3.1493 \(\int \frac{(a+b x)^{5/2}}{\sqrt{c+d x}} \, dx\)

Optimal. Leaf size=148 \[ -\frac{5 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{8 \sqrt{b} d^{7/2}}+\frac{5 \sqrt{a+b x} \sqrt{c+d x} (b c-a d)^2}{8 d^3}-\frac{5 (a+b x)^{3/2} \sqrt{c+d x} (b c-a d)}{12 d^2}+\frac{(a+b x)^{5/2} \sqrt{c+d x}}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.176396, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158 \[ -\frac{5 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{8 \sqrt{b} d^{7/2}}+\frac{5 \sqrt{a+b x} \sqrt{c+d x} (b c-a d)^2}{8 d^3}-\frac{5 (a+b x)^{3/2} \sqrt{c+d x} (b c-a d)}{12 d^2}+\frac{(a+b x)^{5/2} \sqrt{c+d x}}{3 d} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 25.9131, size = 133, normalized size = 0.9 \[ \frac{\left (a + b x\right )^{\frac{5}{2}} \sqrt{c + d x}}{3 d} + \frac{5 \left (a + b x\right )^{\frac{3}{2}} \sqrt{c + d x} \left (a d - b c\right )}{12 d^{2}} + \frac{5 \sqrt{a + b x} \sqrt{c + d x} \left (a d - b c\right )^{2}}{8 d^{3}} + \frac{5 \left (a d - b c\right )^{3} \operatorname{atanh}{\left (\frac{\sqrt{d} \sqrt{a + b x}}{\sqrt{b} \sqrt{c + d x}} \right )}}{8 \sqrt{b} d^{\frac{7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Mathematica [A]  time = 0.127954, size = 138, normalized size = 0.93 \[ \frac{\sqrt{a+b x} \sqrt{c+d x} \left (33 a^2 d^2+2 a b d (13 d x-20 c)+b^2 \left (15 c^2-10 c d x+8 d^2 x^2\right )\right )}{24 d^3}-\frac{5 (b c-a d)^3 \log \left (2 \sqrt{b} \sqrt{d} \sqrt{a+b x} \sqrt{c+d x}+a d+b c+2 b d x\right )}{16 \sqrt{b} d^{7/2}} \]

Antiderivative was successfully verified.

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

[Out]

(Sqrt[a + b*x]*Sqrt[c + d*x]*(33*a^2*d^2 + 2*a*b*d*(-20*c + 13*d*x) + b^2*(15*c^
2 - 10*c*d*x + 8*d^2*x^2)))/(24*d^3) - (5*(b*c - a*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*Sqrt[b]*d^(7/2))

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Maple [B]  time = 0.01, size = 465, normalized size = 3.1 \[{\frac{1}{3\,d} \left ( bx+a \right ) ^{{\frac{5}{2}}}\sqrt{dx+c}}+{\frac{5\,a}{12\,d} \left ( bx+a \right ) ^{{\frac{3}{2}}}\sqrt{dx+c}}-{\frac{5\,bc}{12\,{d}^{2}} \left ( bx+a \right ) ^{{\frac{3}{2}}}\sqrt{dx+c}}+{\frac{5\,{a}^{2}}{8\,d}\sqrt{bx+a}\sqrt{dx+c}}-{\frac{5\,abc}{4\,{d}^{2}}\sqrt{bx+a}\sqrt{dx+c}}+{\frac{5\,{b}^{2}{c}^{2}}{8\,{d}^{3}}\sqrt{bx+a}\sqrt{dx+c}}+{\frac{5\,{a}^{3}}{16}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\ln \left ({1 \left ({\frac{ad}{2}}+{\frac{bc}{2}}+bdx \right ){\frac{1}{\sqrt{bd}}}}+\sqrt{d{x}^{2}b+ \left ( ad+bc \right ) x+ac} \right ){\frac{1}{\sqrt{bx+a}}}{\frac{1}{\sqrt{dx+c}}}{\frac{1}{\sqrt{bd}}}}-{\frac{15\,{a}^{2}bc}{16\,d}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\ln \left ({1 \left ({\frac{ad}{2}}+{\frac{bc}{2}}+bdx \right ){\frac{1}{\sqrt{bd}}}}+\sqrt{d{x}^{2}b+ \left ( ad+bc \right ) x+ac} \right ){\frac{1}{\sqrt{bx+a}}}{\frac{1}{\sqrt{dx+c}}}{\frac{1}{\sqrt{bd}}}}+{\frac{15\,a{b}^{2}{c}^{2}}{16\,{d}^{2}}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\ln \left ({1 \left ({\frac{ad}{2}}+{\frac{bc}{2}}+bdx \right ){\frac{1}{\sqrt{bd}}}}+\sqrt{d{x}^{2}b+ \left ( ad+bc \right ) x+ac} \right ){\frac{1}{\sqrt{bx+a}}}{\frac{1}{\sqrt{dx+c}}}{\frac{1}{\sqrt{bd}}}}-{\frac{5\,{b}^{3}{c}^{3}}{16\,{d}^{3}}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\ln \left ({1 \left ({\frac{ad}{2}}+{\frac{bc}{2}}+bdx \right ){\frac{1}{\sqrt{bd}}}}+\sqrt{d{x}^{2}b+ \left ( ad+bc \right ) x+ac} \right ){\frac{1}{\sqrt{bx+a}}}{\frac{1}{\sqrt{dx+c}}}{\frac{1}{\sqrt{bd}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/3*(b*x+a)^(5/2)*(d*x+c)^(1/2)/d+5/12/d*(b*x+a)^(3/2)*(d*x+c)^(1/2)*a-5/12/d^2*
(b*x+a)^(3/2)*(d*x+c)^(1/2)*b*c+5/8/d*(b*x+a)^(1/2)*(d*x+c)^(1/2)*a^2-5/4/d^2*(b
*x+a)^(1/2)*(d*x+c)^(1/2)*a*b*c+5/8/d^3*(b*x+a)^(1/2)*(d*x+c)^(1/2)*b^2*c^2+5/16
*((b*x+a)*(d*x+c))^(1/2)/(b*x+a)^(1/2)/(d*x+c)^(1/2)*ln((1/2*a*d+1/2*b*c+b*d*x)/
(b*d)^(1/2)+(d*x^2*b+(a*d+b*c)*x+a*c)^(1/2))/(b*d)^(1/2)*a^3-15/16/d*((b*x+a)*(d
*x+c))^(1/2)/(b*x+a)^(1/2)/(d*x+c)^(1/2)*ln((1/2*a*d+1/2*b*c+b*d*x)/(b*d)^(1/2)+
(d*x^2*b+(a*d+b*c)*x+a*c)^(1/2))/(b*d)^(1/2)*a^2*b*c+15/16/d^2*((b*x+a)*(d*x+c))
^(1/2)/(b*x+a)^(1/2)/(d*x+c)^(1/2)*ln((1/2*a*d+1/2*b*c+b*d*x)/(b*d)^(1/2)+(d*x^2
*b+(a*d+b*c)*x+a*c)^(1/2))/(b*d)^(1/2)*a*b^2*c^2-5/16/d^3*((b*x+a)*(d*x+c))^(1/2
)/(b*x+a)^(1/2)/(d*x+c)^(1/2)*ln((1/2*a*d+1/2*b*c+b*d*x)/(b*d)^(1/2)+(d*x^2*b+(a
*d+b*c)*x+a*c)^(1/2))/(b*d)^(1/2)*b^3*c^3

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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, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.249362, size = 1, normalized size = 0.01 \[ \left [\frac{4 \,{\left (8 \, b^{2} d^{2} x^{2} + 15 \, b^{2} c^{2} - 40 \, a b c d + 33 \, a^{2} d^{2} - 2 \,{\left (5 \, b^{2} c d - 13 \, a b d^{2}\right )} x\right )} \sqrt{b d} \sqrt{b x + a} \sqrt{d x + c} - 15 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left (4 \,{\left (2 \, b^{2} d^{2} x + b^{2} c d + a b d^{2}\right )} \sqrt{b x + a} \sqrt{d x + c} +{\left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \,{\left (b^{2} c d + a b d^{2}\right )} x\right )} \sqrt{b d}\right )}{96 \, \sqrt{b d} d^{3}}, \frac{2 \,{\left (8 \, b^{2} d^{2} x^{2} + 15 \, b^{2} c^{2} - 40 \, a b c d + 33 \, a^{2} d^{2} - 2 \,{\left (5 \, b^{2} c d - 13 \, a b d^{2}\right )} x\right )} \sqrt{-b d} \sqrt{b x + a} \sqrt{d x + c} - 15 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac{{\left (2 \, b d x + b c + a d\right )} \sqrt{-b d}}{2 \, \sqrt{b x + a} \sqrt{d x + c} b d}\right )}{48 \, \sqrt{-b d} d^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

[1/96*(4*(8*b^2*d^2*x^2 + 15*b^2*c^2 - 40*a*b*c*d + 33*a^2*d^2 - 2*(5*b^2*c*d -
13*a*b*d^2)*x)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) - 15*(b^3*c^3 - 3*a*b^2*c^2
*d + 3*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)*d^3), 1/48*(2*(8*b^2*d^2*x^2 + 15*b^2*c^2 -
40*a*b*c*d + 33*a^2*d^2 - 2*(5*b^2*c*d - 13*a*b*d^2)*x)*sqrt(-b*d)*sqrt(b*x + a
)*sqrt(d*x + c) - 15*(b^3*c^3 - 3*a*b^2*c^2*d + 3*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)*d^3)]

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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GIAC/XCAS [A]  time = 0.236028, size = 267, normalized size = 1.8 \[ \frac{{\left (\sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} \sqrt{b x + a}{\left (2 \,{\left (b x + a\right )}{\left (\frac{4 \,{\left (b x + a\right )}}{b d} - \frac{5 \,{\left (b c d^{3} - a d^{4}\right )}}{b d^{5}}\right )} + \frac{15 \,{\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )}}{b d^{5}}\right )} + \frac{15 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )}{\rm ln}\left ({\left | -\sqrt{b d} \sqrt{b x + a} + \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt{b d} d^{3}}\right )} b}{24 \,{\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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
)/(b*d) - 5*(b*c*d^3 - a*d^4)/(b*d^5)) + 15*(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4
)/(b*d^5)) + 15*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - 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)*d^3))*b/a
bs(b)