3.1732 \(\int \frac{A+B x}{(a+b x) (d+e x)^{7/2}} \, dx\)

Optimal. Leaf size=151 \[ -\frac{2 b^{3/2} (A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{(b d-a e)^{7/2}}+\frac{2 b (A b-a B)}{\sqrt{d+e x} (b d-a e)^3}+\frac{2 (A b-a B)}{3 (d+e x)^{3/2} (b d-a e)^2}-\frac{2 (B d-A e)}{5 e (d+e x)^{5/2} (b d-a e)} \]

[Out]

(-2*(B*d - A*e))/(5*e*(b*d - a*e)*(d + e*x)^(5/2)) + (2*(A*b - a*B))/(3*(b*d - a
*e)^2*(d + e*x)^(3/2)) + (2*b*(A*b - a*B))/((b*d - a*e)^3*Sqrt[d + e*x]) - (2*b^
(3/2)*(A*b - a*B)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(b*d - a*e)^
(7/2)

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Rubi [A]  time = 0.272204, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182 \[ -\frac{2 b^{3/2} (A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{(b d-a e)^{7/2}}+\frac{2 b (A b-a B)}{\sqrt{d+e x} (b d-a e)^3}+\frac{2 (A b-a B)}{3 (d+e x)^{3/2} (b d-a e)^2}-\frac{2 (B d-A e)}{5 e (d+e x)^{5/2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]  Int[(A + B*x)/((a + b*x)*(d + e*x)^(7/2)),x]

[Out]

(-2*(B*d - A*e))/(5*e*(b*d - a*e)*(d + e*x)^(5/2)) + (2*(A*b - a*B))/(3*(b*d - a
*e)^2*(d + e*x)^(3/2)) + (2*b*(A*b - a*B))/((b*d - a*e)^3*Sqrt[d + e*x]) - (2*b^
(3/2)*(A*b - a*B)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(b*d - a*e)^
(7/2)

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Rubi in Sympy [A]  time = 30.8357, size = 133, normalized size = 0.88 \[ - \frac{2 b^{\frac{3}{2}} \left (A b - B a\right ) \operatorname{atan}{\left (\frac{\sqrt{b} \sqrt{d + e x}}{\sqrt{a e - b d}} \right )}}{\left (a e - b d\right )^{\frac{7}{2}}} - \frac{2 b \left (A b - B a\right )}{\sqrt{d + e x} \left (a e - b d\right )^{3}} + \frac{2 \left (A b - B a\right )}{3 \left (d + e x\right )^{\frac{3}{2}} \left (a e - b d\right )^{2}} - \frac{2 \left (A e - B d\right )}{5 e \left (d + e x\right )^{\frac{5}{2}} \left (a e - b d\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((B*x+A)/(b*x+a)/(e*x+d)**(7/2),x)

[Out]

-2*b**(3/2)*(A*b - B*a)*atan(sqrt(b)*sqrt(d + e*x)/sqrt(a*e - b*d))/(a*e - b*d)*
*(7/2) - 2*b*(A*b - B*a)/(sqrt(d + e*x)*(a*e - b*d)**3) + 2*(A*b - B*a)/(3*(d +
e*x)**(3/2)*(a*e - b*d)**2) - 2*(A*e - B*d)/(5*e*(d + e*x)**(5/2)*(a*e - b*d))

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Mathematica [A]  time = 0.427198, size = 151, normalized size = 1. \[ -\frac{2 b^{3/2} (A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{(b d-a e)^{7/2}}+\frac{2 b (A b-a B)}{\sqrt{d+e x} (b d-a e)^3}+\frac{2 (A b-a B)}{3 (d+e x)^{3/2} (b d-a e)^2}-\frac{2 (A e-B d)}{5 e (d+e x)^{5/2} (a e-b d)} \]

Antiderivative was successfully verified.

[In]  Integrate[(A + B*x)/((a + b*x)*(d + e*x)^(7/2)),x]

[Out]

(-2*(-(B*d) + A*e))/(5*e*(-(b*d) + a*e)*(d + e*x)^(5/2)) + (2*(A*b - a*B))/(3*(b
*d - a*e)^2*(d + e*x)^(3/2)) + (2*b*(A*b - a*B))/((b*d - a*e)^3*Sqrt[d + e*x]) -
(2*b^(3/2)*(A*b - a*B)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(b*d -
a*e)^(7/2)

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Maple [A]  time = 0.02, size = 234, normalized size = 1.6 \[ -{\frac{2\,A}{5\,ae-5\,bd} \left ( ex+d \right ) ^{-{\frac{5}{2}}}}+{\frac{2\,Bd}{5\,e \left ( ae-bd \right ) } \left ( ex+d \right ) ^{-{\frac{5}{2}}}}-2\,{\frac{{b}^{2}A}{ \left ( ae-bd \right ) ^{3}\sqrt{ex+d}}}+2\,{\frac{Bba}{ \left ( ae-bd \right ) ^{3}\sqrt{ex+d}}}+{\frac{2\,Ab}{3\, \left ( ae-bd \right ) ^{2}} \left ( ex+d \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,Ba}{3\, \left ( ae-bd \right ) ^{2}} \left ( ex+d \right ) ^{-{\frac{3}{2}}}}-2\,{\frac{{b}^{3}A}{ \left ( ae-bd \right ) ^{3}\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) }+2\,{\frac{{b}^{2}Ba}{ \left ( ae-bd \right ) ^{3}\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) } \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((B*x+A)/(b*x+a)/(e*x+d)^(7/2),x)

[Out]

-2/5/(a*e-b*d)/(e*x+d)^(5/2)*A+2/5/e/(a*e-b*d)/(e*x+d)^(5/2)*B*d-2/(a*e-b*d)^3*b
^2/(e*x+d)^(1/2)*A+2/(a*e-b*d)^3*b/(e*x+d)^(1/2)*B*a+2/3/(a*e-b*d)^2/(e*x+d)^(3/
2)*A*b-2/3/(a*e-b*d)^2/(e*x+d)^(3/2)*B*a-2*b^3/(a*e-b*d)^3/((a*e-b*d)*b)^(1/2)*a
rctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*A+2*b^2/(a*e-b*d)^3/((a*e-b*d)*b)^(1/
2)*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*B*a

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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*x + a)*(e*x + d)^(7/2)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.231961, size = 1, normalized size = 0.01 \[ \left [-\frac{6 \, B b^{2} d^{3} - 6 \, A a^{2} e^{3} + 30 \,{\left (B a b - A b^{2}\right )} e^{3} x^{2} + 2 \,{\left (14 \, B a b - 23 \, A b^{2}\right )} d^{2} e - 2 \,{\left (2 \, B a^{2} - 11 \, A a b\right )} d e^{2} - 15 \,{\left ({\left (B a b - A b^{2}\right )} e^{3} x^{2} + 2 \,{\left (B a b - A b^{2}\right )} d e^{2} x +{\left (B a b - A b^{2}\right )} d^{2} e\right )} \sqrt{e x + d} \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 ) + 10 \,{\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}{15 \,{\left (b^{3} d^{5} e - 3 \, a b^{2} d^{4} e^{2} + 3 \, a^{2} b d^{3} e^{3} - a^{3} d^{2} e^{4} +{\left (b^{3} d^{3} e^{3} - 3 \, a b^{2} d^{2} e^{4} + 3 \, a^{2} b d e^{5} - a^{3} e^{6}\right )} x^{2} + 2 \,{\left (b^{3} d^{4} e^{2} - 3 \, a b^{2} d^{3} e^{3} + 3 \, a^{2} b d^{2} e^{4} - a^{3} d e^{5}\right )} x\right )} \sqrt{e x + d}}, -\frac{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} - 15 \,{\left ({\left (B a b - A b^{2}\right )} e^{3} x^{2} + 2 \,{\left (B a b - A b^{2}\right )} d e^{2} x +{\left (B a b - A b^{2}\right )} d^{2} e\right )} \sqrt{e x + d} \sqrt{-\frac{b}{b d - a e}} \arctan \left (-\frac{{\left (b d - a e\right )} \sqrt{-\frac{b}{b d - a e}}}{\sqrt{e x + d} b}\right ) + 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 )}}{15 \,{\left (b^{3} d^{5} e - 3 \, a b^{2} d^{4} e^{2} + 3 \, a^{2} b d^{3} e^{3} - a^{3} d^{2} e^{4} +{\left (b^{3} d^{3} e^{3} - 3 \, a b^{2} d^{2} e^{4} + 3 \, a^{2} b d e^{5} - a^{3} e^{6}\right )} x^{2} + 2 \,{\left (b^{3} d^{4} e^{2} - 3 \, a b^{2} d^{3} e^{3} + 3 \, a^{2} b d^{2} e^{4} - a^{3} d e^{5}\right )} x\right )} \sqrt{e x + d}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x + A)/((b*x + a)*(e*x + d)^(7/2)),x, algorithm="fricas")

[Out]

[-1/15*(6*B*b^2*d^3 - 6*A*a^2*e^3 + 30*(B*a*b - A*b^2)*e^3*x^2 + 2*(14*B*a*b - 2
3*A*b^2)*d^2*e - 2*(2*B*a^2 - 11*A*a*b)*d*e^2 - 15*((B*a*b - A*b^2)*e^3*x^2 + 2*
(B*a*b - A*b^2)*d*e^2*x + (B*a*b - A*b^2)*d^2*e)*sqrt(e*x + d)*sqrt(b/(b*d - a*e
))*log((b*e*x + 2*b*d - a*e + 2*(b*d - a*e)*sqrt(e*x + d)*sqrt(b/(b*d - a*e)))/(
b*x + a)) + 10*(7*(B*a*b - A*b^2)*d*e^2 - (B*a^2 - A*a*b)*e^3)*x)/((b^3*d^5*e -
3*a*b^2*d^4*e^2 + 3*a^2*b*d^3*e^3 - a^3*d^2*e^4 + (b^3*d^3*e^3 - 3*a*b^2*d^2*e^4
+ 3*a^2*b*d*e^5 - a^3*e^6)*x^2 + 2*(b^3*d^4*e^2 - 3*a*b^2*d^3*e^3 + 3*a^2*b*d^2
*e^4 - a^3*d*e^5)*x)*sqrt(e*x + d)), -2/15*(3*B*b^2*d^3 - 3*A*a^2*e^3 + 15*(B*a*
b - A*b^2)*e^3*x^2 + (14*B*a*b - 23*A*b^2)*d^2*e - (2*B*a^2 - 11*A*a*b)*d*e^2 -
15*((B*a*b - A*b^2)*e^3*x^2 + 2*(B*a*b - A*b^2)*d*e^2*x + (B*a*b - A*b^2)*d^2*e)
*sqrt(e*x + d)*sqrt(-b/(b*d - a*e))*arctan(-(b*d - a*e)*sqrt(-b/(b*d - a*e))/(sq
rt(e*x + d)*b)) + 5*(7*(B*a*b - A*b^2)*d*e^2 - (B*a^2 - A*a*b)*e^3)*x)/((b^3*d^5
*e - 3*a*b^2*d^4*e^2 + 3*a^2*b*d^3*e^3 - a^3*d^2*e^4 + (b^3*d^3*e^3 - 3*a*b^2*d^
2*e^4 + 3*a^2*b*d*e^5 - a^3*e^6)*x^2 + 2*(b^3*d^4*e^2 - 3*a*b^2*d^3*e^3 + 3*a^2*
b*d^2*e^4 - a^3*d*e^5)*x)*sqrt(e*x + d))]

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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)/(b*x+a)/(e*x+d)**(7/2),x)

[Out]

Timed out

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GIAC/XCAS [A]  time = 0.220733, size = 383, normalized size = 2.54 \[ -\frac{2 \,{\left (B a b^{2} - A b^{3}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \sqrt{-b^{2} d + a b e}} - \frac{2 \,{\left (3 \, B b^{2} d^{3} + 15 \,{\left (x e + d\right )}^{2} B a b e - 15 \,{\left (x e + d\right )}^{2} A b^{2} e + 5 \,{\left (x e + d\right )} B a b d e - 5 \,{\left (x e + d\right )} A b^{2} d e - 6 \, B a b d^{2} e - 3 \, A b^{2} d^{2} e - 5 \,{\left (x e + d\right )} B a^{2} e^{2} + 5 \,{\left (x e + d\right )} A a b e^{2} + 3 \, B a^{2} d e^{2} + 6 \, A a b d e^{2} - 3 \, A a^{2} e^{3}\right )}}{15 \,{\left (b^{3} d^{3} e - 3 \, a b^{2} d^{2} e^{2} + 3 \, a^{2} b d e^{3} - a^{3} e^{4}\right )}{\left (x e + d\right )}^{\frac{5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x + A)/((b*x + a)*(e*x + d)^(7/2)),x, algorithm="giac")

[Out]

-2*(B*a*b^2 - A*b^3)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/((b^3*d^3 - 3*
a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*sqrt(-b^2*d + a*b*e)) - 2/15*(3*B*b^2*d^3
+ 15*(x*e + d)^2*B*a*b*e - 15*(x*e + d)^2*A*b^2*e + 5*(x*e + d)*B*a*b*d*e - 5*(
x*e + d)*A*b^2*d*e - 6*B*a*b*d^2*e - 3*A*b^2*d^2*e - 5*(x*e + d)*B*a^2*e^2 + 5*(
x*e + d)*A*a*b*e^2 + 3*B*a^2*d*e^2 + 6*A*a*b*d*e^2 - 3*A*a^2*e^3)/((b^3*d^3*e -
3*a*b^2*d^2*e^2 + 3*a^2*b*d*e^3 - a^3*e^4)*(x*e + d)^(5/2))