Optimal. Leaf size=404 \[ \frac{13 A b-5 a B}{24 a^2 b x^{5/2} (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A b-a B}{4 a b x^{5/2} (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{231 b^{3/2} (a+b x) (13 A b-5 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{64 a^{15/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{231 b (a+b x) (13 A b-5 a B)}{64 a^7 \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{77 (a+b x) (13 A b-5 a B)}{64 a^6 x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{231 (a+b x) (13 A b-5 a B)}{320 a^5 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{33 (13 A b-5 a B)}{64 a^4 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{11 (13 A b-5 a B)}{96 a^3 b x^{5/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}} \]
[Out]
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Rubi [A] time = 0.535327, antiderivative size = 404, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194 \[ \frac{13 A b-5 a B}{24 a^2 b x^{5/2} (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A b-a B}{4 a b x^{5/2} (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{231 b^{3/2} (a+b x) (13 A b-5 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{64 a^{15/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{231 b (a+b x) (13 A b-5 a B)}{64 a^7 \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{77 (a+b x) (13 A b-5 a B)}{64 a^6 x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{231 (a+b x) (13 A b-5 a B)}{320 a^5 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{33 (13 A b-5 a B)}{64 a^4 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{11 (13 A b-5 a B)}{96 a^3 b x^{5/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}} \]
Antiderivative was successfully verified.
[In] Int[(A + B*x)/(x^(7/2)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2)),x]
[Out]
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Rubi in Sympy [F(-2)] time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: RecursionError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] rubi_integrate((B*x+A)/x**(7/2)/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)
[Out]
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Mathematica [A] time = 0.248517, size = 196, normalized size = 0.49 \[ \frac{\sqrt{a} \left (-128 a^6 (3 A+5 B x)+128 a^5 b x (13 A+55 B x)+11 a^4 b^2 x^2 (4185 B x-1664 A)+33 a^3 b^3 x^3 (2555 B x-3627 A)+231 a^2 b^4 x^4 (275 B x-949 A)+1155 a b^5 x^5 (15 B x-143 A)-45045 A b^6 x^6\right )+3465 b^{3/2} x^{5/2} (a+b x)^4 (5 a B-13 A b) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{960 a^{15/2} x^{5/2} (a+b x)^3 \sqrt{(a+b x)^2}} \]
Antiderivative was successfully verified.
[In] Integrate[(A + B*x)/(x^(7/2)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2)),x]
[Out]
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Maple [A] time = 0.038, size = 449, normalized size = 1.1 \[ -{\frac{bx+a}{960\,{a}^{7}} \left ( -69300\,B\arctan \left ({\frac{b\sqrt{x}}{\sqrt{ab}}} \right ){x}^{7/2}{a}^{4}{b}^{3}-69300\,B\arctan \left ({\frac{b\sqrt{x}}{\sqrt{ab}}} \right ){x}^{11/2}{a}^{2}{b}^{5}+270270\,A\arctan \left ({\frac{b\sqrt{x}}{\sqrt{ab}}} \right ){x}^{9/2}{a}^{2}{b}^{5}-103950\,B\arctan \left ({\frac{b\sqrt{x}}{\sqrt{ab}}} \right ){x}^{9/2}{a}^{3}{b}^{4}+180180\,A\arctan \left ({\frac{b\sqrt{x}}{\sqrt{ab}}} \right ){x}^{7/2}{a}^{3}{b}^{4}-17325\,B\sqrt{ab}{x}^{6}a{b}^{5}+165165\,A\sqrt{ab}{x}^{5}a{b}^{5}+45045\,A\arctan \left ({\frac{b\sqrt{x}}{\sqrt{ab}}} \right ){x}^{13/2}{b}^{7}-63525\,B\sqrt{ab}{x}^{5}{a}^{2}{b}^{4}+219219\,A\sqrt{ab}{x}^{4}{a}^{2}{b}^{4}-84315\,B\sqrt{ab}{x}^{4}{a}^{3}{b}^{3}+119691\,A\sqrt{ab}{x}^{3}{a}^{3}{b}^{3}+45045\,A\arctan \left ({\frac{b\sqrt{x}}{\sqrt{ab}}} \right ){x}^{5/2}{a}^{4}{b}^{3}-17325\,B\arctan \left ({\frac{b\sqrt{x}}{\sqrt{ab}}} \right ){x}^{13/2}a{b}^{6}+180180\,A\arctan \left ({\frac{b\sqrt{x}}{\sqrt{ab}}} \right ){x}^{11/2}a{b}^{6}-46035\,B\sqrt{ab}{x}^{3}{a}^{4}{b}^{2}-17325\,B\arctan \left ({\frac{b\sqrt{x}}{\sqrt{ab}}} \right ){x}^{5/2}{a}^{5}{b}^{2}+18304\,A\sqrt{ab}{x}^{2}{a}^{4}{b}^{2}-7040\,B\sqrt{ab}{x}^{2}{a}^{5}b-1664\,A\sqrt{ab}x{a}^{5}b+45045\,A\sqrt{ab}{x}^{6}{b}^{6}+640\,B\sqrt{ab}x{a}^{6}+384\,A\sqrt{ab}{a}^{6} \right ){\frac{1}{\sqrt{ab}}}{x}^{-{\frac{5}{2}}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{5}{2}}}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] int((B*x+A)/x^(7/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),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)/((b^2*x^2 + 2*a*b*x + a^2)^(5/2)*x^(7/2)),x, algorithm="maxima")
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Fricas [A] time = 0.294672, 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)/((b^2*x^2 + 2*a*b*x + a^2)^(5/2)*x^(7/2)),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)/x**(7/2)/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)
[Out]
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GIAC/XCAS [A] time = 0.27913, size = 279, normalized size = 0.69 \[ \frac{231 \,{\left (5 \, B a b^{2} - 13 \, A b^{3}\right )} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{64 \, \sqrt{a b} a^{7}{\rm sign}\left (b x + a\right )} + \frac{2 \,{\left (75 \, B a b x^{2} - 225 \, A b^{2} x^{2} - 5 \, B a^{2} x + 25 \, A a b x - 3 \, A a^{2}\right )}}{15 \, a^{7} x^{\frac{5}{2}}{\rm sign}\left (b x + a\right )} + \frac{1545 \, B a b^{5} x^{\frac{7}{2}} - 3249 \, A b^{6} x^{\frac{7}{2}} + 5153 \, B a^{2} b^{4} x^{\frac{5}{2}} - 10633 \, A a b^{5} x^{\frac{5}{2}} + 5855 \, B a^{3} b^{3} x^{\frac{3}{2}} - 11767 \, A a^{2} b^{4} x^{\frac{3}{2}} + 2295 \, B a^{4} b^{2} \sqrt{x} - 4431 \, A a^{3} b^{3} \sqrt{x}}{192 \,{\left (b x + a\right )}^{4} a^{7}{\rm sign}\left (b x + a\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate((B*x + A)/((b^2*x^2 + 2*a*b*x + a^2)^(5/2)*x^(7/2)),x, algorithm="giac")
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