Optimal. Leaf size=414 \[ -\frac {7 b e (a+b x) (5 a B e-9 A b e+4 b B d)}{4 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^5}-\frac {7 e (a+b x) (5 a B e-9 A b e+4 b B d)}{12 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^4}+\frac {-5 a B e+9 A b e-4 b B d}{4 b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)^2}-\frac {7 e (a+b x) (5 a B e-9 A b e+4 b B d)}{20 b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)^3}-\frac {A b-a B}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)}+\frac {7 b^{3/2} e (a+b x) (5 a B e-9 A b e+4 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{11/2}} \]
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Rubi [A] time = 0.37, antiderivative size = 414, normalized size of antiderivative = 1.00, number of steps used = 8, 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 {7 b e (a+b x) (5 a B e-9 A b e+4 b B d)}{4 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^5}-\frac {7 e (a+b x) (5 a B e-9 A b e+4 b B d)}{12 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^4}-\frac {5 a B e-9 A b e+4 b B d}{4 b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)^2}-\frac {7 e (a+b x) (5 a B e-9 A b e+4 b B d)}{20 b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)^3}-\frac {A b-a B}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)}+\frac {7 b^{3/2} e (a+b x) (5 a B e-9 A b e+4 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{11/2}} \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} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac {A+B x}{\left (a b+b^2 x\right )^3 (d+e x)^{7/2}} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {A b-a B}{2 b (b d-a e) (a+b x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left ((4 b B d-9 A b e+5 a B e) \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right )^2 (d+e x)^{7/2}} \, dx}{4 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {4 b B d-9 A b e+5 a B e}{4 b (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A b-a B}{2 b (b d-a e) (a+b x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (7 e (4 b B d-9 A b e+5 a B e) \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) (d+e x)^{7/2}} \, dx}{8 b (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {4 b B d-9 A b e+5 a B e}{4 b (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A b-a B}{2 b (b d-a e) (a+b x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (4 b B d-9 A b e+5 a B e) (a+b x)}{20 b (b d-a e)^3 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (7 e (4 b B d-9 A b e+5 a B e) \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) (d+e x)^{5/2}} \, dx}{8 (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {4 b B d-9 A b e+5 a B e}{4 b (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A b-a B}{2 b (b d-a e) (a+b x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (4 b B d-9 A b e+5 a B e) (a+b x)}{20 b (b d-a e)^3 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (4 b B d-9 A b e+5 a B e) (a+b x)}{12 (b d-a e)^4 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (7 b e (4 b B d-9 A b e+5 a B e) \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) (d+e x)^{3/2}} \, dx}{8 (b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {4 b B d-9 A b e+5 a B e}{4 b (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A b-a B}{2 b (b d-a e) (a+b x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (4 b B d-9 A b e+5 a B e) (a+b x)}{20 b (b d-a e)^3 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (4 b B d-9 A b e+5 a B e) (a+b x)}{12 (b d-a e)^4 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 b e (4 b B d-9 A b e+5 a B e) (a+b x)}{4 (b d-a e)^5 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (7 b^2 e (4 b B d-9 A b e+5 a B e) \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) \sqrt {d+e x}} \, dx}{8 (b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {4 b B d-9 A b e+5 a B e}{4 b (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A b-a B}{2 b (b d-a e) (a+b x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (4 b B d-9 A b e+5 a B e) (a+b x)}{20 b (b d-a e)^3 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (4 b B d-9 A b e+5 a B e) (a+b x)}{12 (b d-a e)^4 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 b e (4 b B d-9 A b e+5 a B e) (a+b x)}{4 (b d-a e)^5 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (7 b^2 (4 b B d-9 A b e+5 a B e) \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 )}{4 (b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {4 b B d-9 A b e+5 a B e}{4 b (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A b-a B}{2 b (b d-a e) (a+b x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (4 b B d-9 A b e+5 a B e) (a+b x)}{20 b (b d-a e)^3 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (4 b B d-9 A b e+5 a B e) (a+b x)}{12 (b d-a e)^4 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 b e (4 b B d-9 A b e+5 a B e) (a+b x)}{4 (b d-a e)^5 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 b^{3/2} e (4 b B d-9 A b e+5 a B e) (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 (b d-a e)^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}
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Mathematica [C] time = 0.07, size = 111, normalized size = 0.27 \begin {gather*} \frac {(a+b x) \left (\frac {e (a+b x)^2 (-5 a B e+9 A b e-4 b B d) \, _2F_1\left (-\frac {5}{2},2;-\frac {3}{2};\frac {b (d+e x)}{b d-a e}\right )}{(b d-a e)^2}+5 a B-5 A b\right )}{10 b \left ((a+b x)^2\right )^{3/2} (d+e x)^{5/2} (b d-a e)} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 70.06, size = 687, normalized size = 1.66 \begin {gather*} \frac {(-a e-b e x) \left (-\frac {e \left (24 a^4 A e^5+40 a^4 B e^4 (d+e x)-24 a^4 B d e^4-72 a^3 A b e^4 (d+e x)-96 a^3 A b d e^4+96 a^3 b B d^2 e^3-88 a^3 b B d e^3 (d+e x)-280 a^3 b B e^3 (d+e x)^2+144 a^2 A b^2 d^2 e^3+216 a^2 A b^2 d e^3 (d+e x)+504 a^2 A b^2 e^3 (d+e x)^2-144 a^2 b^2 B d^3 e^2+24 a^2 b^2 B d^2 e^2 (d+e x)+336 a^2 b^2 B d e^2 (d+e x)^2-875 a^2 b^2 B e^2 (d+e x)^3-96 a A b^3 d^3 e^2-216 a A b^3 d^2 e^2 (d+e x)-1008 a A b^3 d e^2 (d+e x)^2+1575 a A b^3 e^2 (d+e x)^3+96 a b^3 B d^4 e+56 a b^3 B d^3 e (d+e x)+168 a b^3 B d^2 e (d+e x)^2+175 a b^3 B d e (d+e x)^3-525 a b^3 B e (d+e x)^4+24 A b^4 d^4 e+72 A b^4 d^3 e (d+e x)+504 A b^4 d^2 e (d+e x)^2-1575 A b^4 d e (d+e x)^3+945 A b^4 e (d+e x)^4-24 b^4 B d^5-32 b^4 B d^4 (d+e x)-224 b^4 B d^3 (d+e x)^2+700 b^4 B d^2 (d+e x)^3-420 b^4 B d (d+e x)^4\right )}{60 (d+e x)^{5/2} (b d-a e)^5 (-a e-b (d+e x)+b d)^2}-\frac {7 \left (5 a b^{3/2} B e^2-9 A b^{5/2} e^2+4 b^{5/2} B d e\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{4 (b d-a e)^5 \sqrt {a e-b d}}\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 = 2675, normalized size = 6.46
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.57, size = 1011, normalized size = 2.44 \begin {gather*} -\frac {7 \, {\left (4 \, B b^{3} d e^{2} + 5 \, B a b^{2} e^{3} - 9 \, A b^{3} e^{3}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{4 \, {\left (b^{5} d^{5} e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 5 \, a b^{4} d^{4} e^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 10 \, a^{2} b^{3} d^{3} e^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 10 \, a^{3} b^{2} d^{2} e^{4} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 5 \, a^{4} b d e^{5} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a^{5} e^{6} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )} \sqrt {-b^{2} d + a b e}} - \frac {4 \, {\left (x e + d\right )}^{\frac {3}{2}} B b^{4} d e^{2} - 4 \, \sqrt {x e + d} B b^{4} d^{2} e^{2} + 11 \, {\left (x e + d\right )}^{\frac {3}{2}} B a b^{3} e^{3} - 15 \, {\left (x e + d\right )}^{\frac {3}{2}} A b^{4} e^{3} - 9 \, \sqrt {x e + d} B a b^{3} d e^{3} + 17 \, \sqrt {x e + d} A b^{4} d e^{3} + 13 \, \sqrt {x e + d} B a^{2} b^{2} e^{4} - 17 \, \sqrt {x e + d} A a b^{3} e^{4}}{4 \, {\left (b^{5} d^{5} e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 5 \, a b^{4} d^{4} e^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 10 \, a^{2} b^{3} d^{3} e^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 10 \, a^{3} b^{2} d^{2} e^{4} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 5 \, a^{4} b d e^{5} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a^{5} e^{6} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )} {\left ({\left (x e + d\right )} b - b d + a e\right )}^{2}} - \frac {2 \, {\left (45 \, {\left (x e + d\right )}^{2} B b^{2} d e^{2} + 10 \, {\left (x e + d\right )} B b^{2} d^{2} e^{2} + 3 \, B b^{2} d^{3} e^{2} + 45 \, {\left (x e + d\right )}^{2} B a b e^{3} - 90 \, {\left (x e + d\right )}^{2} A b^{2} e^{3} - 5 \, {\left (x e + d\right )} B a b d e^{3} - 15 \, {\left (x e + d\right )} A b^{2} d e^{3} - 6 \, B a b d^{2} e^{3} - 3 \, A b^{2} d^{2} e^{3} - 5 \, {\left (x e + d\right )} B a^{2} e^{4} + 15 \, {\left (x e + d\right )} A a b e^{4} + 3 \, B a^{2} d e^{4} + 6 \, A a b d e^{4} - 3 \, A a^{2} e^{5}\right )}}{15 \, {\left (b^{5} d^{5} e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 5 \, a b^{4} d^{4} e^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 10 \, a^{2} b^{3} d^{3} e^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 10 \, a^{3} b^{2} d^{2} e^{4} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 5 \, a^{4} b d e^{5} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a^{5} e^{6} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\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 = 1230, normalized size = 2.97
result too large to display
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}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{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}{{\left (d+e\,x\right )}^{7/2}\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,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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