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

Optimal. Leaf size=251 \[ \frac{(a+b x)^{3/2} \sqrt{c+d x} \left (15 a^2 d^2-4 b d x (5 a d+7 b c)+22 a b c d+35 b^2 c^2\right )}{96 b^3 d^3}-\frac{\sqrt{a+b x} \sqrt{c+d x} \left (5 a^3 d^3+9 a^2 b c d^2+15 a b^2 c^2 d+35 b^3 c^3\right )}{64 b^3 d^4}+\frac{(b c-a d) \left (5 a^3 d^3+9 a^2 b c d^2+15 a b^2 c^2 d+35 b^3 c^3\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{64 b^{7/2} d^{9/2}}+\frac{x^2 (a+b x)^{3/2} \sqrt{c+d x}}{4 b d} \]

[Out]

-((35*b^3*c^3 + 15*a*b^2*c^2*d + 9*a^2*b*c*d^2 + 5*a^3*d^3)*Sqrt[a + b*x]*Sqrt[c
+ d*x])/(64*b^3*d^4) + (x^2*(a + b*x)^(3/2)*Sqrt[c + d*x])/(4*b*d) + ((a + b*x)
^(3/2)*Sqrt[c + d*x]*(35*b^2*c^2 + 22*a*b*c*d + 15*a^2*d^2 - 4*b*d*(7*b*c + 5*a*
d)*x))/(96*b^3*d^3) + ((b*c - a*d)*(35*b^3*c^3 + 15*a*b^2*c^2*d + 9*a^2*b*c*d^2
+ 5*a^3*d^3)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(64*b^(7/
2)*d^(9/2))

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Rubi [A]  time = 0.445664, antiderivative size = 251, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227 \[ \frac{(a+b x)^{3/2} \sqrt{c+d x} \left (15 a^2 d^2-4 b d x (5 a d+7 b c)+22 a b c d+35 b^2 c^2\right )}{96 b^3 d^3}-\frac{\sqrt{a+b x} \sqrt{c+d x} \left (5 a^3 d^3+9 a^2 b c d^2+15 a b^2 c^2 d+35 b^3 c^3\right )}{64 b^3 d^4}+\frac{(b c-a d) \left (5 a^3 d^3+9 a^2 b c d^2+15 a b^2 c^2 d+35 b^3 c^3\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{64 b^{7/2} d^{9/2}}+\frac{x^2 (a+b x)^{3/2} \sqrt{c+d x}}{4 b d} \]

Antiderivative was successfully verified.

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

[Out]

-((35*b^3*c^3 + 15*a*b^2*c^2*d + 9*a^2*b*c*d^2 + 5*a^3*d^3)*Sqrt[a + b*x]*Sqrt[c
+ d*x])/(64*b^3*d^4) + (x^2*(a + b*x)^(3/2)*Sqrt[c + d*x])/(4*b*d) + ((a + b*x)
^(3/2)*Sqrt[c + d*x]*(35*b^2*c^2 + 22*a*b*c*d + 15*a^2*d^2 - 4*b*d*(7*b*c + 5*a*
d)*x))/(96*b^3*d^3) + ((b*c - a*d)*(35*b^3*c^3 + 15*a*b^2*c^2*d + 9*a^2*b*c*d^2
+ 5*a^3*d^3)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(64*b^(7/
2)*d^(9/2))

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Rubi in Sympy [A]  time = 36.3635, size = 250, normalized size = 1. \[ \frac{x^{2} \left (a + b x\right )^{\frac{3}{2}} \sqrt{c + d x}}{4 b d} + \frac{\left (a + b x\right )^{\frac{3}{2}} \sqrt{c + d x} \left (\frac{15 a^{2} d^{2}}{4} + \frac{11 a b c d}{2} + \frac{35 b^{2} c^{2}}{4} - b d x \left (5 a d + 7 b c\right )\right )}{24 b^{3} d^{3}} - \frac{\sqrt{a + b x} \sqrt{c + d x} \left (5 a^{3} d^{3} + 9 a^{2} b c d^{2} + 15 a b^{2} c^{2} d + 35 b^{3} c^{3}\right )}{64 b^{3} d^{4}} - \frac{\left (a d - b c\right ) \left (5 a^{3} d^{3} + 9 a^{2} b c d^{2} + 15 a b^{2} c^{2} d + 35 b^{3} c^{3}\right ) \operatorname{atanh}{\left (\frac{\sqrt{b} \sqrt{c + d x}}{\sqrt{d} \sqrt{a + b x}} \right )}}{64 b^{\frac{7}{2}} d^{\frac{9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

x**2*(a + b*x)**(3/2)*sqrt(c + d*x)/(4*b*d) + (a + b*x)**(3/2)*sqrt(c + d*x)*(15
*a**2*d**2/4 + 11*a*b*c*d/2 + 35*b**2*c**2/4 - b*d*x*(5*a*d + 7*b*c))/(24*b**3*d
**3) - sqrt(a + b*x)*sqrt(c + d*x)*(5*a**3*d**3 + 9*a**2*b*c*d**2 + 15*a*b**2*c*
*2*d + 35*b**3*c**3)/(64*b**3*d**4) - (a*d - b*c)*(5*a**3*d**3 + 9*a**2*b*c*d**2
+ 15*a*b**2*c**2*d + 35*b**3*c**3)*atanh(sqrt(b)*sqrt(c + d*x)/(sqrt(d)*sqrt(a
+ b*x)))/(64*b**(7/2)*d**(9/2))

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Mathematica [A]  time = 0.207696, size = 219, normalized size = 0.87 \[ \frac{3 (b c-a d) \left (5 a^3 d^3+9 a^2 b c d^2+15 a b^2 c^2 d+35 b^3 c^3\right ) \log \left (2 \sqrt{b} \sqrt{d} \sqrt{a+b x} \sqrt{c+d x}+a d+b c+2 b d x\right )-2 \sqrt{b} \sqrt{d} \sqrt{a+b x} \sqrt{c+d x} \left (-15 a^3 d^3+a^2 b d^2 (10 d x-17 c)+a b^2 d \left (-25 c^2+12 c d x-8 d^2 x^2\right )+b^3 \left (105 c^3-70 c^2 d x+56 c d^2 x^2-48 d^3 x^3\right )\right )}{384 b^{7/2} d^{9/2}} \]

Antiderivative was successfully verified.

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

[Out]

(-2*Sqrt[b]*Sqrt[d]*Sqrt[a + b*x]*Sqrt[c + d*x]*(-15*a^3*d^3 + a^2*b*d^2*(-17*c
+ 10*d*x) + a*b^2*d*(-25*c^2 + 12*c*d*x - 8*d^2*x^2) + b^3*(105*c^3 - 70*c^2*d*x
+ 56*c*d^2*x^2 - 48*d^3*x^3)) + 3*(b*c - a*d)*(35*b^3*c^3 + 15*a*b^2*c^2*d + 9*
a^2*b*c*d^2 + 5*a^3*d^3)*Log[b*c + a*d + 2*b*d*x + 2*Sqrt[b]*Sqrt[d]*Sqrt[a + b*
x]*Sqrt[c + d*x]])/(384*b^(7/2)*d^(9/2))

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Maple [B]  time = 0.034, size = 574, normalized size = 2.3 \[ -{\frac{1}{384\,{b}^{3}{d}^{4}}\sqrt{bx+a}\sqrt{dx+c} \left ( -96\,{x}^{3}{b}^{3}{d}^{3}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}-16\,{x}^{2}a{b}^{2}{d}^{3}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+112\,{x}^{2}{b}^{3}c{d}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+15\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{4}{d}^{4}+12\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{3}bc{d}^{3}+18\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{2}{b}^{2}{c}^{2}{d}^{2}+60\,{c}^{3}\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) a{b}^{3}d-105\,{c}^{4}\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){b}^{4}+20\,\sqrt{bd}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }x{a}^{2}b{d}^{3}+24\,\sqrt{bd}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }xa{b}^{2}c{d}^{2}-140\,\sqrt{bd}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }x{b}^{3}{c}^{2}d-30\,\sqrt{bd}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }{a}^{3}{d}^{3}-34\,\sqrt{bd}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }{a}^{2}bc{d}^{2}-50\,\sqrt{bd}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }a{b}^{2}{c}^{2}d+210\,{c}^{3}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }{b}^{3}\sqrt{bd} \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }}}{\frac{1}{\sqrt{bd}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-1/384*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(-96*x^3*b^3*d^3*((b*x+a)*(d*x+c))^(1/2)*(b*d
)^(1/2)-16*x^2*a*b^2*d^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+112*x^2*b^3*c*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^4*d^4+12*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*b*c*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))*a^2*b^2*c^2*d^2+60*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))*a*b^3*d-1
05*c^4*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^4+20*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x*a^2*b*d^3+24*(b*d)^(1/2)*((b*x+a
)*(d*x+c))^(1/2)*x*a*b^2*c*d^2-140*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x*b^3*c^2
*d-30*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^3*d^3-34*(b*d)^(1/2)*((b*x+a)*(d*x+c
))^(1/2)*a^2*b*c*d^2-50*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b^2*c^2*d+210*c^3*
((b*x+a)*(d*x+c))^(1/2)*b^3*(b*d)^(1/2))/((b*x+a)*(d*x+c))^(1/2)/b^3/d^4/(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(sqrt(b*x + a)*x^3/sqrt(d*x + c),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.280097, size = 1, normalized size = 0. \[ \left [\frac{4 \,{\left (48 \, b^{3} d^{3} x^{3} - 105 \, b^{3} c^{3} + 25 \, a b^{2} c^{2} d + 17 \, a^{2} b c d^{2} + 15 \, a^{3} d^{3} - 8 \,{\left (7 \, b^{3} c d^{2} - a b^{2} d^{3}\right )} x^{2} + 2 \,{\left (35 \, b^{3} c^{2} d - 6 \, a b^{2} c d^{2} - 5 \, a^{2} b d^{3}\right )} x\right )} \sqrt{b d} \sqrt{b x + a} \sqrt{d x + c} - 3 \,{\left (35 \, b^{4} c^{4} - 20 \, a b^{3} c^{3} d - 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} - 5 \, a^{4} d^{4}\right )} \log \left (-4 \,{\left (2 \, b^{2} d^{2} x + b^{2} c d + a b d^{2}\right )} \sqrt{b x + a} \sqrt{d x + c} +{\left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \,{\left (b^{2} c d + a b d^{2}\right )} x\right )} \sqrt{b d}\right )}{768 \, \sqrt{b d} b^{3} d^{4}}, \frac{2 \,{\left (48 \, b^{3} d^{3} x^{3} - 105 \, b^{3} c^{3} + 25 \, a b^{2} c^{2} d + 17 \, a^{2} b c d^{2} + 15 \, a^{3} d^{3} - 8 \,{\left (7 \, b^{3} c d^{2} - a b^{2} d^{3}\right )} x^{2} + 2 \,{\left (35 \, b^{3} c^{2} d - 6 \, a b^{2} c d^{2} - 5 \, a^{2} b d^{3}\right )} x\right )} \sqrt{-b d} \sqrt{b x + a} \sqrt{d x + c} + 3 \,{\left (35 \, b^{4} c^{4} - 20 \, a b^{3} c^{3} d - 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} - 5 \, a^{4} d^{4}\right )} \arctan \left (\frac{{\left (2 \, b d x + b c + a d\right )} \sqrt{-b d}}{2 \, \sqrt{b x + a} \sqrt{d x + c} b d}\right )}{384 \, \sqrt{-b d} b^{3} d^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

[1/768*(4*(48*b^3*d^3*x^3 - 105*b^3*c^3 + 25*a*b^2*c^2*d + 17*a^2*b*c*d^2 + 15*a
^3*d^3 - 8*(7*b^3*c*d^2 - a*b^2*d^3)*x^2 + 2*(35*b^3*c^2*d - 6*a*b^2*c*d^2 - 5*a
^2*b*d^3)*x)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) - 3*(35*b^4*c^4 - 20*a*b^3*c^
3*d - 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 - 5*a^4*d^4)*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^4), 1/384*(2*
(48*b^3*d^3*x^3 - 105*b^3*c^3 + 25*a*b^2*c^2*d + 17*a^2*b*c*d^2 + 15*a^3*d^3 - 8
*(7*b^3*c*d^2 - a*b^2*d^3)*x^2 + 2*(35*b^3*c^2*d - 6*a*b^2*c*d^2 - 5*a^2*b*d^3)*
x)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 3*(35*b^4*c^4 - 20*a*b^3*c^3*d - 6*a
^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 - 5*a^4*d^4)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqr
t(-b*d)/(sqrt(b*x + a)*sqrt(d*x + c)*b*d)))/(sqrt(-b*d)*b^3*d^4)]

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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**3*(b*x+a)**(1/2)/(d*x+c)**(1/2),x)

[Out]

Timed out

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GIAC/XCAS [A]  time = 0.259764, size = 393, normalized size = 1.57 \[ \frac{{\left (\sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}{\left (2 \,{\left (b x + a\right )}{\left (4 \,{\left (b x + a\right )}{\left (\frac{6 \,{\left (b x + a\right )}}{b^{4} d} - \frac{7 \, b^{13} c d^{5} + 17 \, a b^{12} d^{6}}{b^{16} d^{7}}\right )} + \frac{35 \, b^{14} c^{2} d^{4} + 50 \, a b^{13} c d^{5} + 59 \, a^{2} b^{12} d^{6}}{b^{16} d^{7}}\right )} - \frac{3 \,{\left (35 \, b^{15} c^{3} d^{3} + 15 \, a b^{14} c^{2} d^{4} + 9 \, a^{2} b^{13} c d^{5} + 5 \, a^{3} b^{12} d^{6}\right )}}{b^{16} d^{7}}\right )} \sqrt{b x + a} - \frac{3 \,{\left (35 \, b^{4} c^{4} - 20 \, a b^{3} c^{3} d - 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} - 5 \, a^{4} d^{4}\right )}{\rm ln}\left ({\left | -\sqrt{b d} \sqrt{b x + a} + \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt{b d} b^{3} d^{4}}\right )} b}{192 \,{\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/192*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(b*x + a)*(4*(b*x + a)*(6*(b*x + a
)/(b^4*d) - (7*b^13*c*d^5 + 17*a*b^12*d^6)/(b^16*d^7)) + (35*b^14*c^2*d^4 + 50*a
*b^13*c*d^5 + 59*a^2*b^12*d^6)/(b^16*d^7)) - 3*(35*b^15*c^3*d^3 + 15*a*b^14*c^2*
d^4 + 9*a^2*b^13*c*d^5 + 5*a^3*b^12*d^6)/(b^16*d^7))*sqrt(b*x + a) - 3*(35*b^4*c
^4 - 20*a*b^3*c^3*d - 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 - 5*a^4*d^4)*ln(abs(-sqr
t(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*b^3*d^4)
)*b/abs(b)