3.667 \(\int \frac{(a+b x)^{5/2}}{x^3 \sqrt{c+d x}} \, dx\)

Optimal. Leaf size=177 \[ -\frac{\sqrt{a} \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 c^{5/2}}+\frac{2 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{\sqrt{d}}-\frac{a \sqrt{a+b x} \sqrt{c+d x} (7 b c-3 a d)}{4 c^2 x}-\frac{a (a+b x)^{3/2} \sqrt{c+d x}}{2 c x^2} \]

[Out]

-(a*(7*b*c - 3*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*c^2*x) - (a*(a + b*x)^(3/2)*
Sqrt[c + d*x])/(2*c*x^2) - (Sqrt[a]*(15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*ArcTan
h[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*c^(5/2)) + (2*b^(5/2)*Arc
Tanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/Sqrt[d]

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Rubi [A]  time = 0.452399, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273 \[ -\frac{\sqrt{a} \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 c^{5/2}}+\frac{2 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{\sqrt{d}}-\frac{a \sqrt{a+b x} \sqrt{c+d x} (7 b c-3 a d)}{4 c^2 x}-\frac{a (a+b x)^{3/2} \sqrt{c+d x}}{2 c x^2} \]

Antiderivative was successfully verified.

[In]  Int[(a + b*x)^(5/2)/(x^3*Sqrt[c + d*x]),x]

[Out]

-(a*(7*b*c - 3*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*c^2*x) - (a*(a + b*x)^(3/2)*
Sqrt[c + d*x])/(2*c*x^2) - (Sqrt[a]*(15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*ArcTan
h[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*c^(5/2)) + (2*b^(5/2)*Arc
Tanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/Sqrt[d]

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Rubi in Sympy [A]  time = 45.0614, size = 167, normalized size = 0.94 \[ - \frac{\sqrt{a} \left (3 a^{2} d^{2} - 10 a b c d + 15 b^{2} c^{2}\right ) \operatorname{atanh}{\left (\frac{\sqrt{c} \sqrt{a + b x}}{\sqrt{a} \sqrt{c + d x}} \right )}}{4 c^{\frac{5}{2}}} - \frac{a \left (a + b x\right )^{\frac{3}{2}} \sqrt{c + d x}}{2 c x^{2}} + \frac{a \sqrt{a + b x} \sqrt{c + d x} \left (3 a d - 7 b c\right )}{4 c^{2} x} + \frac{2 b^{\frac{5}{2}} \operatorname{atanh}{\left (\frac{\sqrt{d} \sqrt{a + b x}}{\sqrt{b} \sqrt{c + d x}} \right )}}{\sqrt{d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((b*x+a)**(5/2)/x**3/(d*x+c)**(1/2),x)

[Out]

-sqrt(a)*(3*a**2*d**2 - 10*a*b*c*d + 15*b**2*c**2)*atanh(sqrt(c)*sqrt(a + b*x)/(
sqrt(a)*sqrt(c + d*x)))/(4*c**(5/2)) - a*(a + b*x)**(3/2)*sqrt(c + d*x)/(2*c*x**
2) + a*sqrt(a + b*x)*sqrt(c + d*x)*(3*a*d - 7*b*c)/(4*c**2*x) + 2*b**(5/2)*atanh
(sqrt(d)*sqrt(a + b*x)/(sqrt(b)*sqrt(c + d*x)))/sqrt(d)

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Mathematica [A]  time = 0.998452, size = 216, normalized size = 1.22 \[ \frac{1}{8} \left (\frac{\sqrt{a} \log (x) \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right )}{c^{5/2}}+\frac{\sqrt{a} \left (-3 a^2 d^2+10 a b c d-15 b^2 c^2\right ) \log \left (2 \sqrt{a} \sqrt{c} \sqrt{a+b x} \sqrt{c+d x}+2 a c+a d x+b c x\right )}{c^{5/2}}+\frac{8 b^{5/2} \log \left (2 \sqrt{b} \sqrt{d} \sqrt{a+b x} \sqrt{c+d x}+a d+b c+2 b d x\right )}{\sqrt{d}}+\frac{2 a \sqrt{a+b x} \sqrt{c+d x} (-2 a c+3 a d x-9 b c x)}{c^2 x^2}\right ) \]

Antiderivative was successfully verified.

[In]  Integrate[(a + b*x)^(5/2)/(x^3*Sqrt[c + d*x]),x]

[Out]

((2*a*Sqrt[a + b*x]*Sqrt[c + d*x]*(-2*a*c - 9*b*c*x + 3*a*d*x))/(c^2*x^2) + (Sqr
t[a]*(15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*Log[x])/c^(5/2) + (Sqrt[a]*(-15*b^2*c
^2 + 10*a*b*c*d - 3*a^2*d^2)*Log[2*a*c + b*c*x + a*d*x + 2*Sqrt[a]*Sqrt[c]*Sqrt[
a + b*x]*Sqrt[c + d*x]])/c^(5/2) + (8*b^(5/2)*Log[b*c + a*d + 2*b*d*x + 2*Sqrt[b
]*Sqrt[d]*Sqrt[a + b*x]*Sqrt[c + d*x]])/Sqrt[d])/8

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Maple [B]  time = 0.033, size = 354, normalized size = 2. \[ -{\frac{1}{8\,{c}^{2}{x}^{2}}\sqrt{bx+a}\sqrt{dx+c} \left ( 3\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{2}{a}^{3}{d}^{2}\sqrt{bd}-10\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{2}{a}^{2}bcd\sqrt{bd}+15\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{2}a{b}^{2}{c}^{2}\sqrt{bd}-8\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){x}^{2}{b}^{3}{c}^{2}\sqrt{ac}-6\,x{a}^{2}d\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+18\,xabc\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+4\,{a}^{2}c\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd} \right ){\frac{1}{\sqrt{ac}}}{\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }}}{\frac{1}{\sqrt{bd}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((b*x+a)^(5/2)/x^3/(d*x+c)^(1/2),x)

[Out]

-1/8*(b*x+a)^(1/2)*(d*x+c)^(1/2)/c^2*(3*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(
d*x+c))^(1/2)+2*a*c)/x)*x^2*a^3*d^2*(b*d)^(1/2)-10*ln((a*d*x+b*c*x+2*(a*c)^(1/2)
*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^2*a^2*b*c*d*(b*d)^(1/2)+15*ln((a*d*x+b*c*x+
2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^2*a*b^2*c^2*(b*d)^(1/2)-8*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^2*b^3*
c^2*(a*c)^(1/2)-6*x*a^2*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+18*x*a
*b*c*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+4*a^2*c*(a*c)^(1/2)*((b*x+a
)*(d*x+c))^(1/2)*(b*d)^(1/2))/((b*x+a)*(d*x+c))^(1/2)/x^2/(a*c)^(1/2)/(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((b*x + a)^(5/2)/(sqrt(d*x + c)*x^3),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.92938, size = 1, normalized size = 0.01 \[ \text{result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x + a)^(5/2)/(sqrt(d*x + c)*x^3),x, algorithm="fricas")

[Out]

[1/16*(8*b^2*c^2*x^2*sqrt(b/d)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2
+ 4*(2*b*d^2*x + b*c*d + a*d^2)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(b/d) + 8*(b^2*
c*d + a*b*d^2)*x) + (15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*x^2*sqrt(a/c)*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^2 + (b*c^2 + a*c*d)*x)*s
qrt(b*x + a)*sqrt(d*x + c)*sqrt(a/c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(2*a^2*
c + 3*(3*a*b*c - a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c^2*x^2), 1/16*(16*b^2*
c^2*x^2*sqrt(-b/d)*arctan(1/2*(2*b*d*x + b*c + a*d)/(sqrt(b*x + a)*sqrt(d*x + c)
*d*sqrt(-b/d))) + (15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*x^2*sqrt(a/c)*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^2 + (b*c^2 + a*c*d)*x)*sqr
t(b*x + a)*sqrt(d*x + c)*sqrt(a/c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(2*a^2*c
+ 3*(3*a*b*c - a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c^2*x^2), 1/8*(4*b^2*c^2*
x^2*sqrt(b/d)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b*d^2*x +
b*c*d + a*d^2)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(b/d) + 8*(b^2*c*d + a*b*d^2)*x)
- (15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*x^2*sqrt(-a/c)*arctan(1/2*(2*a*c + (b*c
+ a*d)*x)/(sqrt(b*x + a)*sqrt(d*x + c)*c*sqrt(-a/c))) - 2*(2*a^2*c + 3*(3*a*b*c
- a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c^2*x^2), 1/8*(8*b^2*c^2*x^2*sqrt(-b/
d)*arctan(1/2*(2*b*d*x + b*c + a*d)/(sqrt(b*x + a)*sqrt(d*x + c)*d*sqrt(-b/d)))
- (15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*x^2*sqrt(-a/c)*arctan(1/2*(2*a*c + (b*c
+ a*d)*x)/(sqrt(b*x + a)*sqrt(d*x + c)*c*sqrt(-a/c))) - 2*(2*a^2*c + 3*(3*a*b*c
- a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c^2*x^2)]

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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((b*x+a)**(5/2)/x**3/(d*x+c)**(1/2),x)

[Out]

Timed out

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GIAC/XCAS [A]  time = 0.591425, size = 4, normalized size = 0.02 \[ \mathit{sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x + a)^(5/2)/(sqrt(d*x + c)*x^3),x, algorithm="giac")

[Out]

sage0*x