Optimal. Leaf size=164 \[ -\frac {2 (A b-a B) (b d-a e)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{9/2}}+\frac {2 \sqrt {d+e x} (A b-a B) (b d-a e)^2}{b^4}+\frac {2 (d+e x)^{3/2} (A b-a B) (b d-a e)}{3 b^3}+\frac {2 (d+e x)^{5/2} (A b-a B)}{5 b^2}+\frac {2 B (d+e x)^{7/2}}{7 b e} \]
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Rubi [A] time = 0.10, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {80, 50, 63, 208} \[ \frac {2 (d+e x)^{5/2} (A b-a B)}{5 b^2}+\frac {2 (d+e x)^{3/2} (A b-a B) (b d-a e)}{3 b^3}+\frac {2 \sqrt {d+e x} (A b-a B) (b d-a e)^2}{b^4}-\frac {2 (A b-a B) (b d-a e)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{9/2}}+\frac {2 B (d+e x)^{7/2}}{7 b e} \]
Antiderivative was successfully verified.
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Rule 50
Rule 63
Rule 80
Rule 208
Rubi steps
\begin {align*} \int \frac {(A+B x) (d+e x)^{5/2}}{a+b x} \, dx &=\frac {2 B (d+e x)^{7/2}}{7 b e}+\frac {\left (2 \left (\frac {7 A b e}{2}-\frac {7 a B e}{2}\right )\right ) \int \frac {(d+e x)^{5/2}}{a+b x} \, dx}{7 b e}\\ &=\frac {2 (A b-a B) (d+e x)^{5/2}}{5 b^2}+\frac {2 B (d+e x)^{7/2}}{7 b e}+\frac {((A b-a B) (b d-a e)) \int \frac {(d+e x)^{3/2}}{a+b x} \, dx}{b^2}\\ &=\frac {2 (A b-a B) (b d-a e) (d+e x)^{3/2}}{3 b^3}+\frac {2 (A b-a B) (d+e x)^{5/2}}{5 b^2}+\frac {2 B (d+e x)^{7/2}}{7 b e}+\frac {\left ((A b-a B) (b d-a e)^2\right ) \int \frac {\sqrt {d+e x}}{a+b x} \, dx}{b^3}\\ &=\frac {2 (A b-a B) (b d-a e)^2 \sqrt {d+e x}}{b^4}+\frac {2 (A b-a B) (b d-a e) (d+e x)^{3/2}}{3 b^3}+\frac {2 (A b-a B) (d+e x)^{5/2}}{5 b^2}+\frac {2 B (d+e x)^{7/2}}{7 b e}+\frac {\left ((A b-a B) (b d-a e)^3\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{b^4}\\ &=\frac {2 (A b-a B) (b d-a e)^2 \sqrt {d+e x}}{b^4}+\frac {2 (A b-a B) (b d-a e) (d+e x)^{3/2}}{3 b^3}+\frac {2 (A b-a B) (d+e x)^{5/2}}{5 b^2}+\frac {2 B (d+e x)^{7/2}}{7 b e}+\frac {\left (2 (A b-a B) (b d-a e)^3\right ) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b d}{e}+\frac {b x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{b^4 e}\\ &=\frac {2 (A b-a B) (b d-a e)^2 \sqrt {d+e x}}{b^4}+\frac {2 (A b-a B) (b d-a e) (d+e x)^{3/2}}{3 b^3}+\frac {2 (A b-a B) (d+e x)^{5/2}}{5 b^2}+\frac {2 B (d+e x)^{7/2}}{7 b e}-\frac {2 (A b-a B) (b d-a e)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{9/2}}\\ \end {align*}
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Mathematica [A] time = 0.27, size = 136, normalized size = 0.83 \[ \frac {2 (A b-a B) \left (5 (b d-a e) \left (\sqrt {b} \sqrt {d+e x} (-3 a e+4 b d+b e x)-3 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )\right )+3 b^{5/2} (d+e x)^{5/2}\right )}{15 b^{9/2}}+\frac {2 B (d+e x)^{7/2}}{7 b e} \]
Antiderivative was successfully verified.
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fricas [B] time = 1.05, size = 591, normalized size = 3.60 \[ \left [\frac {105 \, {\left ({\left (B a b^{2} - A b^{3}\right )} d^{2} e - 2 \, {\left (B a^{2} b - A a b^{2}\right )} d e^{2} + {\left (B a^{3} - A a^{2} b\right )} e^{3}\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e + 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) + 2 \, {\left (15 \, B b^{3} e^{3} x^{3} + 15 \, B b^{3} d^{3} - 161 \, {\left (B a b^{2} - A b^{3}\right )} d^{2} e + 245 \, {\left (B a^{2} b - A a b^{2}\right )} d e^{2} - 105 \, {\left (B a^{3} - A a^{2} b\right )} e^{3} + 3 \, {\left (15 \, B b^{3} d e^{2} - 7 \, {\left (B a b^{2} - A b^{3}\right )} e^{3}\right )} x^{2} + {\left (45 \, B b^{3} d^{2} e - 77 \, {\left (B a b^{2} - A b^{3}\right )} d e^{2} + 35 \, {\left (B a^{2} b - A a b^{2}\right )} e^{3}\right )} x\right )} \sqrt {e x + d}}{105 \, b^{4} e}, \frac {2 \, {\left (105 \, {\left ({\left (B a b^{2} - A b^{3}\right )} d^{2} e - 2 \, {\left (B a^{2} b - A a b^{2}\right )} d e^{2} + {\left (B a^{3} - A a^{2} b\right )} e^{3}\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) + {\left (15 \, B b^{3} e^{3} x^{3} + 15 \, B b^{3} d^{3} - 161 \, {\left (B a b^{2} - A b^{3}\right )} d^{2} e + 245 \, {\left (B a^{2} b - A a b^{2}\right )} d e^{2} - 105 \, {\left (B a^{3} - A a^{2} b\right )} e^{3} + 3 \, {\left (15 \, B b^{3} d e^{2} - 7 \, {\left (B a b^{2} - A b^{3}\right )} e^{3}\right )} x^{2} + {\left (45 \, B b^{3} d^{2} e - 77 \, {\left (B a b^{2} - A b^{3}\right )} d e^{2} + 35 \, {\left (B a^{2} b - A a b^{2}\right )} e^{3}\right )} x\right )} \sqrt {e x + d}\right )}}{105 \, b^{4} e}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 1.30, size = 371, normalized size = 2.26 \[ -\frac {2 \, {\left (B a b^{3} d^{3} - A b^{4} d^{3} - 3 \, B a^{2} b^{2} d^{2} e + 3 \, A a b^{3} d^{2} e + 3 \, B a^{3} b d e^{2} - 3 \, A a^{2} b^{2} d e^{2} - B a^{4} e^{3} + A a^{3} b e^{3}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b^{4}} + \frac {2 \, {\left (15 \, {\left (x e + d\right )}^{\frac {7}{2}} B b^{6} e^{6} - 21 \, {\left (x e + d\right )}^{\frac {5}{2}} B a b^{5} e^{7} + 21 \, {\left (x e + d\right )}^{\frac {5}{2}} A b^{6} e^{7} - 35 \, {\left (x e + d\right )}^{\frac {3}{2}} B a b^{5} d e^{7} + 35 \, {\left (x e + d\right )}^{\frac {3}{2}} A b^{6} d e^{7} - 105 \, \sqrt {x e + d} B a b^{5} d^{2} e^{7} + 105 \, \sqrt {x e + d} A b^{6} d^{2} e^{7} + 35 \, {\left (x e + d\right )}^{\frac {3}{2}} B a^{2} b^{4} e^{8} - 35 \, {\left (x e + d\right )}^{\frac {3}{2}} A a b^{5} e^{8} + 210 \, \sqrt {x e + d} B a^{2} b^{4} d e^{8} - 210 \, \sqrt {x e + d} A a b^{5} d e^{8} - 105 \, \sqrt {x e + d} B a^{3} b^{3} e^{9} + 105 \, \sqrt {x e + d} A a^{2} b^{4} e^{9}\right )} e^{\left (-7\right )}}{105 \, b^{7}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.01, size = 573, normalized size = 3.49 \[ -\frac {2 A \,a^{3} e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{3}}+\frac {6 A \,a^{2} d \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{2}}-\frac {6 A a \,d^{2} e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b}+\frac {2 A \,d^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}}+\frac {2 B \,a^{4} e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{4}}-\frac {6 B \,a^{3} d \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{3}}+\frac {6 B \,a^{2} d^{2} e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{2}}-\frac {2 B a \,d^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b}+\frac {2 \sqrt {e x +d}\, A \,a^{2} e^{2}}{b^{3}}-\frac {4 \sqrt {e x +d}\, A a d e}{b^{2}}+\frac {2 \sqrt {e x +d}\, A \,d^{2}}{b}-\frac {2 \sqrt {e x +d}\, B \,a^{3} e^{2}}{b^{4}}+\frac {4 \sqrt {e x +d}\, B \,a^{2} d e}{b^{3}}-\frac {2 \sqrt {e x +d}\, B a \,d^{2}}{b^{2}}-\frac {2 \left (e x +d \right )^{\frac {3}{2}} A a e}{3 b^{2}}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} A d}{3 b}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} B \,a^{2} e}{3 b^{3}}-\frac {2 \left (e x +d \right )^{\frac {3}{2}} B a d}{3 b^{2}}+\frac {2 \left (e x +d \right )^{\frac {5}{2}} A}{5 b}-\frac {2 \left (e x +d \right )^{\frac {5}{2}} B a}{5 b^{2}}+\frac {2 \left (e x +d \right )^{\frac {7}{2}} B}{7 b e} \]
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 = 1.26, size = 330, normalized size = 2.01 \[ \left (\frac {2\,A\,e-2\,B\,d}{5\,b\,e}-\frac {2\,B\,\left (a\,e^2-b\,d\,e\right )}{5\,b^2\,e^2}\right )\,{\left (d+e\,x\right )}^{5/2}+\frac {2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\left (A\,b-B\,a\right )\,{\left (a\,e-b\,d\right )}^{5/2}\,\sqrt {d+e\,x}}{B\,a^4\,e^3-3\,B\,a^3\,b\,d\,e^2-A\,a^3\,b\,e^3+3\,B\,a^2\,b^2\,d^2\,e+3\,A\,a^2\,b^2\,d\,e^2-B\,a\,b^3\,d^3-3\,A\,a\,b^3\,d^2\,e+A\,b^4\,d^3}\right )\,\left (A\,b-B\,a\right )\,{\left (a\,e-b\,d\right )}^{5/2}}{b^{9/2}}+\frac {2\,B\,{\left (d+e\,x\right )}^{7/2}}{7\,b\,e}+\frac {\left (\frac {2\,A\,e-2\,B\,d}{b\,e}-\frac {2\,B\,\left (a\,e^2-b\,d\,e\right )}{b^2\,e^2}\right )\,{\left (a\,e^2-b\,d\,e\right )}^2\,\sqrt {d+e\,x}}{b^2\,e^2}-\frac {\left (\frac {2\,A\,e-2\,B\,d}{b\,e}-\frac {2\,B\,\left (a\,e^2-b\,d\,e\right )}{b^2\,e^2}\right )\,\left (a\,e^2-b\,d\,e\right )\,{\left (d+e\,x\right )}^{3/2}}{3\,b\,e} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 70.74, size = 221, normalized size = 1.35 \[ \frac {2 B \left (d + e x\right )^{\frac {7}{2}}}{7 b e} + \frac {\left (d + e x\right )^{\frac {5}{2}} \left (2 A b - 2 B a\right )}{5 b^{2}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (- 2 A a b e + 2 A b^{2} d + 2 B a^{2} e - 2 B a b d\right )}{3 b^{3}} + \frac {\sqrt {d + e x} \left (2 A a^{2} b e^{2} - 4 A a b^{2} d e + 2 A b^{3} d^{2} - 2 B a^{3} e^{2} + 4 B a^{2} b d e - 2 B a b^{2} d^{2}\right )}{b^{4}} + \frac {2 \left (- A b + B a\right ) \left (a e - b d\right )^{3} \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {a e - b d}{b}}} \right )}}{b^{5} \sqrt {\frac {a e - b d}{b}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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