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

Optimal. Leaf size=171 \[ -\frac{(b c-a d)^2 (5 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{8 b^{7/2} d^{3/2}}-\frac{\sqrt{a+b x} \sqrt{c+d x} (b c-a d) (5 a d+b c)}{8 b^3 d}-\frac{\sqrt{a+b x} (c+d x)^{3/2} (5 a d+b c)}{12 b^2 d}+\frac{\sqrt{a+b x} (c+d x)^{5/2}}{3 b d} \]

[Out]

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

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Rubi [A]  time = 0.240188, antiderivative size = 171, 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 \[ -\frac{(b c-a d)^2 (5 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{8 b^{7/2} d^{3/2}}-\frac{\sqrt{a+b x} \sqrt{c+d x} (b c-a d) (5 a d+b c)}{8 b^3 d}-\frac{\sqrt{a+b x} (c+d x)^{3/2} (5 a d+b c)}{12 b^2 d}+\frac{\sqrt{a+b x} (c+d x)^{5/2}}{3 b d} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 23.8336, size = 150, normalized size = 0.88 \[ \frac{\sqrt{a + b x} \left (c + d x\right )^{\frac{5}{2}}}{3 b d} - \frac{\sqrt{a + b x} \left (c + d x\right )^{\frac{3}{2}} \left (5 a d + b c\right )}{12 b^{2} d} + \frac{\sqrt{a + b x} \sqrt{c + d x} \left (a d - b c\right ) \left (5 a d + b c\right )}{8 b^{3} d} - \frac{\left (a d - b c\right )^{2} \left (5 a d + b c\right ) \operatorname{atanh}{\left (\frac{\sqrt{d} \sqrt{a + b x}}{\sqrt{b} \sqrt{c + d x}} \right )}}{8 b^{\frac{7}{2}} d^{\frac{3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

sqrt(a + b*x)*(c + d*x)**(5/2)/(3*b*d) - sqrt(a + b*x)*(c + d*x)**(3/2)*(5*a*d +
b*c)/(12*b**2*d) + sqrt(a + b*x)*sqrt(c + d*x)*(a*d - b*c)*(5*a*d + b*c)/(8*b**
3*d) - (a*d - b*c)**2*(5*a*d + b*c)*atanh(sqrt(d)*sqrt(a + b*x)/(sqrt(b)*sqrt(c
+ d*x)))/(8*b**(7/2)*d**(3/2))

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Mathematica [A]  time = 0.130253, size = 149, normalized size = 0.87 \[ \frac{\sqrt{a+b x} \sqrt{c+d x} \left (15 a^2 d^2-2 a b d (11 c+5 d x)+b^2 \left (3 c^2+14 c d x+8 d^2 x^2\right )\right )}{24 b^3 d}-\frac{(b c-a d)^2 (5 a d+b c) \log \left (2 \sqrt{b} \sqrt{d} \sqrt{a+b x} \sqrt{c+d x}+a d+b c+2 b d x\right )}{16 b^{7/2} d^{3/2}} \]

Antiderivative was successfully verified.

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

[Out]

(Sqrt[a + b*x]*Sqrt[c + d*x]*(15*a^2*d^2 - 2*a*b*d*(11*c + 5*d*x) + b^2*(3*c^2 +
14*c*d*x + 8*d^2*x^2)))/(24*b^3*d) - ((b*c - a*d)^2*(b*c + 5*a*d)*Log[b*c + a*d
+ 2*b*d*x + 2*Sqrt[b]*Sqrt[d]*Sqrt[a + b*x]*Sqrt[c + d*x]])/(16*b^(7/2)*d^(3/2)
)

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Maple [B]  time = 0.029, size = 395, normalized size = 2.3 \[ -{\frac{1}{48\,{b}^{3}d}\sqrt{bx+a}\sqrt{dx+c} \left ( -16\,{x}^{2}{b}^{2}{d}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+15\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{3}{d}^{3}-27\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{2}bc{d}^{2}+9\,{c}^{2}\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) a{b}^{2}d+3\,{c}^{3}\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){b}^{3}+20\,\sqrt{bd}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }xab{d}^{2}-28\,\sqrt{bd}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }x{b}^{2}cd-30\,\sqrt{bd}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }{a}^{2}{d}^{2}+44\,\sqrt{bd}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }abcd-6\,{c}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }{b}^{2}\sqrt{bd} \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }}}{\frac{1}{\sqrt{bd}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-1/48*(d*x+c)^(1/2)*(b*x+a)^(1/2)*(-16*x^2*b^2*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)
^(1/2)+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))*a^3*d^3-27*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*b*c*d^2+9*c^2*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^2*d+3*c^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))*b^3+20*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*
x*a*b*d^2-28*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x*b^2*c*d-30*(b*d)^(1/2)*((b*x+
a)*(d*x+c))^(1/2)*a^2*d^2+44*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b*c*d-6*c^2*(
(b*x+a)*(d*x+c))^(1/2)*b^2*(b*d)^(1/2))/((b*x+a)*(d*x+c))^(1/2)/b^3/d/(b*d)^(1/2
)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.258849, size = 1, normalized size = 0.01 \[ \left [\frac{4 \,{\left (8 \, b^{2} d^{2} x^{2} + 3 \, b^{2} c^{2} - 22 \, a b c d + 15 \, a^{2} d^{2} + 2 \,{\left (7 \, b^{2} c d - 5 \, a b d^{2}\right )} x\right )} \sqrt{b d} \sqrt{b x + a} \sqrt{d x + c} + 3 \,{\left (b^{3} c^{3} + 3 \, a b^{2} c^{2} d - 9 \, a^{2} b c d^{2} + 5 \, a^{3} d^{3}\right )} \log \left (-4 \,{\left (2 \, b^{2} d^{2} x + b^{2} c d + a b d^{2}\right )} \sqrt{b x + a} \sqrt{d x + c} +{\left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \,{\left (b^{2} c d + a b d^{2}\right )} x\right )} \sqrt{b d}\right )}{96 \, \sqrt{b d} b^{3} d}, \frac{2 \,{\left (8 \, b^{2} d^{2} x^{2} + 3 \, b^{2} c^{2} - 22 \, a b c d + 15 \, a^{2} d^{2} + 2 \,{\left (7 \, b^{2} c d - 5 \, a b d^{2}\right )} x\right )} \sqrt{-b d} \sqrt{b x + a} \sqrt{d x + c} - 3 \,{\left (b^{3} c^{3} + 3 \, a b^{2} c^{2} d - 9 \, a^{2} b c d^{2} + 5 \, a^{3} d^{3}\right )} \arctan \left (\frac{{\left (2 \, b d x + b c + a d\right )} \sqrt{-b d}}{2 \, \sqrt{b x + a} \sqrt{d x + c} b d}\right )}{48 \, \sqrt{-b d} b^{3} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

[1/96*(4*(8*b^2*d^2*x^2 + 3*b^2*c^2 - 22*a*b*c*d + 15*a^2*d^2 + 2*(7*b^2*c*d - 5
*a*b*d^2)*x)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 3*(b^3*c^3 + 3*a*b^2*c^2*d
- 9*a^2*b*c*d^2 + 5*a^3*d^3)*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)))/(sqrt(b*d)*b^3*d), 1/48*(2*(8*b^2*d^2*x^2 + 3*b^2*c^2
- 22*a*b*c*d + 15*a^2*d^2 + 2*(7*b^2*c*d - 5*a*b*d^2)*x)*sqrt(-b*d)*sqrt(b*x + a
)*sqrt(d*x + c) - 3*(b^3*c^3 + 3*a*b^2*c^2*d - 9*a^2*b*c*d^2 + 5*a^3*d^3)*arctan
(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)/(sqrt(b*x + a)*sqrt(d*x + c)*b*d)))/(sqrt(
-b*d)*b^3*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(x*(d*x+c)**(3/2)/(b*x+a)**(1/2),x)

[Out]

Timed out

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GIAC/XCAS [A]  time = 0.254252, size = 479, normalized size = 2.8 \[ \frac{\frac{2 \,{\left (\sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} \sqrt{b x + a}{\left (2 \,{\left (b x + a\right )}{\left (\frac{4 \,{\left (b x + a\right )}}{b^{2}} + \frac{b^{6} c d^{3} - 13 \, a b^{5} d^{4}}{b^{7} d^{4}}\right )} - \frac{3 \,{\left (b^{7} c^{2} d^{2} + 2 \, a b^{6} c d^{3} - 11 \, a^{2} b^{5} d^{4}\right )}}{b^{7} d^{4}}\right )} - \frac{3 \,{\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )}{\rm ln}\left ({\left | -\sqrt{b d} \sqrt{b x + a} + \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt{b d} b d^{2}}\right )} d{\left | b \right |}}{b^{2}} + \frac{{\left (\sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} \sqrt{b x + a}{\left (\frac{2 \,{\left (b x + a\right )}}{b^{4} d^{2}} + \frac{b c d - 5 \, a d^{2}}{b^{4} d^{4}}\right )} + \frac{{\left (b^{2} c^{2} + 2 \, a b c d - 3 \, a^{2} d^{2}\right )}{\rm ln}\left ({\left | -\sqrt{b d} \sqrt{b x + a} + \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt{b d} b^{3} d^{3}}\right )} c{\left | b \right |}}{b^{3}}}{48 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/48*(2*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x
+ a)/b^2 + (b^6*c*d^3 - 13*a*b^5*d^4)/(b^7*d^4)) - 3*(b^7*c^2*d^2 + 2*a*b^6*c*d^
3 - 11*a^2*b^5*d^4)/(b^7*d^4)) - 3*(b^3*c^3 + a*b^2*c^2*d + 3*a^2*b*c*d^2 - 5*a^
3*d^3)*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*d^2))*d*abs(b)/b^2 + (sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x +
a)*(2*(b*x + a)/(b^4*d^2) + (b*c*d - 5*a*d^2)/(b^4*d^4)) + (b^2*c^2 + 2*a*b*c*d
- 3*a^2*d^2)*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^3*d^3))*c*abs(b)/b^3)/b