{\left (b -c \right )^{3} b^{3}}-\frac {2 \left (\frac {\left (c x +a \right )^{\frac {3}{2}} a^{2}}{3}-\frac {2 \left (c x +a \right )^{\frac {5}{2}} a}{5}+\frac {\left (c x +a \right )^{\frac {7}{2}}}{7}\right )}{\left (b -c \right )^{3} c^{2}} \]
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 \[ \int \frac {x^{4}}{{\left (\sqrt {b x + a} + \sqrt {c x + a}\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.34, size = 429, normalized size = 1.55 \[ \frac {x^2\,\left (\frac {12\,a\,\left (3\,b+c\right )}{7\,{\left (b-c\right )}^3}-\frac {2\,a\,\left (3\,b+5\,c\right )}{{\left (b-c\right )}^3}\right )\,\sqrt {a+c\,x}}{5\,c}-\frac {2\,a\,\left (\frac {8\,a^2}{{\left (b-c\right )}^3}-\frac {4\,a\,\left (\frac {2\,a\,\left (5\,b+3\,c\right )}{{\left (b-c\right )}^3}-\frac {12\,a\,\left (b^2+3\,c\,b\right )}{7\,b\,{\left (b-c\right )}^3}\right )}{5\,b}\right )\,\sqrt {a+b\,x}}{3\,b^2}+\frac {x\,\left (\frac {8\,a^2}{{\left (b-c\right )}^3}-\frac {4\,a\,\left (\frac {2\,a\,\left (5\,b+3\,c\right )}{{\left (b-c\right )}^3}-\frac {12\,a\,\left (b^2+3\,c\,b\right )}{7\,b\,{\left (b-c\right )}^3}\right )}{5\,b}\right )\,\sqrt {a+b\,x}}{3\,b}+\frac {2\,a\,\left (\frac {8\,a^2}{{\left (b-c\right )}^3}+\frac {4\,a\,\left (\frac {12\,a\,\left (3\,b+c\right )}{7\,{\left (b-c\right )}^3}-\frac {2\,a\,\left (3\,b+5\,c\right )}{{\left (b-c\right )}^3}\right )}{5\,c}\right )\,\sqrt {a+c\,x}}{3\,c^2}+\frac {x^2\,\left (\frac {2\,a\,\left (5\,b+3\,c\right )}{{\left (b-c\right )}^3}-\frac {12\,a\,\left (b^2+3\,c\,b\right )}{7\,b\,{\left (b-c\right )}^3}\right )\,\sqrt {a+b\,x}}{5\,b}-\frac {x\,\left (\frac {8\,a^2}{{\left (b-c\right )}^3}+\frac {4\,a\,\left (\frac {12\,a\,\left (3\,b+c\right )}{7\,{\left (b-c\right )}^3}-\frac {2\,a\,\left (3\,b+5\,c\right )}{{\left (b-c\right )}^3}\right )}{5\,c}\right )\,\sqrt {a+c\,x}}{3\,c}-\frac {2\,x^3\,\left (3\,b+c\right )\,\sqrt {a+c\,x}}{7\,{\left (b-c\right )}^3}+\frac {2\,x^3\,\left (b^2+3\,c\,b\right )\,\sqrt {a+b\,x}}{7\,b\,{\left (b-c\right )}^3} \]
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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