3.2197 \(\int \frac{(a+b x)^{3/2} (A+B x)}{\sqrt{d+e x}} \, dx\)

Optimal. Leaf size=193 \[ -\frac{(b d-a e)^2 (a B e-6 A b e+5 b B d) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{8 b^{3/2} e^{7/2}}+\frac{\sqrt{a+b x} \sqrt{d+e x} (b d-a e) (a B e-6 A b e+5 b B d)}{8 b e^3}-\frac{(a+b x)^{3/2} \sqrt{d+e x} (a B e-6 A b e+5 b B d)}{12 b e^2}+\frac{B (a+b x)^{5/2} \sqrt{d+e x}}{3 b e} \]

[Out]

((b*d - a*e)*(5*b*B*d - 6*A*b*e + a*B*e)*Sqrt[a + b*x]*Sqrt[d + e*x])/(8*b*e^3)
- ((5*b*B*d - 6*A*b*e + a*B*e)*(a + b*x)^(3/2)*Sqrt[d + e*x])/(12*b*e^2) + (B*(a
+ b*x)^(5/2)*Sqrt[d + e*x])/(3*b*e) - ((b*d - a*e)^2*(5*b*B*d - 6*A*b*e + a*B*e
)*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(8*b^(3/2)*e^(7/2))

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Rubi [A]  time = 0.385366, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167 \[ -\frac{(b d-a e)^2 (a B e-6 A b e+5 b B d) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{8 b^{3/2} e^{7/2}}+\frac{\sqrt{a+b x} \sqrt{d+e x} (b d-a e) (a B e-6 A b e+5 b B d)}{8 b e^3}-\frac{(a+b x)^{3/2} \sqrt{d+e x} (a B e-6 A b e+5 b B d)}{12 b e^2}+\frac{B (a+b x)^{5/2} \sqrt{d+e x}}{3 b e} \]

Antiderivative was successfully verified.

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

[Out]

((b*d - a*e)*(5*b*B*d - 6*A*b*e + a*B*e)*Sqrt[a + b*x]*Sqrt[d + e*x])/(8*b*e^3)
- ((5*b*B*d - 6*A*b*e + a*B*e)*(a + b*x)^(3/2)*Sqrt[d + e*x])/(12*b*e^2) + (B*(a
+ b*x)^(5/2)*Sqrt[d + e*x])/(3*b*e) - ((b*d - a*e)^2*(5*b*B*d - 6*A*b*e + a*B*e
)*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(8*b^(3/2)*e^(7/2))

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Rubi in Sympy [A]  time = 29.2749, size = 182, normalized size = 0.94 \[ \frac{B \left (a + b x\right )^{\frac{5}{2}} \sqrt{d + e x}}{3 b e} + \frac{\left (a + b x\right )^{\frac{3}{2}} \sqrt{d + e x} \left (6 A b e - B a e - 5 B b d\right )}{12 b e^{2}} + \frac{\sqrt{a + b x} \sqrt{d + e x} \left (a e - b d\right ) \left (6 A b e - B a e - 5 B b d\right )}{8 b e^{3}} + \frac{\left (a e - b d\right )^{2} \left (6 A b e - B a e - 5 B b d\right ) \operatorname{atanh}{\left (\frac{\sqrt{e} \sqrt{a + b x}}{\sqrt{b} \sqrt{d + e x}} \right )}}{8 b^{\frac{3}{2}} e^{\frac{7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((b*x+a)**(3/2)*(B*x+A)/(e*x+d)**(1/2),x)

[Out]

B*(a + b*x)**(5/2)*sqrt(d + e*x)/(3*b*e) + (a + b*x)**(3/2)*sqrt(d + e*x)*(6*A*b
*e - B*a*e - 5*B*b*d)/(12*b*e**2) + sqrt(a + b*x)*sqrt(d + e*x)*(a*e - b*d)*(6*A
*b*e - B*a*e - 5*B*b*d)/(8*b*e**3) + (a*e - b*d)**2*(6*A*b*e - B*a*e - 5*B*b*d)*
atanh(sqrt(e)*sqrt(a + b*x)/(sqrt(b)*sqrt(d + e*x)))/(8*b**(3/2)*e**(7/2))

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Mathematica [A]  time = 0.232473, size = 178, normalized size = 0.92 \[ \frac{\sqrt{a+b x} \sqrt{d+e x} \left (3 a^2 B e^2+2 a b e (15 A e-11 B d+7 B e x)+b^2 \left (6 A e (2 e x-3 d)+B \left (15 d^2-10 d e x+8 e^2 x^2\right )\right )\right )}{24 b e^3}-\frac{(b d-a e)^2 (a B e-6 A b e+5 b B d) \log \left (2 \sqrt{b} \sqrt{e} \sqrt{a+b x} \sqrt{d+e x}+a e+b d+2 b e x\right )}{16 b^{3/2} e^{7/2}} \]

Antiderivative was successfully verified.

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

[Out]

(Sqrt[a + b*x]*Sqrt[d + e*x]*(3*a^2*B*e^2 + 2*a*b*e*(-11*B*d + 15*A*e + 7*B*e*x)
+ b^2*(6*A*e*(-3*d + 2*e*x) + B*(15*d^2 - 10*d*e*x + 8*e^2*x^2))))/(24*b*e^3) -
((b*d - a*e)^2*(5*b*B*d - 6*A*b*e + 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]])/(16*b^(3/2)*e^(7/2))

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Maple [B]  time = 0.033, size = 636, normalized size = 3.3 \[{\frac{1}{48\,{e}^{3}b}\sqrt{bx+a}\sqrt{ex+d} \left ( 16\,B{x}^{2}{b}^{2}{e}^{2}\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+18\,\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){a}^{2}A{e}^{3}b-36\,\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ) aA{b}^{2}d{e}^{2}+18\,\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){b}^{3}{d}^{2}Ae+24\,A\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }x{b}^{2}{e}^{2}\sqrt{be}-3\,B{e}^{3}\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){a}^{3}-9\,\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){a}^{2}Bd{e}^{2}b+27\,\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ) aB{b}^{2}{d}^{2}e-15\,\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){b}^{3}{d}^{3}B+28\,B\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }xab{e}^{2}\sqrt{be}-20\,B\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }xd{b}^{2}e\sqrt{be}+60\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }Aa{e}^{2}\sqrt{be}b-36\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }A{b}^{2}de\sqrt{be}+6\,B\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }{a}^{2}{e}^{2}\sqrt{be}-44\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }Bade\sqrt{be}b+30\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }B{b}^{2}{d}^{2}\sqrt{be} \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }}}{\frac{1}{\sqrt{be}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((b*x+a)^(3/2)*(B*x+A)/(e*x+d)^(1/2),x)

[Out]

1/48*(b*x+a)^(1/2)*(e*x+d)^(1/2)*(16*B*x^2*b^2*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)
^(1/2)+18*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^2*A*e^3*b-36*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*A*b^2*d*e^2+18*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^3*d^2*A*e+24*A*((b*x+a)*(e*x+d))^(1/2)*x*b^2*e^2
*(b*e)^(1/2)-3*B*e^3*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^3-9*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^2*B*d*e^2*b+27*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*b^2*d^2*e-15*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^3*d^3*B+28*B*((b*x+a)*(e*x+d))^(1
/2)*x*a*b*e^2*(b*e)^(1/2)-20*B*((b*x+a)*(e*x+d))^(1/2)*x*d*b^2*e*(b*e)^(1/2)+60*
((b*x+a)*(e*x+d))^(1/2)*A*a*e^2*(b*e)^(1/2)*b-36*((b*x+a)*(e*x+d))^(1/2)*A*b^2*d
*e*(b*e)^(1/2)+6*B*((b*x+a)*(e*x+d))^(1/2)*a^2*e^2*(b*e)^(1/2)-44*((b*x+a)*(e*x+
d))^(1/2)*B*a*d*e*(b*e)^(1/2)*b+30*((b*x+a)*(e*x+d))^(1/2)*B*b^2*d^2*(b*e)^(1/2)
)/((b*x+a)*(e*x+d))^(1/2)/e^3/(b*e)^(1/2)/b

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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)/sqrt(e*x + d),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.530251, size = 1, normalized size = 0.01 \[ \left [\frac{4 \,{\left (8 \, B b^{2} e^{2} x^{2} + 15 \, B b^{2} d^{2} - 2 \,{\left (11 \, B a b + 9 \, A b^{2}\right )} d e + 3 \,{\left (B a^{2} + 10 \, A a b\right )} e^{2} - 2 \,{\left (5 \, B b^{2} d e -{\left (7 \, B a b + 6 \, A b^{2}\right )} e^{2}\right )} x\right )} \sqrt{b e} \sqrt{b x + a} \sqrt{e x + d} - 3 \,{\left (5 \, B b^{3} d^{3} - 3 \,{\left (3 \, B a b^{2} + 2 \, A b^{3}\right )} d^{2} e + 3 \,{\left (B a^{2} b + 4 \, A a b^{2}\right )} d e^{2} +{\left (B a^{3} - 6 \, A a^{2} b\right )} e^{3}\right )} \log \left (4 \,{\left (2 \, b^{2} e^{2} x + b^{2} d e + a b e^{2}\right )} \sqrt{b x + a} \sqrt{e x + d} +{\left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} + 8 \,{\left (b^{2} d e + a b e^{2}\right )} x\right )} \sqrt{b e}\right )}{96 \, \sqrt{b e} b e^{3}}, \frac{2 \,{\left (8 \, B b^{2} e^{2} x^{2} + 15 \, B b^{2} d^{2} - 2 \,{\left (11 \, B a b + 9 \, A b^{2}\right )} d e + 3 \,{\left (B a^{2} + 10 \, A a b\right )} e^{2} - 2 \,{\left (5 \, B b^{2} d e -{\left (7 \, B a b + 6 \, A b^{2}\right )} e^{2}\right )} x\right )} \sqrt{-b e} \sqrt{b x + a} \sqrt{e x + d} - 3 \,{\left (5 \, B b^{3} d^{3} - 3 \,{\left (3 \, B a b^{2} + 2 \, A b^{3}\right )} d^{2} e + 3 \,{\left (B a^{2} b + 4 \, A a b^{2}\right )} d e^{2} +{\left (B a^{3} - 6 \, A a^{2} b\right )} e^{3}\right )} \arctan \left (\frac{{\left (2 \, b e x + b d + a e\right )} \sqrt{-b e}}{2 \, \sqrt{b x + a} \sqrt{e x + d} b e}\right )}{48 \, \sqrt{-b e} b e^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

[1/96*(4*(8*B*b^2*e^2*x^2 + 15*B*b^2*d^2 - 2*(11*B*a*b + 9*A*b^2)*d*e + 3*(B*a^2
+ 10*A*a*b)*e^2 - 2*(5*B*b^2*d*e - (7*B*a*b + 6*A*b^2)*e^2)*x)*sqrt(b*e)*sqrt(b
*x + a)*sqrt(e*x + d) - 3*(5*B*b^3*d^3 - 3*(3*B*a*b^2 + 2*A*b^3)*d^2*e + 3*(B*a^
2*b + 4*A*a*b^2)*d*e^2 + (B*a^3 - 6*A*a^2*b)*e^3)*log(4*(2*b^2*e^2*x + b^2*d*e +
a*b*e^2)*sqrt(b*x + a)*sqrt(e*x + d) + (8*b^2*e^2*x^2 + b^2*d^2 + 6*a*b*d*e + a
^2*e^2 + 8*(b^2*d*e + a*b*e^2)*x)*sqrt(b*e)))/(sqrt(b*e)*b*e^3), 1/48*(2*(8*B*b^
2*e^2*x^2 + 15*B*b^2*d^2 - 2*(11*B*a*b + 9*A*b^2)*d*e + 3*(B*a^2 + 10*A*a*b)*e^2
- 2*(5*B*b^2*d*e - (7*B*a*b + 6*A*b^2)*e^2)*x)*sqrt(-b*e)*sqrt(b*x + a)*sqrt(e*
x + d) - 3*(5*B*b^3*d^3 - 3*(3*B*a*b^2 + 2*A*b^3)*d^2*e + 3*(B*a^2*b + 4*A*a*b^2
)*d*e^2 + (B*a^3 - 6*A*a^2*b)*e^3)*arctan(1/2*(2*b*e*x + b*d + a*e)*sqrt(-b*e)/(
sqrt(b*x + a)*sqrt(e*x + d)*b*e)))/(sqrt(-b*e)*b*e^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{\left (A + B x\right ) \left (a + b x\right )^{\frac{3}{2}}}{\sqrt{d + e x}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x+a)**(3/2)*(B*x+A)/(e*x+d)**(1/2),x)

[Out]

Integral((A + B*x)*(a + b*x)**(3/2)/sqrt(d + e*x), x)

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GIAC/XCAS [A]  time = 0.236843, size = 362, normalized size = 1.88 \[ \frac{{\left (\sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e} \sqrt{b x + a}{\left (2 \,{\left (b x + a\right )}{\left (\frac{4 \,{\left (b x + a\right )} B e^{\left (-1\right )}}{b^{2}} - \frac{{\left (5 \, B b^{3} d e^{3} + B a b^{2} e^{4} - 6 \, A b^{3} e^{4}\right )} e^{\left (-5\right )}}{b^{4}}\right )} + \frac{3 \,{\left (5 \, B b^{4} d^{2} e^{2} - 4 \, B a b^{3} d e^{3} - 6 \, A b^{4} d e^{3} - B a^{2} b^{2} e^{4} + 6 \, A a b^{3} e^{4}\right )} e^{\left (-5\right )}}{b^{4}}\right )} + \frac{3 \,{\left (5 \, B b^{3} d^{3} - 9 \, B a b^{2} d^{2} e - 6 \, A b^{3} d^{2} e + 3 \, B a^{2} b d e^{2} + 12 \, A a b^{2} d e^{2} + B a^{3} e^{3} - 6 \, A a^{2} b e^{3}\right )} e^{\left (-\frac{7}{2}\right )}{\rm ln}\left ({\left | -\sqrt{b x + a} \sqrt{b} e^{\frac{1}{2}} + \sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e} \right |}\right )}{b^{\frac{3}{2}}}\right )} b}{24 \,{\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/24*(sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a
)*B*e^(-1)/b^2 - (5*B*b^3*d*e^3 + B*a*b^2*e^4 - 6*A*b^3*e^4)*e^(-5)/b^4) + 3*(5*
B*b^4*d^2*e^2 - 4*B*a*b^3*d*e^3 - 6*A*b^4*d*e^3 - B*a^2*b^2*e^4 + 6*A*a*b^3*e^4)
*e^(-5)/b^4) + 3*(5*B*b^3*d^3 - 9*B*a*b^2*d^2*e - 6*A*b^3*d^2*e + 3*B*a^2*b*d*e^
2 + 12*A*a*b^2*d*e^2 + B*a^3*e^3 - 6*A*a^2*b*e^3)*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)))/b^(3/2))*b/abs(b)