Optimal. Leaf size=251 \[ \frac {2 b (a+b x) (A b-a B)}{\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^3}+\frac {2 (a+b x) (A b-a B)}{3 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^2}-\frac {2 (a+b x) (B d-A e)}{5 e \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)}-\frac {2 b^{3/2} (a+b x) (A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{7/2}} \]
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Rubi [A] time = 0.16, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {770, 78, 51, 63, 208} \begin {gather*} \frac {2 b (a+b x) (A b-a B)}{\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^3}+\frac {2 (a+b x) (A b-a B)}{3 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^2}-\frac {2 (a+b x) (B d-A e)}{5 e \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)}-\frac {2 b^{3/2} (a+b x) (A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{7/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 51
Rule 63
Rule 78
Rule 208
Rule 770
Rubi steps
\begin {align*} \int \frac {A+B x}{(d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx &=\frac {\left (a b+b^2 x\right ) \int \frac {A+B x}{\left (a b+b^2 x\right ) (d+e x)^{7/2}} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {2 (B d-A e) (a+b x)}{5 e (b d-a e) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left ((A b-a B) \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) (d+e x)^{5/2}} \, dx}{(b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {2 (B d-A e) (a+b x)}{5 e (b d-a e) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) (a+b x)}{3 (b d-a e)^2 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (b (A b-a B) \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) (d+e x)^{3/2}} \, dx}{(b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {2 (B d-A e) (a+b x)}{5 e (b d-a e) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) (a+b x)}{3 (b d-a e)^2 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 b (A b-a B) (a+b x)}{(b d-a e)^3 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (b^2 (A b-a B) \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) \sqrt {d+e x}} \, dx}{(b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {2 (B d-A e) (a+b x)}{5 e (b d-a e) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) (a+b x)}{3 (b d-a e)^2 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 b (A b-a B) (a+b x)}{(b d-a e)^3 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (2 b^2 (A b-a B) \left (a b+b^2 x\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a b-\frac {b^2 d}{e}+\frac {b^2 x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{e (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {2 (B d-A e) (a+b x)}{5 e (b d-a e) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) (a+b x)}{3 (b d-a e)^2 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 b (A b-a B) (a+b x)}{(b d-a e)^3 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 b^{3/2} (A b-a B) (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}
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Mathematica [C] time = 0.05, size = 102, normalized size = 0.41 \begin {gather*} \frac {2 (a+b x) \left (5 e (d+e x) (A b-a B) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {b (d+e x)}{b d-a e}\right )-3 (b d-a e) (B d-A e)\right )}{15 e \sqrt {(a+b x)^2} (d+e x)^{5/2} (b d-a e)^2} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 60.56, size = 264, normalized size = 1.05 \begin {gather*} \frac {(-a e-b e x) \left (\frac {2 \left (3 a^2 A e^3+5 a^2 B e^2 (d+e x)-3 a^2 B d e^2-5 a A b e^2 (d+e x)-6 a A b d e^2+6 a b B d^2 e-5 a b B d e (d+e x)-15 a b B e (d+e x)^2+3 A b^2 d^2 e+5 A b^2 d e (d+e x)+15 A b^2 e (d+e x)^2-3 b^2 B d^3\right )}{15 e (d+e x)^{5/2} (a e-b d)^3}-\frac {2 \left (A b^{5/2}-a b^{3/2} B\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{(a e-b d)^{7/2}}\right )}{e \sqrt {\frac {(a e+b e x)^2}{e^2}}} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.49, size = 902, normalized size = 3.59 \begin {gather*} \left [\frac {15 \, {\left ({\left (B a b - A b^{2}\right )} e^{4} x^{3} + 3 \, {\left (B a b - A b^{2}\right )} d e^{3} x^{2} + 3 \, {\left (B a b - A b^{2}\right )} d^{2} e^{2} x + {\left (B a b - A b^{2}\right )} d^{3} e\right )} \sqrt {\frac {b}{b d - a e}} \log \left (\frac {b e x + 2 \, b d - a e + 2 \, {\left (b d - a e\right )} \sqrt {e x + d} \sqrt {\frac {b}{b d - a e}}}{b x + a}\right ) - 2 \, {\left (3 \, B b^{2} d^{3} - 3 \, A a^{2} e^{3} + 15 \, {\left (B a b - A b^{2}\right )} e^{3} x^{2} + {\left (14 \, B a b - 23 \, A b^{2}\right )} d^{2} e - {\left (2 \, B a^{2} - 11 \, A a b\right )} d e^{2} + 5 \, {\left (7 \, {\left (B a b - A b^{2}\right )} d e^{2} - {\left (B a^{2} - A a b\right )} e^{3}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (b^{3} d^{6} e - 3 \, a b^{2} d^{5} e^{2} + 3 \, a^{2} b d^{4} e^{3} - a^{3} d^{3} e^{4} + {\left (b^{3} d^{3} e^{4} - 3 \, a b^{2} d^{2} e^{5} + 3 \, a^{2} b d e^{6} - a^{3} e^{7}\right )} x^{3} + 3 \, {\left (b^{3} d^{4} e^{3} - 3 \, a b^{2} d^{3} e^{4} + 3 \, a^{2} b d^{2} e^{5} - a^{3} d e^{6}\right )} x^{2} + 3 \, {\left (b^{3} d^{5} e^{2} - 3 \, a b^{2} d^{4} e^{3} + 3 \, a^{2} b d^{3} e^{4} - a^{3} d^{2} e^{5}\right )} x\right )}}, \frac {2 \, {\left (15 \, {\left ({\left (B a b - A b^{2}\right )} e^{4} x^{3} + 3 \, {\left (B a b - A b^{2}\right )} d e^{3} x^{2} + 3 \, {\left (B a b - A b^{2}\right )} d^{2} e^{2} x + {\left (B a b - A b^{2}\right )} d^{3} e\right )} \sqrt {-\frac {b}{b d - a e}} \arctan \left (-\frac {{\left (b d - a e\right )} \sqrt {e x + d} \sqrt {-\frac {b}{b d - a e}}}{b e x + b d}\right ) - {\left (3 \, B b^{2} d^{3} - 3 \, A a^{2} e^{3} + 15 \, {\left (B a b - A b^{2}\right )} e^{3} x^{2} + {\left (14 \, B a b - 23 \, A b^{2}\right )} d^{2} e - {\left (2 \, B a^{2} - 11 \, A a b\right )} d e^{2} + 5 \, {\left (7 \, {\left (B a b - A b^{2}\right )} d e^{2} - {\left (B a^{2} - A a b\right )} e^{3}\right )} x\right )} \sqrt {e x + d}\right )}}{15 \, {\left (b^{3} d^{6} e - 3 \, a b^{2} d^{5} e^{2} + 3 \, a^{2} b d^{4} e^{3} - a^{3} d^{3} e^{4} + {\left (b^{3} d^{3} e^{4} - 3 \, a b^{2} d^{2} e^{5} + 3 \, a^{2} b d e^{6} - a^{3} e^{7}\right )} x^{3} + 3 \, {\left (b^{3} d^{4} e^{3} - 3 \, a b^{2} d^{3} e^{4} + 3 \, a^{2} b d^{2} e^{5} - a^{3} d e^{6}\right )} x^{2} + 3 \, {\left (b^{3} d^{5} e^{2} - 3 \, a b^{2} d^{4} e^{3} + 3 \, a^{2} b d^{3} e^{4} - a^{3} d^{2} e^{5}\right )} x\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.27, size = 368, normalized size = 1.47 \begin {gather*} -\frac {2 \, {\left (B a b^{2} \mathrm {sgn}\left (b x + a\right ) - A b^{3} \mathrm {sgn}\left (b x + a\right )\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \sqrt {-b^{2} d + a b e}} - \frac {2 \, {\left (3 \, B b^{2} d^{3} \mathrm {sgn}\left (b x + a\right ) + 15 \, {\left (x e + d\right )}^{2} B a b e \mathrm {sgn}\left (b x + a\right ) - 15 \, {\left (x e + d\right )}^{2} A b^{2} e \mathrm {sgn}\left (b x + a\right ) + 5 \, {\left (x e + d\right )} B a b d e \mathrm {sgn}\left (b x + a\right ) - 5 \, {\left (x e + d\right )} A b^{2} d e \mathrm {sgn}\left (b x + a\right ) - 6 \, B a b d^{2} e \mathrm {sgn}\left (b x + a\right ) - 3 \, A b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) - 5 \, {\left (x e + d\right )} B a^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, {\left (x e + d\right )} A a b e^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, B a^{2} d e^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, A a b d e^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, A a^{2} e^{3} \mathrm {sgn}\left (b x + a\right )\right )}}{15 \, {\left (b^{3} d^{3} e - 3 \, a b^{2} d^{2} e^{2} + 3 \, a^{2} b d e^{3} - a^{3} e^{4}\right )} {\left (x e + d\right )}^{\frac {5}{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.08, size = 386, normalized size = 1.54 \begin {gather*} -\frac {2 \left (b x +a \right ) \left (15 \sqrt {\left (a e -b d \right ) b}\, A \,b^{2} e^{3} x^{2}-15 \sqrt {\left (a e -b d \right ) b}\, B a b \,e^{3} x^{2}-5 \sqrt {\left (a e -b d \right ) b}\, A a b \,e^{3} x +35 \sqrt {\left (a e -b d \right ) b}\, A \,b^{2} d \,e^{2} x +5 \sqrt {\left (a e -b d \right ) b}\, B \,a^{2} e^{3} x -35 \sqrt {\left (a e -b d \right ) b}\, B a b d \,e^{2} x +3 \sqrt {\left (a e -b d \right ) b}\, A \,a^{2} e^{3}-11 \sqrt {\left (a e -b d \right ) b}\, A a b d \,e^{2}+15 \left (e x +d \right )^{\frac {5}{2}} A \,b^{3} e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+23 \sqrt {\left (a e -b d \right ) b}\, A \,b^{2} d^{2} e +2 \sqrt {\left (a e -b d \right ) b}\, B \,a^{2} d \,e^{2}-15 \left (e x +d \right )^{\frac {5}{2}} B a \,b^{2} e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-14 \sqrt {\left (a e -b d \right ) b}\, B a b \,d^{2} e -3 \sqrt {\left (a e -b d \right ) b}\, B \,b^{2} d^{3}\right )}{15 \sqrt {\left (b x +a \right )^{2}}\, \left (a e -b d \right )^{3} \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {5}{2}} e} \end {gather*}
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 \begin {gather*} \int \frac {B x + A}{\sqrt {{\left (b x + a\right )}^{2}} {\left (e x + d\right )}^{\frac {7}{2}}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {A+B\,x}{\sqrt {{\left (a+b\,x\right )}^2}\,{\left (d+e\,x\right )}^{7/2}} \,d x \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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