3.623 \(\int \frac{x (a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=174 \[ \frac{3 (b c-a d) (5 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 \sqrt{b} d^{7/2}}-\frac{3 \sqrt{a+b x} \sqrt{c+d x} (5 b c-a d)}{4 d^3}+\frac{(a+b x)^{3/2} \sqrt{c+d x} (5 b c-a d)}{2 d^2 (b c-a d)}-\frac{2 c (a+b x)^{5/2}}{d \sqrt{c+d x} (b c-a d)} \]

[Out]

(-2*c*(a + b*x)^(5/2))/(d*(b*c - a*d)*Sqrt[c + d*x]) - (3*(5*b*c - a*d)*Sqrt[a +
b*x]*Sqrt[c + d*x])/(4*d^3) + ((5*b*c - a*d)*(a + b*x)^(3/2)*Sqrt[c + d*x])/(2*
d^2*(b*c - a*d)) + (3*(b*c - a*d)*(5*b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/
(Sqrt[b]*Sqrt[c + d*x])])/(4*Sqrt[b]*d^(7/2))

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Rubi [A]  time = 0.247069, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2 \[ \frac{3 (b c-a d) (5 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 \sqrt{b} d^{7/2}}-\frac{3 \sqrt{a+b x} \sqrt{c+d x} (5 b c-a d)}{4 d^3}+\frac{(a+b x)^{3/2} \sqrt{c+d x} (5 b c-a d)}{2 d^2 (b c-a d)}-\frac{2 c (a+b x)^{5/2}}{d \sqrt{c+d x} (b c-a d)} \]

Antiderivative was successfully verified.

[In]  Int[(x*(a + b*x)^(3/2))/(c + d*x)^(3/2),x]

[Out]

(-2*c*(a + b*x)^(5/2))/(d*(b*c - a*d)*Sqrt[c + d*x]) - (3*(5*b*c - a*d)*Sqrt[a +
b*x]*Sqrt[c + d*x])/(4*d^3) + ((5*b*c - a*d)*(a + b*x)^(3/2)*Sqrt[c + d*x])/(2*
d^2*(b*c - a*d)) + (3*(b*c - a*d)*(5*b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/
(Sqrt[b]*Sqrt[c + d*x])])/(4*Sqrt[b]*d^(7/2))

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Rubi in Sympy [A]  time = 26.012, size = 155, normalized size = 0.89 \[ \frac{2 c \left (a + b x\right )^{\frac{5}{2}}}{d \sqrt{c + d x} \left (a d - b c\right )} + \frac{\left (a + b x\right )^{\frac{3}{2}} \sqrt{c + d x} \left (a d - 5 b c\right )}{2 d^{2} \left (a d - b c\right )} + \frac{3 \sqrt{a + b x} \sqrt{c + d x} \left (a d - 5 b c\right )}{4 d^{3}} + \frac{3 \left (a d - 5 b c\right ) \left (a d - b c\right ) \operatorname{atanh}{\left (\frac{\sqrt{d} \sqrt{a + b x}}{\sqrt{b} \sqrt{c + d x}} \right )}}{4 \sqrt{b} d^{\frac{7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(x*(b*x+a)**(3/2)/(d*x+c)**(3/2),x)

[Out]

2*c*(a + b*x)**(5/2)/(d*sqrt(c + d*x)*(a*d - b*c)) + (a + b*x)**(3/2)*sqrt(c + d
*x)*(a*d - 5*b*c)/(2*d**2*(a*d - b*c)) + 3*sqrt(a + b*x)*sqrt(c + d*x)*(a*d - 5*
b*c)/(4*d**3) + 3*(a*d - 5*b*c)*(a*d - b*c)*atanh(sqrt(d)*sqrt(a + b*x)/(sqrt(b)
*sqrt(c + d*x)))/(4*sqrt(b)*d**(7/2))

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Mathematica [A]  time = 0.140792, size = 132, normalized size = 0.76 \[ \frac{\sqrt{a+b x} \left (a d (13 c+5 d x)+b \left (-15 c^2-5 c d x+2 d^2 x^2\right )\right )}{4 d^3 \sqrt{c+d x}}+\frac{3 (a d-5 b c) (a d-b c) \log \left (2 \sqrt{b} \sqrt{d} \sqrt{a+b x} \sqrt{c+d x}+a d+b c+2 b d x\right )}{8 \sqrt{b} d^{7/2}} \]

Antiderivative was successfully verified.

[In]  Integrate[(x*(a + b*x)^(3/2))/(c + d*x)^(3/2),x]

[Out]

(Sqrt[a + b*x]*(a*d*(13*c + 5*d*x) + b*(-15*c^2 - 5*c*d*x + 2*d^2*x^2)))/(4*d^3*
Sqrt[c + d*x]) + (3*(-5*b*c + a*d)*(-(b*c) + a*d)*Log[b*c + a*d + 2*b*d*x + 2*Sq
rt[b]*Sqrt[d]*Sqrt[a + b*x]*Sqrt[c + d*x]])/(8*Sqrt[b]*d^(7/2))

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Maple [B]  time = 0.03, size = 455, normalized size = 2.6 \[{\frac{1}{8\,{d}^{3}}\sqrt{bx+a} \left ( 3\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) x{a}^{2}{d}^{3}-18\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) xabc{d}^{2}+15\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) x{b}^{2}{c}^{2}d+4\,{x}^{2}b{d}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+3\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{2}c{d}^{2}-18\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) ab{c}^{2}d+15\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){b}^{2}{c}^{3}+10\,xa{d}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}-10\,xbcd\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+26\,acd\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}-30\,b{c}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd} \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }}}{\frac{1}{\sqrt{bd}}}{\frac{1}{\sqrt{dx+c}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(x*(b*x+a)^(3/2)/(d*x+c)^(3/2),x)

[Out]

1/8*(b*x+a)^(1/2)*(3*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b
*c)/(b*d)^(1/2))*x*a^2*d^3-18*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1
/2)+a*d+b*c)/(b*d)^(1/2))*x*a*b*c*d^2+15*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/
2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x*b^2*c^2*d+4*x^2*b*d^2*((b*x+a)*(d*x+c))^(
1/2)*(b*d)^(1/2)+3*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c
)/(b*d)^(1/2))*a^2*c*d^2-18*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2
)+a*d+b*c)/(b*d)^(1/2))*a*b*c^2*d+15*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(
b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*b^2*c^3+10*x*a*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d
)^(1/2)-10*x*b*c*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+26*a*c*d*((b*x+a)*(d*x+c)
)^(1/2)*(b*d)^(1/2)-30*b*c^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/((b*x+a)*(d*x+
c))^(1/2)/(b*d)^(1/2)/(d*x+c)^(1/2)/d^3

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.389422, size = 1, normalized size = 0.01 \[ \left [\frac{4 \,{\left (2 \, b d^{2} x^{2} - 15 \, b c^{2} + 13 \, a c d - 5 \,{\left (b c d - a d^{2}\right )} x\right )} \sqrt{b d} \sqrt{b x + a} \sqrt{d x + c} + 3 \,{\left (5 \, b^{2} c^{3} - 6 \, a b c^{2} d + a^{2} c d^{2} +{\left (5 \, b^{2} c^{2} d - 6 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \log \left (4 \,{\left (2 \, b^{2} d^{2} x + b^{2} c d + a b d^{2}\right )} \sqrt{b x + a} \sqrt{d x + c} +{\left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \,{\left (b^{2} c d + a b d^{2}\right )} x\right )} \sqrt{b d}\right )}{16 \,{\left (d^{4} x + c d^{3}\right )} \sqrt{b d}}, \frac{2 \,{\left (2 \, b d^{2} x^{2} - 15 \, b c^{2} + 13 \, a c d - 5 \,{\left (b c d - a d^{2}\right )} x\right )} \sqrt{-b d} \sqrt{b x + a} \sqrt{d x + c} + 3 \,{\left (5 \, b^{2} c^{3} - 6 \, a b c^{2} d + a^{2} c d^{2} +{\left (5 \, b^{2} c^{2} d - 6 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \arctan \left (\frac{{\left (2 \, b d x + b c + a d\right )} \sqrt{-b d}}{2 \, \sqrt{b x + a} \sqrt{d x + c} b d}\right )}{8 \,{\left (d^{4} x + c d^{3}\right )} \sqrt{-b d}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

[1/16*(4*(2*b*d^2*x^2 - 15*b*c^2 + 13*a*c*d - 5*(b*c*d - a*d^2)*x)*sqrt(b*d)*sqr
t(b*x + a)*sqrt(d*x + c) + 3*(5*b^2*c^3 - 6*a*b*c^2*d + a^2*c*d^2 + (5*b^2*c^2*d
- 6*a*b*c*d^2 + a^2*d^3)*x)*log(4*(2*b^2*d^2*x + b^2*c*d + a*b*d^2)*sqrt(b*x +
a)*sqrt(d*x + c) + (8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 8*(b^2*c*d +
a*b*d^2)*x)*sqrt(b*d)))/((d^4*x + c*d^3)*sqrt(b*d)), 1/8*(2*(2*b*d^2*x^2 - 15*b
*c^2 + 13*a*c*d - 5*(b*c*d - a*d^2)*x)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c) +
3*(5*b^2*c^3 - 6*a*b*c^2*d + a^2*c*d^2 + (5*b^2*c^2*d - 6*a*b*c*d^2 + a^2*d^3)*x
)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)/(sqrt(b*x + a)*sqrt(d*x + c)*b*d))
)/((d^4*x + c*d^3)*sqrt(-b*d))]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x*(b*x+a)**(3/2)/(d*x+c)**(3/2),x)

[Out]

Integral(x*(a + b*x)**(3/2)/(c + d*x)**(3/2), x)

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GIAC/XCAS [A]  time = 0.243377, size = 320, normalized size = 1.84 \[ \frac{{\left ({\left (\frac{2 \,{\left (b x + a\right )} b d^{4}{\left | b \right |}}{b^{8} c d^{6} - a b^{7} d^{7}} - \frac{5 \, b^{2} c d^{3}{\left | b \right |} - a b d^{4}{\left | b \right |}}{b^{8} c d^{6} - a b^{7} d^{7}}\right )}{\left (b x + a\right )} - \frac{3 \,{\left (5 \, b^{3} c^{2} d^{2}{\left | b \right |} - 6 \, a b^{2} c d^{3}{\left | b \right |} + a^{2} b d^{4}{\left | b \right |}\right )}}{b^{8} c d^{6} - a b^{7} d^{7}}\right )} \sqrt{b x + a}}{1536 \, \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}} - \frac{{\left (5 \, b c{\left | b \right |} - a d{\left | b \right |}\right )}{\rm ln}\left ({\left | -\sqrt{b d} \sqrt{b x + a} + \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} \right |}\right )}{512 \, \sqrt{b d} b^{6} d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/1536*((2*(b*x + a)*b*d^4*abs(b)/(b^8*c*d^6 - a*b^7*d^7) - (5*b^2*c*d^3*abs(b)
- a*b*d^4*abs(b))/(b^8*c*d^6 - a*b^7*d^7))*(b*x + a) - 3*(5*b^3*c^2*d^2*abs(b) -
6*a*b^2*c*d^3*abs(b) + a^2*b*d^4*abs(b))/(b^8*c*d^6 - a*b^7*d^7))*sqrt(b*x + a)
/sqrt(b^2*c + (b*x + a)*b*d - a*b*d) - 1/512*(5*b*c*abs(b) - a*d*abs(b))*ln(abs(
-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*b^6*
d^4)