3.361 \(\int \frac{(d+e x)^{7/2}}{b x+c x^2} \, dx\)

Optimal. Leaf size=157 \[ \frac{2 e \sqrt{d+e x} \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )}{c^3}+\frac{2 (c d-b e)^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b c^{7/2}}+\frac{2 e (d+e x)^{3/2} (2 c d-b e)}{3 c^2}-\frac{2 d^{7/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b}+\frac{2 e (d+e x)^{5/2}}{5 c} \]

[Out]

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

_______________________________________________________________________________________

Rubi [A]  time = 0.782993, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333 \[ \frac{2 e \sqrt{d+e x} \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )}{c^3}+\frac{2 (c d-b e)^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b c^{7/2}}+\frac{2 e (d+e x)^{3/2} (2 c d-b e)}{3 c^2}-\frac{2 d^{7/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b}+\frac{2 e (d+e x)^{5/2}}{5 c} \]

Antiderivative was successfully verified.

[In]  Int[(d + e*x)^(7/2)/(b*x + c*x^2),x]

[Out]

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

_______________________________________________________________________________________

Rubi in Sympy [A]  time = 81.1854, size = 146, normalized size = 0.93 \[ \frac{2 e \left (d + e x\right )^{\frac{5}{2}}}{5 c} - \frac{2 e \left (d + e x\right )^{\frac{3}{2}} \left (b e - 2 c d\right )}{3 c^{2}} + \frac{2 e \sqrt{d + e x} \left (b^{2} e^{2} - 3 b c d e + 3 c^{2} d^{2}\right )}{c^{3}} - \frac{2 d^{\frac{7}{2}} \operatorname{atanh}{\left (\frac{\sqrt{d + e x}}{\sqrt{d}} \right )}}{b} - \frac{2 \left (b e - c d\right )^{\frac{7}{2}} \operatorname{atan}{\left (\frac{\sqrt{c} \sqrt{d + e x}}{\sqrt{b e - c d}} \right )}}{b c^{\frac{7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((e*x+d)**(7/2)/(c*x**2+b*x),x)

[Out]

2*e*(d + e*x)**(5/2)/(5*c) - 2*e*(d + e*x)**(3/2)*(b*e - 2*c*d)/(3*c**2) + 2*e*s
qrt(d + e*x)*(b**2*e**2 - 3*b*c*d*e + 3*c**2*d**2)/c**3 - 2*d**(7/2)*atanh(sqrt(
d + e*x)/sqrt(d))/b - 2*(b*e - c*d)**(7/2)*atan(sqrt(c)*sqrt(d + e*x)/sqrt(b*e -
c*d))/(b*c**(7/2))

_______________________________________________________________________________________

Mathematica [A]  time = 0.189219, size = 138, normalized size = 0.88 \[ \frac{2 e \sqrt{d+e x} \left (15 b^2 e^2-5 b c e (10 d+e x)+c^2 \left (58 d^2+16 d e x+3 e^2 x^2\right )\right )}{15 c^3}+\frac{2 (c d-b e)^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b c^{7/2}}-\frac{2 d^{7/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b} \]

Antiderivative was successfully verified.

[In]  Integrate[(d + e*x)^(7/2)/(b*x + c*x^2),x]

[Out]

(2*e*Sqrt[d + e*x]*(15*b^2*e^2 - 5*b*c*e*(10*d + e*x) + c^2*(58*d^2 + 16*d*e*x +
3*e^2*x^2)))/(15*c^3) - (2*d^(7/2)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/b + (2*(c*d
- b*e)^(7/2)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(b*c^(7/2))

_______________________________________________________________________________________

Maple [B]  time = 0.028, size = 336, normalized size = 2.1 \[{\frac{2\,e}{5\,c} \left ( ex+d \right ) ^{{\frac{5}{2}}}}-{\frac{2\,b{e}^{2}}{3\,{c}^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+{\frac{4\,de}{3\,c} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+2\,{\frac{{b}^{2}{e}^{3}\sqrt{ex+d}}{{c}^{3}}}-6\,{\frac{bd{e}^{2}\sqrt{ex+d}}{{c}^{2}}}+6\,{\frac{e{d}^{2}\sqrt{ex+d}}{c}}-2\,{\frac{{b}^{3}{e}^{4}}{{c}^{3}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{c\sqrt{ex+d}}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+8\,{\frac{{b}^{2}{e}^{3}d}{{c}^{2}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{c\sqrt{ex+d}}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-12\,{\frac{b{e}^{2}{d}^{2}}{c\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{c\sqrt{ex+d}}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+8\,{\frac{e{d}^{3}}{\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{c\sqrt{ex+d}}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-2\,{\frac{c{d}^{4}}{b\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{c\sqrt{ex+d}}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-2\,{\frac{{d}^{7/2}}{b}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) } \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((e*x+d)^(7/2)/(c*x^2+b*x),x)

[Out]

2/5*e*(e*x+d)^(5/2)/c-2/3/c^2*(e*x+d)^(3/2)*b*e^2+4/3*e/c*(e*x+d)^(3/2)*d+2/c^3*
b^2*e^3*(e*x+d)^(1/2)-6/c^2*b*d*e^2*(e*x+d)^(1/2)+6*e/c*d^2*(e*x+d)^(1/2)-2/c^3*
b^3*e^4/((b*e-c*d)*c)^(1/2)*arctan(c*(e*x+d)^(1/2)/((b*e-c*d)*c)^(1/2))+8/c^2*b^
2*e^3/((b*e-c*d)*c)^(1/2)*arctan(c*(e*x+d)^(1/2)/((b*e-c*d)*c)^(1/2))*d-12/c*b*e
^2/((b*e-c*d)*c)^(1/2)*arctan(c*(e*x+d)^(1/2)/((b*e-c*d)*c)^(1/2))*d^2+8*e/((b*e
-c*d)*c)^(1/2)*arctan(c*(e*x+d)^(1/2)/((b*e-c*d)*c)^(1/2))*d^3-2*c/b/((b*e-c*d)*
c)^(1/2)*arctan(c*(e*x+d)^(1/2)/((b*e-c*d)*c)^(1/2))*d^4-2*d^(7/2)*arctanh((e*x+
d)^(1/2)/d^(1/2))/b

_______________________________________________________________________________________

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((e*x + d)^(7/2)/(c*x^2 + b*x),x, algorithm="maxima")

[Out]

Exception raised: ValueError

_______________________________________________________________________________________

Fricas [A]  time = 1.08893, size = 1, normalized size = 0.01 \[ \text{result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((e*x + d)^(7/2)/(c*x^2 + b*x),x, algorithm="fricas")

[Out]

[1/15*(15*c^3*d^(7/2)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - 15*(c^3*d^3
- 3*b*c^2*d^2*e + 3*b^2*c*d*e^2 - b^3*e^3)*sqrt((c*d - b*e)/c)*log((c*e*x + 2*c
*d - b*e - 2*sqrt(e*x + d)*c*sqrt((c*d - b*e)/c))/(c*x + b)) + 2*(3*b*c^2*e^3*x^
2 + 58*b*c^2*d^2*e - 50*b^2*c*d*e^2 + 15*b^3*e^3 + (16*b*c^2*d*e^2 - 5*b^2*c*e^3
)*x)*sqrt(e*x + d))/(b*c^3), 1/15*(15*c^3*d^(7/2)*log((e*x - 2*sqrt(e*x + d)*sqr
t(d) + 2*d)/x) + 30*(c^3*d^3 - 3*b*c^2*d^2*e + 3*b^2*c*d*e^2 - b^3*e^3)*sqrt(-(c
*d - b*e)/c)*arctan(sqrt(e*x + d)/sqrt(-(c*d - b*e)/c)) + 2*(3*b*c^2*e^3*x^2 + 5
8*b*c^2*d^2*e - 50*b^2*c*d*e^2 + 15*b^3*e^3 + (16*b*c^2*d*e^2 - 5*b^2*c*e^3)*x)*
sqrt(e*x + d))/(b*c^3), -1/15*(30*c^3*sqrt(-d)*d^3*arctan(sqrt(e*x + d)/sqrt(-d)
) + 15*(c^3*d^3 - 3*b*c^2*d^2*e + 3*b^2*c*d*e^2 - b^3*e^3)*sqrt((c*d - b*e)/c)*l
og((c*e*x + 2*c*d - b*e - 2*sqrt(e*x + d)*c*sqrt((c*d - b*e)/c))/(c*x + b)) - 2*
(3*b*c^2*e^3*x^2 + 58*b*c^2*d^2*e - 50*b^2*c*d*e^2 + 15*b^3*e^3 + (16*b*c^2*d*e^
2 - 5*b^2*c*e^3)*x)*sqrt(e*x + d))/(b*c^3), -2/15*(15*c^3*sqrt(-d)*d^3*arctan(sq
rt(e*x + d)/sqrt(-d)) - 15*(c^3*d^3 - 3*b*c^2*d^2*e + 3*b^2*c*d*e^2 - b^3*e^3)*s
qrt(-(c*d - b*e)/c)*arctan(sqrt(e*x + d)/sqrt(-(c*d - b*e)/c)) - (3*b*c^2*e^3*x^
2 + 58*b*c^2*d^2*e - 50*b^2*c*d*e^2 + 15*b^3*e^3 + (16*b*c^2*d*e^2 - 5*b^2*c*e^3
)*x)*sqrt(e*x + d))/(b*c^3)]

_______________________________________________________________________________________

Sympy [A]  time = 81.362, size = 337, normalized size = 2.15 \[ \frac{2 e \left (d + e x\right )^{\frac{5}{2}}}{5 c} + \frac{\left (d + e x\right )^{\frac{3}{2}} \left (- 2 b e^{2} + 4 c d e\right )}{3 c^{2}} + \frac{\sqrt{d + e x} \left (2 b^{2} e^{3} - 6 b c d e^{2} + 6 c^{2} d^{2} e\right )}{c^{3}} - \frac{2 d^{4} \left (\begin{cases} - \frac{\operatorname{atan}{\left (\frac{\sqrt{d + e x}}{\sqrt{- d}} \right )}}{\sqrt{- d}} & \text{for}\: - d > 0 \\\frac{\operatorname{acoth}{\left (\frac{\sqrt{d + e x}}{\sqrt{d}} \right )}}{\sqrt{d}} & \text{for}\: - d < 0 \wedge d < d + e x \\\frac{\operatorname{atanh}{\left (\frac{\sqrt{d + e x}}{\sqrt{d}} \right )}}{\sqrt{d}} & \text{for}\: d > d + e x \wedge - d < 0 \end{cases}\right )}{b} - \frac{2 \left (b e - c d\right )^{4} \left (\begin{cases} \frac{\operatorname{atan}{\left (\frac{\sqrt{d + e x}}{\sqrt{\frac{b e - c d}{c}}} \right )}}{c \sqrt{\frac{b e - c d}{c}}} & \text{for}\: \frac{b e - c d}{c} > 0 \\- \frac{\operatorname{acoth}{\left (\frac{\sqrt{d + e x}}{\sqrt{\frac{- b e + c d}{c}}} \right )}}{c \sqrt{\frac{- b e + c d}{c}}} & \text{for}\: d + e x > \frac{- b e + c d}{c} \wedge \frac{b e - c d}{c} < 0 \\- \frac{\operatorname{atanh}{\left (\frac{\sqrt{d + e x}}{\sqrt{\frac{- b e + c d}{c}}} \right )}}{c \sqrt{\frac{- b e + c d}{c}}} & \text{for}\: \frac{b e - c d}{c} < 0 \wedge d + e x < \frac{- b e + c d}{c} \end{cases}\right )}{b c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((e*x+d)**(7/2)/(c*x**2+b*x),x)

[Out]

2*e*(d + e*x)**(5/2)/(5*c) + (d + e*x)**(3/2)*(-2*b*e**2 + 4*c*d*e)/(3*c**2) + s
qrt(d + e*x)*(2*b**2*e**3 - 6*b*c*d*e**2 + 6*c**2*d**2*e)/c**3 - 2*d**4*Piecewis
e((-atan(sqrt(d + e*x)/sqrt(-d))/sqrt(-d), -d > 0), (acoth(sqrt(d + e*x)/sqrt(d)
)/sqrt(d), (-d < 0) & (d < d + e*x)), (atanh(sqrt(d + e*x)/sqrt(d))/sqrt(d), (-d
< 0) & (d > d + e*x)))/b - 2*(b*e - c*d)**4*Piecewise((atan(sqrt(d + e*x)/sqrt(
(b*e - c*d)/c))/(c*sqrt((b*e - c*d)/c)), (b*e - c*d)/c > 0), (-acoth(sqrt(d + e*
x)/sqrt((-b*e + c*d)/c))/(c*sqrt((-b*e + c*d)/c)), ((b*e - c*d)/c < 0) & (d + e*
x > (-b*e + c*d)/c)), (-atanh(sqrt(d + e*x)/sqrt((-b*e + c*d)/c))/(c*sqrt((-b*e
+ c*d)/c)), ((b*e - c*d)/c < 0) & (d + e*x < (-b*e + c*d)/c)))/(b*c**3)

_______________________________________________________________________________________

GIAC/XCAS [A]  time = 0.215887, size = 309, normalized size = 1.97 \[ \frac{2 \, d^{4} \arctan \left (\frac{\sqrt{x e + d}}{\sqrt{-d}}\right )}{b \sqrt{-d}} - \frac{2 \,{\left (c^{4} d^{4} - 4 \, b c^{3} d^{3} e + 6 \, b^{2} c^{2} d^{2} e^{2} - 4 \, b^{3} c d e^{3} + b^{4} e^{4}\right )} \arctan \left (\frac{\sqrt{x e + d} c}{\sqrt{-c^{2} d + b c e}}\right )}{\sqrt{-c^{2} d + b c e} b c^{3}} + \frac{2 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} c^{4} e + 10 \,{\left (x e + d\right )}^{\frac{3}{2}} c^{4} d e + 45 \, \sqrt{x e + d} c^{4} d^{2} e - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} b c^{3} e^{2} - 45 \, \sqrt{x e + d} b c^{3} d e^{2} + 15 \, \sqrt{x e + d} b^{2} c^{2} e^{3}\right )}}{15 \, c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((e*x + d)^(7/2)/(c*x^2 + b*x),x, algorithm="giac")

[Out]

2*d^4*arctan(sqrt(x*e + d)/sqrt(-d))/(b*sqrt(-d)) - 2*(c^4*d^4 - 4*b*c^3*d^3*e +
6*b^2*c^2*d^2*e^2 - 4*b^3*c*d*e^3 + b^4*e^4)*arctan(sqrt(x*e + d)*c/sqrt(-c^2*d
+ b*c*e))/(sqrt(-c^2*d + b*c*e)*b*c^3) + 2/15*(3*(x*e + d)^(5/2)*c^4*e + 10*(x*
e + d)^(3/2)*c^4*d*e + 45*sqrt(x*e + d)*c^4*d^2*e - 5*(x*e + d)^(3/2)*b*c^3*e^2
- 45*sqrt(x*e + d)*b*c^3*d*e^2 + 15*sqrt(x*e + d)*b^2*c^2*e^3)/c^5