3.557 \(\int \frac{\sqrt{a+b x} (c+d x)^{5/2}}{x} \, dx\)

Optimal. Leaf size=219 \[ \frac{\left (a^3 d^3-5 a^2 b c d^2+15 a b^2 c^2 d+5 b^3 c^3\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{8 b^{5/2} \sqrt{d}}+\frac{\sqrt{a+b x} \sqrt{c+d x} (5 b c-a d) (a d+b c)}{8 b^2}-2 \sqrt{a} c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )+\frac{1}{3} \sqrt{a+b x} (c+d x)^{5/2}+\frac{\sqrt{a+b x} (c+d x)^{3/2} (a d+5 b c)}{12 b} \]

[Out]

((5*b*c - a*d)*(b*c + a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*b^2) + ((5*b*c + a*d)
*Sqrt[a + b*x]*(c + d*x)^(3/2))/(12*b) + (Sqrt[a + b*x]*(c + d*x)^(5/2))/3 - 2*S
qrt[a]*c^(5/2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])] + ((5*b^
3*c^3 + 15*a*b^2*c^2*d - 5*a^2*b*c*d^2 + a^3*d^3)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x]
)/(Sqrt[b]*Sqrt[c + d*x])])/(8*b^(5/2)*Sqrt[d])

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Rubi [A]  time = 0.720676, antiderivative size = 219, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273 \[ \frac{\left (a^3 d^3-5 a^2 b c d^2+15 a b^2 c^2 d+5 b^3 c^3\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{8 b^{5/2} \sqrt{d}}+\frac{\sqrt{a+b x} \sqrt{c+d x} (5 b c-a d) (a d+b c)}{8 b^2}-2 \sqrt{a} c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )+\frac{1}{3} \sqrt{a+b x} (c+d x)^{5/2}+\frac{\sqrt{a+b x} (c+d x)^{3/2} (a d+5 b c)}{12 b} \]

Antiderivative was successfully verified.

[In]  Int[(Sqrt[a + b*x]*(c + d*x)^(5/2))/x,x]

[Out]

((5*b*c - a*d)*(b*c + a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*b^2) + ((5*b*c + a*d)
*Sqrt[a + b*x]*(c + d*x)^(3/2))/(12*b) + (Sqrt[a + b*x]*(c + d*x)^(5/2))/3 - 2*S
qrt[a]*c^(5/2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])] + ((5*b^
3*c^3 + 15*a*b^2*c^2*d - 5*a^2*b*c*d^2 + a^3*d^3)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x]
)/(Sqrt[b]*Sqrt[c + d*x])])/(8*b^(5/2)*Sqrt[d])

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Rubi in Sympy [A]  time = 63.0574, size = 199, normalized size = 0.91 \[ - 2 \sqrt{a} c^{\frac{5}{2}} \operatorname{atanh}{\left (\frac{\sqrt{c} \sqrt{a + b x}}{\sqrt{a} \sqrt{c + d x}} \right )} + \frac{\sqrt{a + b x} \left (c + d x\right )^{\frac{5}{2}}}{3} + \frac{\sqrt{a + b x} \left (c + d x\right )^{\frac{3}{2}} \left (a d + 5 b c\right )}{12 b} - \frac{\sqrt{a + b x} \sqrt{c + d x} \left (a d - 5 b c\right ) \left (a d + b c\right )}{8 b^{2}} + \frac{\left (16 a b^{2} c^{2} d + \left (a d - 5 b c\right ) \left (a d - b c\right ) \left (a d + b c\right )\right ) \operatorname{atanh}{\left (\frac{\sqrt{b} \sqrt{c + d x}}{\sqrt{d} \sqrt{a + b x}} \right )}}{8 b^{\frac{5}{2}} \sqrt{d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((d*x+c)**(5/2)*(b*x+a)**(1/2)/x,x)

[Out]

-2*sqrt(a)*c**(5/2)*atanh(sqrt(c)*sqrt(a + b*x)/(sqrt(a)*sqrt(c + d*x))) + sqrt(
a + b*x)*(c + d*x)**(5/2)/3 + sqrt(a + b*x)*(c + d*x)**(3/2)*(a*d + 5*b*c)/(12*b
) - sqrt(a + b*x)*sqrt(c + d*x)*(a*d - 5*b*c)*(a*d + b*c)/(8*b**2) + (16*a*b**2*
c**2*d + (a*d - 5*b*c)*(a*d - b*c)*(a*d + b*c))*atanh(sqrt(b)*sqrt(c + d*x)/(sqr
t(d)*sqrt(a + b*x)))/(8*b**(5/2)*sqrt(d))

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Mathematica [A]  time = 0.158314, size = 232, normalized size = 1.06 \[ \frac{\sqrt{a+b x} \sqrt{c+d x} \left (-3 a^2 d^2+2 a b d (7 c+d x)+b^2 \left (33 c^2+26 c d x+8 d^2 x^2\right )\right )}{24 b^2}+\frac{\left (a^3 d^3-5 a^2 b c d^2+15 a b^2 c^2 d+5 b^3 c^3\right ) \log \left (2 \sqrt{b} \sqrt{d} \sqrt{a+b x} \sqrt{c+d x}+a d+b c+2 b d x\right )}{16 b^{5/2} \sqrt{d}}-\sqrt{a} c^{5/2} \log \left (2 \sqrt{a} \sqrt{c} \sqrt{a+b x} \sqrt{c+d x}+2 a c+a d x+b c x\right )+\sqrt{a} c^{5/2} \log (x) \]

Antiderivative was successfully verified.

[In]  Integrate[(Sqrt[a + b*x]*(c + d*x)^(5/2))/x,x]

[Out]

(Sqrt[a + b*x]*Sqrt[c + d*x]*(-3*a^2*d^2 + 2*a*b*d*(7*c + d*x) + b^2*(33*c^2 + 2
6*c*d*x + 8*d^2*x^2)))/(24*b^2) + Sqrt[a]*c^(5/2)*Log[x] - Sqrt[a]*c^(5/2)*Log[2
*a*c + b*c*x + a*d*x + 2*Sqrt[a]*Sqrt[c]*Sqrt[a + b*x]*Sqrt[c + d*x]] + ((5*b^3*
c^3 + 15*a*b^2*c^2*d - 5*a^2*b*c*d^2 + 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]])/(16*b^(5/2)*Sqrt[d])

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Maple [B]  time = 0.02, size = 583, normalized size = 2.7 \[{\frac{1}{48\,{b}^{2}}\sqrt{bx+a}\sqrt{dx+c} \left ( 16\,{x}^{2}{b}^{2}{d}^{2}\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+3\,{d}^{3}\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{3}\sqrt{ac}-15\,{d}^{2}\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{2}c\sqrt{ac}b+45\,{c}^{2}\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) ad\sqrt{ac}{b}^{2}+15\,{c}^{3}\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){b}^{3}\sqrt{ac}-48\,{c}^{3}a\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac}{x}} \right ) \sqrt{bd}{b}^{2}+4\,{d}^{2}\sqrt{d{x}^{2}b+adx+bcx+ac}xa\sqrt{bd}\sqrt{ac}b+52\,d\sqrt{d{x}^{2}b+adx+bcx+ac}xc\sqrt{bd}\sqrt{ac}{b}^{2}-6\,{d}^{2}\sqrt{d{x}^{2}b+adx+bcx+ac}{a}^{2}\sqrt{bd}\sqrt{ac}+28\,d\sqrt{d{x}^{2}b+adx+bcx+ac}ac\sqrt{bd}\sqrt{ac}b+66\,{c}^{2}\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}\sqrt{ac}{b}^{2} \right ){\frac{1}{\sqrt{d{x}^{2}b+adx+bcx+ac}}}{\frac{1}{\sqrt{bd}}}{\frac{1}{\sqrt{ac}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((d*x+c)^(5/2)*(b*x+a)^(1/2)/x,x)

[Out]

1/48*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(16*x^2*b^2*d^2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*
x+a*c)^(1/2)*(b*d)^(1/2)+3*d^3*ln(1/2*(2*b*d*x+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)
*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^3*(a*c)^(1/2)-15*d^2*ln(1/2*(2*b*d*x+2*(b*d
*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^2*c*(a*c)^(1/2)*
b+45*c^2*ln(1/2*(2*b*d*x+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/
(b*d)^(1/2))*a*d*(a*c)^(1/2)*b^2+15*c^3*ln(1/2*(2*b*d*x+2*(b*d*x^2+a*d*x+b*c*x+a
*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*b^3*(a*c)^(1/2)-48*c^3*a*ln((a*d*x+b
*c*x+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)+2*a*c)/x)*(b*d)^(1/2)*b^2+4*d
^2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x*a*(b*d)^(1/2)*(a*c)^(1/2)*b+52*d*(b*d*x^2+a
*d*x+b*c*x+a*c)^(1/2)*x*c*(b*d)^(1/2)*(a*c)^(1/2)*b^2-6*d^2*(b*d*x^2+a*d*x+b*c*x
+a*c)^(1/2)*a^2*(b*d)^(1/2)*(a*c)^(1/2)+28*d*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a*c
*(b*d)^(1/2)*(a*c)^(1/2)*b+66*c^2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a
*c)^(1/2)*b^2)/(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)/(b*d)^(1/2)/(a*c)^(1/2)/b^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)*(d*x + c)^(5/2)/x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 5.96798, size = 1, normalized size = 0. \[ \text{result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(sqrt(b*x + a)*(d*x + c)^(5/2)/x,x, algorithm="fricas")

[Out]

[1/96*(48*sqrt(a*c)*sqrt(b*d)*b^2*c^2*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^
2*d^2)*x^2 - 4*(2*a*c + (b*c + a*d)*x)*sqrt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8
*(a*b*c^2 + a^2*c*d)*x)/x^2) + 4*(8*b^2*d^2*x^2 + 33*b^2*c^2 + 14*a*b*c*d - 3*a^
2*d^2 + 2*(13*b^2*c*d + a*b*d^2)*x)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 3*(5
*b^3*c^3 + 15*a*b^2*c^2*d - 5*a^2*b*c*d^2 + 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^2), 1/48*(24*sqrt(
a*c)*sqrt(-b*d)*b^2*c^2*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4
*(2*a*c + (b*c + a*d)*x)*sqrt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^
2*c*d)*x)/x^2) + 2*(8*b^2*d^2*x^2 + 33*b^2*c^2 + 14*a*b*c*d - 3*a^2*d^2 + 2*(13*
b^2*c*d + a*b*d^2)*x)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 3*(5*b^3*c^3 + 15
*a*b^2*c^2*d - 5*a^2*b*c*d^2 + 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^2), -1/96*(96*sqrt(-a*c)*s
qrt(b*d)*b^2*c^2*arctan(1/2*(2*a*c + (b*c + a*d)*x)/(sqrt(-a*c)*sqrt(b*x + a)*sq
rt(d*x + c))) - 4*(8*b^2*d^2*x^2 + 33*b^2*c^2 + 14*a*b*c*d - 3*a^2*d^2 + 2*(13*b
^2*c*d + a*b*d^2)*x)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) - 3*(5*b^3*c^3 + 15*a
*b^2*c^2*d - 5*a^2*b*c*d^2 + a^3*d^3)*log(4*(2*b^2*d^2*x + b^2*c*d + a*b*d^2)*sq
rt(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^2), -1/48*(48*sqrt(-a*c)*sqrt(-b*
d)*b^2*c^2*arctan(1/2*(2*a*c + (b*c + a*d)*x)/(sqrt(-a*c)*sqrt(b*x + a)*sqrt(d*x
+ c))) - 2*(8*b^2*d^2*x^2 + 33*b^2*c^2 + 14*a*b*c*d - 3*a^2*d^2 + 2*(13*b^2*c*d
+ a*b*d^2)*x)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c) - 3*(5*b^3*c^3 + 15*a*b^2*
c^2*d - 5*a^2*b*c*d^2 + a^3*d^3)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)/(sq
rt(b*x + a)*sqrt(d*x + c)*b*d)))/(sqrt(-b*d)*b^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{\sqrt{a + b x} \left (c + d x\right )^{\frac{5}{2}}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((d*x+c)**(5/2)*(b*x+a)**(1/2)/x,x)

[Out]

Integral(sqrt(a + b*x)*(c + d*x)**(5/2)/x, x)

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GIAC/XCAS [A]  time = 0.298736, size = 448, normalized size = 2.05 \[ -\frac{2 \, \sqrt{b d} a c^{3}{\left | b \right |} \arctan \left (-\frac{b^{2} c + a b d -{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt{-a b c d} b}\right )}{\sqrt{-a b c d} b} + \frac{1}{24} \, \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} \sqrt{b x + a}{\left (2 \,{\left (b x + a\right )}{\left (\frac{4 \,{\left (b x + a\right )} d^{2}{\left | b \right |}}{b^{4}} + \frac{13 \, b^{9} c d^{5}{\left | b \right |} - 7 \, a b^{8} d^{6}{\left | b \right |}}{b^{12} d^{4}}\right )} + \frac{3 \,{\left (11 \, b^{10} c^{2} d^{4}{\left | b \right |} - 4 \, a b^{9} c d^{5}{\left | b \right |} + a^{2} b^{8} d^{6}{\left | b \right |}\right )}}{b^{12} d^{4}}\right )} - \frac{{\left (5 \, \sqrt{b d} b^{3} c^{3}{\left | b \right |} + 15 \, \sqrt{b d} a b^{2} c^{2} d{\left | b \right |} - 5 \, \sqrt{b d} a^{2} b c d^{2}{\left | b \right |} + \sqrt{b d} a^{3} d^{3}{\left | b \right |}\right )}{\rm ln}\left ({\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{16 \, b^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(sqrt(b*x + a)*(d*x + c)^(5/2)/x,x, algorithm="giac")

[Out]

-2*sqrt(b*d)*a*c^3*abs(b)*arctan(-1/2*(b^2*c + a*b*d - (sqrt(b*d)*sqrt(b*x + a)
- sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)*b)
+ 1/24*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x +
a)*d^2*abs(b)/b^4 + (13*b^9*c*d^5*abs(b) - 7*a*b^8*d^6*abs(b))/(b^12*d^4)) + 3*
(11*b^10*c^2*d^4*abs(b) - 4*a*b^9*c*d^5*abs(b) + a^2*b^8*d^6*abs(b))/(b^12*d^4))
- 1/16*(5*sqrt(b*d)*b^3*c^3*abs(b) + 15*sqrt(b*d)*a*b^2*c^2*d*abs(b) - 5*sqrt(b
*d)*a^2*b*c*d^2*abs(b) + sqrt(b*d)*a^3*d^3*abs(b))*ln((sqrt(b*d)*sqrt(b*x + a) -
sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(b^4*d)