Optimal. Leaf size=214 \[ -\frac {(b d-a e)^{3/2} (-7 a B e+5 A b e+2 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{9/2}}+\frac {\sqrt {d+e x} (b d-a e) (-7 a B e+5 A b e+2 b B d)}{b^4}+\frac {(d+e x)^{3/2} (-7 a B e+5 A b e+2 b B d)}{3 b^3}+\frac {(d+e x)^{5/2} (-7 a B e+5 A b e+2 b B d)}{5 b^2 (b d-a e)}-\frac {(d+e x)^{7/2} (A b-a B)}{b (a+b x) (b d-a e)} \]
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Rubi [A] time = 0.19, antiderivative size = 214, 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 = {78, 50, 63, 208} \begin {gather*} \frac {(d+e x)^{5/2} (-7 a B e+5 A b e+2 b B d)}{5 b^2 (b d-a e)}+\frac {(d+e x)^{3/2} (-7 a B e+5 A b e+2 b B d)}{3 b^3}+\frac {\sqrt {d+e x} (b d-a e) (-7 a B e+5 A b e+2 b B d)}{b^4}-\frac {(b d-a e)^{3/2} (-7 a B e+5 A b e+2 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{9/2}}-\frac {(d+e x)^{7/2} (A b-a B)}{b (a+b x) (b d-a e)} \end {gather*}
Antiderivative was successfully verified.
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Rule 50
Rule 63
Rule 78
Rule 208
Rubi steps
\begin {align*} \int \frac {(A+B x) (d+e x)^{5/2}}{(a+b x)^2} \, dx &=-\frac {(A b-a B) (d+e x)^{7/2}}{b (b d-a e) (a+b x)}+\frac {(2 b B d+5 A b e-7 a B e) \int \frac {(d+e x)^{5/2}}{a+b x} \, dx}{2 b (b d-a e)}\\ &=\frac {(2 b B d+5 A b e-7 a B e) (d+e x)^{5/2}}{5 b^2 (b d-a e)}-\frac {(A b-a B) (d+e x)^{7/2}}{b (b d-a e) (a+b x)}+\frac {(2 b B d+5 A b e-7 a B e) \int \frac {(d+e x)^{3/2}}{a+b x} \, dx}{2 b^2}\\ &=\frac {(2 b B d+5 A b e-7 a B e) (d+e x)^{3/2}}{3 b^3}+\frac {(2 b B d+5 A b e-7 a B e) (d+e x)^{5/2}}{5 b^2 (b d-a e)}-\frac {(A b-a B) (d+e x)^{7/2}}{b (b d-a e) (a+b x)}+\frac {((b d-a e) (2 b B d+5 A b e-7 a B e)) \int \frac {\sqrt {d+e x}}{a+b x} \, dx}{2 b^3}\\ &=\frac {(b d-a e) (2 b B d+5 A b e-7 a B e) \sqrt {d+e x}}{b^4}+\frac {(2 b B d+5 A b e-7 a B e) (d+e x)^{3/2}}{3 b^3}+\frac {(2 b B d+5 A b e-7 a B e) (d+e x)^{5/2}}{5 b^2 (b d-a e)}-\frac {(A b-a B) (d+e x)^{7/2}}{b (b d-a e) (a+b x)}+\frac {\left ((b d-a e)^2 (2 b B d+5 A b e-7 a B e)\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{2 b^4}\\ &=\frac {(b d-a e) (2 b B d+5 A b e-7 a B e) \sqrt {d+e x}}{b^4}+\frac {(2 b B d+5 A b e-7 a B e) (d+e x)^{3/2}}{3 b^3}+\frac {(2 b B d+5 A b e-7 a B e) (d+e x)^{5/2}}{5 b^2 (b d-a e)}-\frac {(A b-a B) (d+e x)^{7/2}}{b (b d-a e) (a+b x)}+\frac {\left ((b d-a e)^2 (2 b B d+5 A b e-7 a B e)\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 {(b d-a e) (2 b B d+5 A b e-7 a B e) \sqrt {d+e x}}{b^4}+\frac {(2 b B d+5 A b e-7 a B e) (d+e x)^{3/2}}{3 b^3}+\frac {(2 b B d+5 A b e-7 a B e) (d+e x)^{5/2}}{5 b^2 (b d-a e)}-\frac {(A b-a B) (d+e x)^{7/2}}{b (b d-a e) (a+b x)}-\frac {(b d-a e)^{3/2} (2 b B d+5 A b e-7 a B e) \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 = 0.50, size = 166, normalized size = 0.78 \begin {gather*} \frac {\frac {2 \left (-\frac {7 a B e}{2}+\frac {5 A b e}{2}+b B d\right ) \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^{7/2}}+\frac {(d+e x)^{7/2} (a B-A b)}{a+b x}}{b (b d-a e)} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.57, size = 391, normalized size = 1.83 \begin {gather*} \frac {\sqrt {d+e x} \left (105 a^3 B e^3-75 a^2 A b e^3+70 a^2 b B e^2 (d+e x)-240 a^2 b B d e^2-50 a A b^2 e^2 (d+e x)+150 a A b^2 d e^2+165 a b^2 B d^2 e-14 a b^2 B e (d+e x)^2-90 a b^2 B d e (d+e x)-75 A b^3 d^2 e+10 A b^3 e (d+e x)^2+50 A b^3 d e (d+e x)-30 b^3 B d^3+20 b^3 B d^2 (d+e x)+6 b^3 B (d+e x)^3+4 b^3 B d (d+e x)^2\right )}{15 b^4 (a e+b (d+e x)-b d)}+\frac {\left (7 a^4 B e^4-5 a^3 A b e^4-23 a^3 b B d e^3+15 a^2 A b^2 d e^3+27 a^2 b^2 B d^2 e^2-15 a A b^3 d^2 e^2-13 a b^3 B d^3 e+5 A b^4 d^3 e+2 b^4 B d^4\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{b^{9/2} (a e-b d)^{3/2}} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.39, size = 666, normalized size = 3.11 \begin {gather*} \left [-\frac {15 \, {\left (2 \, B a b^{2} d^{2} - {\left (9 \, B a^{2} b - 5 \, A a b^{2}\right )} d e + {\left (7 \, B a^{3} - 5 \, A a^{2} b\right )} e^{2} + {\left (2 \, B b^{3} d^{2} - {\left (9 \, B a b^{2} - 5 \, A b^{3}\right )} d e + {\left (7 \, B a^{2} b - 5 \, A a b^{2}\right )} e^{2}\right )} x\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 (6 \, B b^{3} e^{2} x^{3} + {\left (61 \, B a b^{2} - 15 \, A b^{3}\right )} d^{2} - 10 \, {\left (17 \, B a^{2} b - 10 \, A a b^{2}\right )} d e + 15 \, {\left (7 \, B a^{3} - 5 \, A a^{2} b\right )} e^{2} + 2 \, {\left (11 \, B b^{3} d e - {\left (7 \, B a b^{2} - 5 \, A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (23 \, B b^{3} d^{2} - {\left (59 \, B a b^{2} - 35 \, A b^{3}\right )} d e + 5 \, {\left (7 \, B a^{2} b - 5 \, A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{30 \, {\left (b^{5} x + a b^{4}\right )}}, -\frac {15 \, {\left (2 \, B a b^{2} d^{2} - {\left (9 \, B a^{2} b - 5 \, A a b^{2}\right )} d e + {\left (7 \, B a^{3} - 5 \, A a^{2} b\right )} e^{2} + {\left (2 \, B b^{3} d^{2} - {\left (9 \, B a b^{2} - 5 \, A b^{3}\right )} d e + {\left (7 \, B a^{2} b - 5 \, A a b^{2}\right )} e^{2}\right )} x\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 (6 \, B b^{3} e^{2} x^{3} + {\left (61 \, B a b^{2} - 15 \, A b^{3}\right )} d^{2} - 10 \, {\left (17 \, B a^{2} b - 10 \, A a b^{2}\right )} d e + 15 \, {\left (7 \, B a^{3} - 5 \, A a^{2} b\right )} e^{2} + 2 \, {\left (11 \, B b^{3} d e - {\left (7 \, B a b^{2} - 5 \, A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (23 \, B b^{3} d^{2} - {\left (59 \, B a b^{2} - 35 \, A b^{3}\right )} d e + 5 \, {\left (7 \, B a^{2} b - 5 \, A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (b^{5} x + a b^{4}\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 1.29, size = 400, normalized size = 1.87 \begin {gather*} \frac {{\left (2 \, B b^{3} d^{3} - 11 \, B a b^{2} d^{2} e + 5 \, A b^{3} d^{2} e + 16 \, B a^{2} b d e^{2} - 10 \, A a b^{2} d e^{2} - 7 \, B a^{3} e^{3} + 5 \, A a^{2} 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 {\sqrt {x e + d} B a b^{2} d^{2} e - \sqrt {x e + d} A b^{3} d^{2} e - 2 \, \sqrt {x e + d} B a^{2} b d e^{2} + 2 \, \sqrt {x e + d} A a b^{2} d e^{2} + \sqrt {x e + d} B a^{3} e^{3} - \sqrt {x e + d} A a^{2} b e^{3}}{{\left ({\left (x e + d\right )} b - b d + a e\right )} b^{4}} + \frac {2 \, {\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} B b^{8} + 5 \, {\left (x e + d\right )}^{\frac {3}{2}} B b^{8} d + 15 \, \sqrt {x e + d} B b^{8} d^{2} - 10 \, {\left (x e + d\right )}^{\frac {3}{2}} B a b^{7} e + 5 \, {\left (x e + d\right )}^{\frac {3}{2}} A b^{8} e - 60 \, \sqrt {x e + d} B a b^{7} d e + 30 \, \sqrt {x e + d} A b^{8} d e + 45 \, \sqrt {x e + d} B a^{2} b^{6} e^{2} - 30 \, \sqrt {x e + d} A a b^{7} e^{2}\right )}}{15 \, b^{10}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.02, size = 626, normalized size = 2.93 \begin {gather*} \frac {5 A \,a^{2} 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 {10 A a 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 {5 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 {7 B \,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^{4}}+\frac {16 B \,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^{3}}-\frac {11 B 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^{2}}+\frac {2 B \,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 {\sqrt {e x +d}\, A \,a^{2} e^{3}}{\left (b x e +a e \right ) b^{3}}+\frac {2 \sqrt {e x +d}\, A a d \,e^{2}}{\left (b x e +a e \right ) b^{2}}-\frac {\sqrt {e x +d}\, A \,d^{2} e}{\left (b x e +a e \right ) b}+\frac {\sqrt {e x +d}\, B \,a^{3} e^{3}}{\left (b x e +a e \right ) b^{4}}-\frac {2 \sqrt {e x +d}\, B \,a^{2} d \,e^{2}}{\left (b x e +a e \right ) b^{3}}+\frac {\sqrt {e x +d}\, B a \,d^{2} e}{\left (b x e +a e \right ) b^{2}}-\frac {4 \sqrt {e x +d}\, A a \,e^{2}}{b^{3}}+\frac {4 \sqrt {e x +d}\, A d e}{b^{2}}+\frac {6 \sqrt {e x +d}\, B \,a^{2} e^{2}}{b^{4}}-\frac {8 \sqrt {e x +d}\, B a d e}{b^{3}}+\frac {2 \sqrt {e x +d}\, B \,d^{2}}{b^{2}}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} A e}{3 b^{2}}-\frac {4 \left (e x +d \right )^{\frac {3}{2}} B a e}{3 b^{3}}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} B d}{3 b^{2}}+\frac {2 \left (e x +d \right )^{\frac {5}{2}} B}{5 b^{2}} \end {gather*}
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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.13, size = 363, normalized size = 1.70 \begin {gather*} \left (\frac {2\,A\,e-2\,B\,d}{3\,b^2}+\frac {2\,B\,\left (2\,b^2\,d-2\,a\,b\,e\right )}{3\,b^4}\right )\,{\left (d+e\,x\right )}^{3/2}+\left (\frac {\left (2\,b^2\,d-2\,a\,b\,e\right )\,\left (\frac {2\,A\,e-2\,B\,d}{b^2}+\frac {2\,B\,\left (2\,b^2\,d-2\,a\,b\,e\right )}{b^4}\right )}{b^2}-\frac {2\,B\,{\left (a\,e-b\,d\right )}^2}{b^4}\right )\,\sqrt {d+e\,x}+\frac {\sqrt {d+e\,x}\,\left (B\,a^3\,e^3-2\,B\,a^2\,b\,d\,e^2-A\,a^2\,b\,e^3+B\,a\,b^2\,d^2\,e+2\,A\,a\,b^2\,d\,e^2-A\,b^3\,d^2\,e\right )}{b^5\,\left (d+e\,x\right )-b^5\,d+a\,b^4\,e}+\frac {2\,B\,{\left (d+e\,x\right )}^{5/2}}{5\,b^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,{\left (a\,e-b\,d\right )}^{3/2}\,\sqrt {d+e\,x}\,\left (5\,A\,b\,e-7\,B\,a\,e+2\,B\,b\,d\right )}{-7\,B\,a^3\,e^3+16\,B\,a^2\,b\,d\,e^2+5\,A\,a^2\,b\,e^3-11\,B\,a\,b^2\,d^2\,e-10\,A\,a\,b^2\,d\,e^2+2\,B\,b^3\,d^3+5\,A\,b^3\,d^2\,e}\right )\,{\left (a\,e-b\,d\right )}^{3/2}\,\left (5\,A\,b\,e-7\,B\,a\,e+2\,B\,b\,d\right )}{b^{9/2}} \end {gather*}
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 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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