3.2199 \(\int \frac{(a+b x)^{3/2} (A+B x)}{(d+e x)^{5/2}} \, dx\)

Optimal. Leaf size=202 \[ -\frac{\sqrt{b} (-3 a B e-2 A b e+5 b B d) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{e^{7/2}}+\frac{b \sqrt{a+b x} \sqrt{d+e x} (-3 a B e-2 A b e+5 b B d)}{e^3 (b d-a e)}-\frac{2 (a+b x)^{3/2} (-3 a B e-2 A b e+5 b B d)}{3 e^2 \sqrt{d+e x} (b d-a e)}-\frac{2 (a+b x)^{5/2} (B d-A e)}{3 e (d+e x)^{3/2} (b d-a e)} \]

[Out]

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

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Rubi [A]  time = 0.383194, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208 \[ -\frac{\sqrt{b} (-3 a B e-2 A b e+5 b B d) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{e^{7/2}}+\frac{b \sqrt{a+b x} \sqrt{d+e x} (-3 a B e-2 A b e+5 b B d)}{e^3 (b d-a e)}-\frac{2 (a+b x)^{3/2} (-3 a B e-2 A b e+5 b B d)}{3 e^2 \sqrt{d+e x} (b d-a e)}-\frac{2 (a+b x)^{5/2} (B d-A e)}{3 e (d+e x)^{3/2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]  Int[((a + b*x)^(3/2)*(A + B*x))/(d + e*x)^(5/2),x]

[Out]

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

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Rubi in Sympy [A]  time = 35.9896, size = 192, normalized size = 0.95 \[ \frac{2 \sqrt{b} \left (A b e + \frac{B \left (3 a e - 5 b d\right )}{2}\right ) \operatorname{atanh}{\left (\frac{\sqrt{b} \sqrt{d + e x}}{\sqrt{e} \sqrt{a + b x}} \right )}}{e^{\frac{7}{2}}} + \frac{b \sqrt{a + b x} \sqrt{d + e x} \left (2 A b e + 3 B a e - 5 B b d\right )}{e^{3} \left (a e - b d\right )} - \frac{2 \left (a + b x\right )^{\frac{5}{2}} \left (A e - B d\right )}{3 e \left (d + e x\right )^{\frac{3}{2}} \left (a e - b d\right )} - \frac{4 \left (a + b x\right )^{\frac{3}{2}} \left (A b e + \frac{B \left (3 a e - 5 b d\right )}{2}\right )}{3 e^{2} \sqrt{d + e x} \left (a e - b d\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((b*x+a)**(3/2)*(B*x+A)/(e*x+d)**(5/2),x)

[Out]

2*sqrt(b)*(A*b*e + B*(3*a*e - 5*b*d)/2)*atanh(sqrt(b)*sqrt(d + e*x)/(sqrt(e)*sqr
t(a + b*x)))/e**(7/2) + b*sqrt(a + b*x)*sqrt(d + e*x)*(2*A*b*e + 3*B*a*e - 5*B*b
*d)/(e**3*(a*e - b*d)) - 2*(a + b*x)**(5/2)*(A*e - B*d)/(3*e*(d + e*x)**(3/2)*(a
*e - b*d)) - 4*(a + b*x)**(3/2)*(A*b*e + B*(3*a*e - 5*b*d)/2)/(3*e**2*sqrt(d + e
*x)*(a*e - b*d))

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Mathematica [A]  time = 0.295369, size = 152, normalized size = 0.75 \[ \frac{\sqrt{a+b x} \left (-2 a e (A e+2 B d+3 B e x)-2 A b e (3 d+4 e x)+b B \left (15 d^2+20 d e x+3 e^2 x^2\right )\right )}{3 e^3 (d+e x)^{3/2}}+\frac{\sqrt{b} (3 a B e+2 A b e-5 b B d) \log \left (2 \sqrt{b} \sqrt{e} \sqrt{a+b x} \sqrt{d+e x}+a e+b d+2 b e x\right )}{2 e^{7/2}} \]

Antiderivative was successfully verified.

[In]  Integrate[((a + b*x)^(3/2)*(A + B*x))/(d + e*x)^(5/2),x]

[Out]

(Sqrt[a + b*x]*(-2*A*b*e*(3*d + 4*e*x) - 2*a*e*(2*B*d + A*e + 3*B*e*x) + b*B*(15
*d^2 + 20*d*e*x + 3*e^2*x^2)))/(3*e^3*(d + e*x)^(3/2)) + (Sqrt[b]*(-5*b*B*d + 2*
A*b*e + 3*a*B*e)*Log[b*d + a*e + 2*b*e*x + 2*Sqrt[b]*Sqrt[e]*Sqrt[a + b*x]*Sqrt[
d + e*x]])/(2*e^(7/2))

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Maple [B]  time = 0.034, size = 698, normalized size = 3.5 \[ \text{result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((b*x+a)^(3/2)*(B*x+A)/(e*x+d)^(5/2),x)

[Out]

1/6*(b*x+a)^(1/2)*(6*A*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e
+b*d)/(b*e)^(1/2))*x^2*b^2*e^3+9*B*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*
e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*x^2*a*b*e^3-15*B*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+
d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*x^2*b^2*d*e^2+12*A*ln(1/2*(2*b*x*e+2
*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*x*b^2*d*e^2+18*B*ln(1
/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*x*a*b*d*
e^2-30*B*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1
/2))*x*b^2*d^2*e+6*B*x^2*b*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+6*A*ln(1/2*(2
*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b^2*d^2*e-16*
A*x*b*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+9*B*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*
x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*b*d^2*e-15*B*ln(1/2*(2*b*x*e+2*(
(b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b^2*d^3-12*B*x*a*e^2*((
b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+40*B*x*b*d*e*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/
2)-4*A*a*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-12*A*b*d*e*((b*x+a)*(e*x+d))^(1
/2)*(b*e)^(1/2)-8*B*a*d*e*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+30*B*b*d^2*((b*x+a
)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2)/((b*x+a)*(e*x+d))^(1/2)/e^3/(e*x+d)^(3
/2)

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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)^(3/2)/(e*x + d)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.848446, size = 1, normalized size = 0. \[ \left [-\frac{3 \,{\left (5 \, B b d^{3} -{\left (3 \, B a + 2 \, A b\right )} d^{2} e +{\left (5 \, B b d e^{2} -{\left (3 \, B a + 2 \, A b\right )} e^{3}\right )} x^{2} + 2 \,{\left (5 \, B b d^{2} e -{\left (3 \, B a + 2 \, A b\right )} d e^{2}\right )} x\right )} \sqrt{\frac{b}{e}} \log \left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} + 4 \,{\left (2 \, b e^{2} x + b d e + a e^{2}\right )} \sqrt{b x + a} \sqrt{e x + d} \sqrt{\frac{b}{e}} + 8 \,{\left (b^{2} d e + a b e^{2}\right )} x\right ) - 4 \,{\left (3 \, B b e^{2} x^{2} + 15 \, B b d^{2} - 2 \, A a e^{2} - 2 \,{\left (2 \, B a + 3 \, A b\right )} d e + 2 \,{\left (10 \, B b d e -{\left (3 \, B a + 4 \, A b\right )} e^{2}\right )} x\right )} \sqrt{b x + a} \sqrt{e x + d}}{12 \,{\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}}, -\frac{3 \,{\left (5 \, B b d^{3} -{\left (3 \, B a + 2 \, A b\right )} d^{2} e +{\left (5 \, B b d e^{2} -{\left (3 \, B a + 2 \, A b\right )} e^{3}\right )} x^{2} + 2 \,{\left (5 \, B b d^{2} e -{\left (3 \, B a + 2 \, A b\right )} d e^{2}\right )} x\right )} \sqrt{-\frac{b}{e}} \arctan \left (\frac{2 \, b e x + b d + a e}{2 \, \sqrt{b x + a} \sqrt{e x + d} e \sqrt{-\frac{b}{e}}}\right ) - 2 \,{\left (3 \, B b e^{2} x^{2} + 15 \, B b d^{2} - 2 \, A a e^{2} - 2 \,{\left (2 \, B a + 3 \, A b\right )} d e + 2 \,{\left (10 \, B b d e -{\left (3 \, B a + 4 \, A b\right )} e^{2}\right )} x\right )} \sqrt{b x + a} \sqrt{e x + d}}{6 \,{\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x + A)*(b*x + a)^(3/2)/(e*x + d)^(5/2),x, algorithm="fricas")

[Out]

[-1/12*(3*(5*B*b*d^3 - (3*B*a + 2*A*b)*d^2*e + (5*B*b*d*e^2 - (3*B*a + 2*A*b)*e^
3)*x^2 + 2*(5*B*b*d^2*e - (3*B*a + 2*A*b)*d*e^2)*x)*sqrt(b/e)*log(8*b^2*e^2*x^2
+ b^2*d^2 + 6*a*b*d*e + a^2*e^2 + 4*(2*b*e^2*x + b*d*e + a*e^2)*sqrt(b*x + a)*sq
rt(e*x + d)*sqrt(b/e) + 8*(b^2*d*e + a*b*e^2)*x) - 4*(3*B*b*e^2*x^2 + 15*B*b*d^2
- 2*A*a*e^2 - 2*(2*B*a + 3*A*b)*d*e + 2*(10*B*b*d*e - (3*B*a + 4*A*b)*e^2)*x)*s
qrt(b*x + a)*sqrt(e*x + d))/(e^5*x^2 + 2*d*e^4*x + d^2*e^3), -1/6*(3*(5*B*b*d^3
- (3*B*a + 2*A*b)*d^2*e + (5*B*b*d*e^2 - (3*B*a + 2*A*b)*e^3)*x^2 + 2*(5*B*b*d^2
*e - (3*B*a + 2*A*b)*d*e^2)*x)*sqrt(-b/e)*arctan(1/2*(2*b*e*x + b*d + a*e)/(sqrt
(b*x + a)*sqrt(e*x + d)*e*sqrt(-b/e))) - 2*(3*B*b*e^2*x^2 + 15*B*b*d^2 - 2*A*a*e
^2 - 2*(2*B*a + 3*A*b)*d*e + 2*(10*B*b*d*e - (3*B*a + 4*A*b)*e^2)*x)*sqrt(b*x +
a)*sqrt(e*x + d))/(e^5*x^2 + 2*d*e^4*x + d^2*e^3)]

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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)**(3/2)*(B*x+A)/(e*x+d)**(5/2),x)

[Out]

Timed out

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GIAC/XCAS [A]  time = 0.265444, size = 475, normalized size = 2.35 \[ \frac{{\left (5 \, B b d{\left | b \right |} - 3 \, B a{\left | b \right |} e - 2 \, A b{\left | b \right |} e\right )} e^{\left (-\frac{7}{2}\right )}{\rm ln}\left ({\left | -\sqrt{b x + a} \sqrt{b} e^{\frac{1}{2}} + \sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e} \right |}\right )}{\sqrt{b}} + \frac{{\left ({\left (b x + a\right )}{\left (\frac{3 \,{\left (B b^{5} d{\left | b \right |} e^{4} - B a b^{4}{\left | b \right |} e^{5}\right )}{\left (b x + a\right )}}{b^{4} d e^{5} - a b^{3} e^{6}} + \frac{4 \,{\left (5 \, B b^{6} d^{2}{\left | b \right |} e^{3} - 8 \, B a b^{5} d{\left | b \right |} e^{4} - 2 \, A b^{6} d{\left | b \right |} e^{4} + 3 \, B a^{2} b^{4}{\left | b \right |} e^{5} + 2 \, A a b^{5}{\left | b \right |} e^{5}\right )}}{b^{4} d e^{5} - a b^{3} e^{6}}\right )} + \frac{3 \,{\left (5 \, B b^{7} d^{3}{\left | b \right |} e^{2} - 13 \, B a b^{6} d^{2}{\left | b \right |} e^{3} - 2 \, A b^{7} d^{2}{\left | b \right |} e^{3} + 11 \, B a^{2} b^{5} d{\left | b \right |} e^{4} + 4 \, A a b^{6} d{\left | b \right |} e^{4} - 3 \, B a^{3} b^{4}{\left | b \right |} e^{5} - 2 \, A a^{2} b^{5}{\left | b \right |} e^{5}\right )}}{b^{4} d e^{5} - a b^{3} e^{6}}\right )} \sqrt{b x + a}}{3 \,{\left (b^{2} d +{\left (b x + a\right )} b e - a b e\right )}^{\frac{3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x + A)*(b*x + a)^(3/2)/(e*x + d)^(5/2),x, algorithm="giac")

[Out]

(5*B*b*d*abs(b) - 3*B*a*abs(b)*e - 2*A*b*abs(b)*e)*e^(-7/2)*ln(abs(-sqrt(b*x + a
)*sqrt(b)*e^(1/2) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)))/sqrt(b) + 1/3*((b*x +
a)*(3*(B*b^5*d*abs(b)*e^4 - B*a*b^4*abs(b)*e^5)*(b*x + a)/(b^4*d*e^5 - a*b^3*e^6
) + 4*(5*B*b^6*d^2*abs(b)*e^3 - 8*B*a*b^5*d*abs(b)*e^4 - 2*A*b^6*d*abs(b)*e^4 +
3*B*a^2*b^4*abs(b)*e^5 + 2*A*a*b^5*abs(b)*e^5)/(b^4*d*e^5 - a*b^3*e^6)) + 3*(5*B
*b^7*d^3*abs(b)*e^2 - 13*B*a*b^6*d^2*abs(b)*e^3 - 2*A*b^7*d^2*abs(b)*e^3 + 11*B*
a^2*b^5*d*abs(b)*e^4 + 4*A*a*b^6*d*abs(b)*e^4 - 3*B*a^3*b^4*abs(b)*e^5 - 2*A*a^2
*b^5*abs(b)*e^5)/(b^4*d*e^5 - a*b^3*e^6))*sqrt(b*x + a)/(b^2*d + (b*x + a)*b*e -
a*b*e)^(3/2)