Optimal. Leaf size=274 \[ -\frac{7 e (b d-a e)^{3/2} (-9 a B e+5 A b e+4 b B d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{11/2}}+\frac{7 e \sqrt{d+e x} (b d-a e) (-9 a B e+5 A b e+4 b B d)}{4 b^5}+\frac{7 e (d+e x)^{3/2} (-9 a B e+5 A b e+4 b B d)}{12 b^4}+\frac{7 e (d+e x)^{5/2} (-9 a B e+5 A b e+4 b B d)}{20 b^3 (b d-a e)}-\frac{(d+e x)^{7/2} (-9 a B e+5 A b e+4 b B d)}{4 b^2 (a+b x) (b d-a e)}-\frac{(d+e x)^{9/2} (A b-a B)}{2 b (a+b x)^2 (b d-a e)} \]
[Out]
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Rubi [A] time = 0.593197, antiderivative size = 274, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227 \[ -\frac{7 e (b d-a e)^{3/2} (-9 a B e+5 A b e+4 b B d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{11/2}}+\frac{7 e \sqrt{d+e x} (b d-a e) (-9 a B e+5 A b e+4 b B d)}{4 b^5}+\frac{7 e (d+e x)^{3/2} (-9 a B e+5 A b e+4 b B d)}{12 b^4}+\frac{7 e (d+e x)^{5/2} (-9 a B e+5 A b e+4 b B d)}{20 b^3 (b d-a e)}-\frac{(d+e x)^{7/2} (-9 a B e+5 A b e+4 b B d)}{4 b^2 (a+b x) (b d-a e)}-\frac{(d+e x)^{9/2} (A b-a B)}{2 b (a+b x)^2 (b d-a e)} \]
Antiderivative was successfully verified.
[In] Int[((A + B*x)*(d + e*x)^(7/2))/(a + b*x)^3,x]
[Out]
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Rubi in Sympy [A] time = 60.7287, size = 269, normalized size = 0.98 \[ \frac{\left (d + e x\right )^{\frac{9}{2}} \left (A b - B a\right )}{2 b \left (a + b x\right )^{2} \left (a e - b d\right )} + \frac{\left (d + e x\right )^{\frac{7}{2}} \left (5 A b e - 9 B a e + 4 B b d\right )}{4 b^{2} \left (a + b x\right ) \left (a e - b d\right )} - \frac{7 e \left (d + e x\right )^{\frac{5}{2}} \left (5 A b e - 9 B a e + 4 B b d\right )}{20 b^{3} \left (a e - b d\right )} + \frac{7 e \left (d + e x\right )^{\frac{3}{2}} \left (5 A b e - 9 B a e + 4 B b d\right )}{12 b^{4}} - \frac{7 e \sqrt{d + e x} \left (a e - b d\right ) \left (5 A b e - 9 B a e + 4 B b d\right )}{4 b^{5}} + \frac{7 e \left (a e - b d\right )^{\frac{3}{2}} \left (5 A b e - 9 B a e + 4 B b d\right ) \operatorname{atan}{\left (\frac{\sqrt{b} \sqrt{d + e x}}{\sqrt{a e - b d}} \right )}}{4 b^{\frac{11}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] rubi_integrate((B*x+A)*(e*x+d)**(7/2)/(b*x+a)**3,x)
[Out]
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Mathematica [A] time = 0.805434, size = 223, normalized size = 0.81 \[ \frac{\sqrt{d+e x} \left (8 e \left (90 a^2 B e^2-15 a b e (3 A e+10 B d)+2 b^2 d (25 A e+29 B d)\right )+8 b e^2 x (-15 a B e+5 A b e+16 b B d)-\frac{15 (b d-a e)^2 (-17 a B e+13 A b e+4 b B d)}{a+b x}-\frac{30 (A b-a B) (b d-a e)^3}{(a+b x)^2}+24 b^2 B e^3 x^2\right )}{60 b^5}-\frac{7 e (b d-a e)^{3/2} (-9 a B e+5 A b e+4 b B d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{11/2}} \]
Antiderivative was successfully verified.
[In] Integrate[((A + B*x)*(d + e*x)^(7/2))/(a + b*x)^3,x]
[Out]
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Maple [B] time = 0.036, size = 940, normalized size = 3.4 \[ \text{result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] int((B*x+A)*(e*x+d)^(7/2)/(b*x+a)^3,x)
[Out]
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Maxima [F] time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate((B*x + A)*(e*x + d)^(7/2)/(b*x + a)^3,x, algorithm="maxima")
[Out]
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Fricas [A] time = 0.232953, size = 1, normalized size = 0. \[ \text{result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate((B*x + A)*(e*x + d)^(7/2)/(b*x + a)^3,x, algorithm="fricas")
[Out]
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate((B*x+A)*(e*x+d)**(7/2)/(b*x+a)**3,x)
[Out]
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GIAC/XCAS [A] time = 0.234489, size = 819, normalized size = 2.99 \[ \frac{7 \,{\left (4 \, B b^{3} d^{3} e - 17 \, B a b^{2} d^{2} e^{2} + 5 \, A b^{3} d^{2} e^{2} + 22 \, B a^{2} b d e^{3} - 10 \, A a b^{2} d e^{3} - 9 \, B a^{3} e^{4} + 5 \, A a^{2} b e^{4}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{4 \, \sqrt{-b^{2} d + a b e} b^{5}} - \frac{4 \,{\left (x e + d\right )}^{\frac{3}{2}} B b^{4} d^{3} e - 4 \, \sqrt{x e + d} B b^{4} d^{4} e - 25 \,{\left (x e + d\right )}^{\frac{3}{2}} B a b^{3} d^{2} e^{2} + 13 \,{\left (x e + d\right )}^{\frac{3}{2}} A b^{4} d^{2} e^{2} + 27 \, \sqrt{x e + d} B a b^{3} d^{3} e^{2} - 11 \, \sqrt{x e + d} A b^{4} d^{3} e^{2} + 38 \,{\left (x e + d\right )}^{\frac{3}{2}} B a^{2} b^{2} d e^{3} - 26 \,{\left (x e + d\right )}^{\frac{3}{2}} A a b^{3} d e^{3} - 57 \, \sqrt{x e + d} B a^{2} b^{2} d^{2} e^{3} + 33 \, \sqrt{x e + d} A a b^{3} d^{2} e^{3} - 17 \,{\left (x e + d\right )}^{\frac{3}{2}} B a^{3} b e^{4} + 13 \,{\left (x e + d\right )}^{\frac{3}{2}} A a^{2} b^{2} e^{4} + 49 \, \sqrt{x e + d} B a^{3} b d e^{4} - 33 \, \sqrt{x e + d} A a^{2} b^{2} d e^{4} - 15 \, \sqrt{x e + d} B a^{4} e^{5} + 11 \, \sqrt{x e + d} A a^{3} b e^{5}}{4 \,{\left ({\left (x e + d\right )} b - b d + a e\right )}^{2} b^{5}} + \frac{2 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} B b^{12} e + 10 \,{\left (x e + d\right )}^{\frac{3}{2}} B b^{12} d e + 45 \, \sqrt{x e + d} B b^{12} d^{2} e - 15 \,{\left (x e + d\right )}^{\frac{3}{2}} B a b^{11} e^{2} + 5 \,{\left (x e + d\right )}^{\frac{3}{2}} A b^{12} e^{2} - 135 \, \sqrt{x e + d} B a b^{11} d e^{2} + 45 \, \sqrt{x e + d} A b^{12} d e^{2} + 90 \, \sqrt{x e + d} B a^{2} b^{10} e^{3} - 45 \, \sqrt{x e + d} A a b^{11} e^{3}\right )}}{15 \, b^{15}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate((B*x + A)*(e*x + d)^(7/2)/(b*x + a)^3,x, algorithm="giac")
[Out]