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

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

[Out]

(c*(3*b*c - 7*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*a^2*x) - (c*Sqrt[a + b*x]*(c
+ d*x)^(3/2))/(2*a*x^2) - (Sqrt[c]*(3*b^2*c^2 - 10*a*b*c*d + 15*a^2*d^2)*ArcTanh
[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*a^(5/2)) + (2*d^(5/2)*ArcT
anh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/Sqrt[b]

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Rubi [A]  time = 0.449387, 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 \[ \frac{c \sqrt{a+b x} \sqrt{c+d x} (3 b c-7 a d)}{4 a^2 x}-\frac{\sqrt{c} \left (15 a^2 d^2-10 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 a^{5/2}}+\frac{2 d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{\sqrt{b}}-\frac{c \sqrt{a+b x} (c+d x)^{3/2}}{2 a x^2} \]

Antiderivative was successfully verified.

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

[Out]

(c*(3*b*c - 7*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*a^2*x) - (c*Sqrt[a + b*x]*(c
+ d*x)^(3/2))/(2*a*x^2) - (Sqrt[c]*(3*b^2*c^2 - 10*a*b*c*d + 15*a^2*d^2)*ArcTanh
[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*a^(5/2)) + (2*d^(5/2)*ArcT
anh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/Sqrt[b]

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Rubi in Sympy [A]  time = 45.2708, size = 167, normalized size = 0.94 \[ \frac{2 d^{\frac{5}{2}} \operatorname{atanh}{\left (\frac{\sqrt{d} \sqrt{a + b x}}{\sqrt{b} \sqrt{c + d x}} \right )}}{\sqrt{b}} - \frac{c \sqrt{a + b x} \left (c + d x\right )^{\frac{3}{2}}}{2 a x^{2}} - \frac{c \sqrt{a + b x} \sqrt{c + d x} \left (7 a d - 3 b c\right )}{4 a^{2} x} - \frac{\sqrt{c} \left (15 a^{2} d^{2} - 10 a b c d + 3 b^{2} c^{2}\right ) \operatorname{atanh}{\left (\frac{\sqrt{c} \sqrt{a + b x}}{\sqrt{a} \sqrt{c + d x}} \right )}}{4 a^{\frac{5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Mathematica [A]  time = 0.850479, size = 216, normalized size = 1.22 \[ \frac{1}{8} \left (-\frac{2 c \sqrt{a+b x} \sqrt{c+d x} (2 a c+9 a d x-3 b c x)}{a^2 x^2}+\frac{\sqrt{c} \log (x) \left (15 a^2 d^2-10 a b c d+3 b^2 c^2\right )}{a^{5/2}}+\frac{\sqrt{c} \left (-15 a^2 d^2+10 a b c d-3 b^2 c^2\right ) \log \left (2 \sqrt{a} \sqrt{c} \sqrt{a+b x} \sqrt{c+d x}+2 a c+a d x+b c x\right )}{a^{5/2}}+\frac{8 d^{5/2} \log \left (2 \sqrt{b} \sqrt{d} \sqrt{a+b x} \sqrt{c+d x}+a d+b c+2 b d x\right )}{\sqrt{b}}\right ) \]

Antiderivative was successfully verified.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.69528, size = 1, normalized size = 0.01 \[ \text{result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

[1/16*(8*a^2*d^2*x^2*sqrt(d/b)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2
 + 4*(2*b^2*d*x + b^2*c + a*b*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(d/b) + 8*(b^2*
c*d + a*b*d^2)*x) + (3*b^2*c^2 - 10*a*b*c*d + 15*a^2*d^2)*x^2*sqrt(c/a)*log((8*a
^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a^2*c + (a*b*c + a^2*d)*x)*s
qrt(b*x + a)*sqrt(d*x + c)*sqrt(c/a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(2*a*c^
2 - 3*(b*c^2 - 3*a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*x^2), 1/16*(16*a^2*
d^2*x^2*sqrt(-d/b)*arctan(1/2*(2*b*d*x + b*c + a*d)/(sqrt(b*x + a)*sqrt(d*x + c)
*b*sqrt(-d/b))) + (3*b^2*c^2 - 10*a*b*c*d + 15*a^2*d^2)*x^2*sqrt(c/a)*log((8*a^2
*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a^2*c + (a*b*c + a^2*d)*x)*sqr
t(b*x + a)*sqrt(d*x + c)*sqrt(c/a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(2*a*c^2
- 3*(b*c^2 - 3*a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*x^2), 1/8*(4*a^2*d^2*
x^2*sqrt(d/b)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b^2*d*x +
 b^2*c + a*b*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(d/b) + 8*(b^2*c*d + a*b*d^2)*x)
 - (3*b^2*c^2 - 10*a*b*c*d + 15*a^2*d^2)*x^2*sqrt(-c/a)*arctan(1/2*(2*a*c + (b*c
 + a*d)*x)/(sqrt(b*x + a)*sqrt(d*x + c)*a*sqrt(-c/a))) - 2*(2*a*c^2 - 3*(b*c^2 -
 3*a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*x^2), 1/8*(8*a^2*d^2*x^2*sqrt(-d/
b)*arctan(1/2*(2*b*d*x + b*c + a*d)/(sqrt(b*x + a)*sqrt(d*x + c)*b*sqrt(-d/b)))
- (3*b^2*c^2 - 10*a*b*c*d + 15*a^2*d^2)*x^2*sqrt(-c/a)*arctan(1/2*(2*a*c + (b*c
+ a*d)*x)/(sqrt(b*x + a)*sqrt(d*x + c)*a*sqrt(-c/a))) - 2*(2*a*c^2 - 3*(b*c^2 -
3*a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^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((d*x+c)**(5/2)/x**3/(b*x+a)**(1/2),x)

[Out]

Timed out

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GIAC/XCAS [A]  time = 0.646393, size = 4, normalized size = 0.02 \[ \mathit{sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

sage0*x