\(\int \frac {(d+e x)^{7/2}}{b x+c x^2} \, dx\) [105]

Optimal result

Integrand size = 21, antiderivative size = 157 \[ \int \frac {(d+e x)^{7/2}}{b x+c x^2} \, dx=\frac {2 e \left (3 c^2 d^2-3 b c d e+b^2 e^2\right ) \sqrt {d+e x}}{c^3}+\frac {2 e (2 c d-b e) (d+e x)^{3/2}}{3 c^2}+\frac {2 e (d+e x)^{5/2}}{5 c}-\frac {2 d^{7/2} \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b}+\frac {2 (c d-b e)^{7/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b c^{7/2}} \] Output:

2*e*(b^2*e^2-3*b*c*d*e+3*c^2*d^2)*(e*x+d)^(1/2)/c^3+2/3*e*(-b*e+2*c*d)*(e* 
x+d)^(3/2)/c^2+2/5*e*(e*x+d)^(5/2)/c-2*d^(7/2)*arctanh((e*x+d)^(1/2)/d^(1/ 
2))/b+2*(-b*e+c*d)^(7/2)*arctanh(c^(1/2)*(e*x+d)^(1/2)/(-b*e+c*d)^(1/2))/b 
/c^(7/2)
 

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.88 \[ \int \frac {(d+e x)^{7/2}}{b x+c x^2} \, dx=\frac {2 e \sqrt {d+e x} \left (15 b^2 e^2-5 b c e (10 d+e x)+c^2 \left (58 d^2+16 d e x+3 e^2 x^2\right )\right )}{15 c^3}-\frac {2 (-c d+b e)^{7/2} \arctan \left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {-c d+b e}}\right )}{b c^{7/2}}-\frac {2 d^{7/2} \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b} \] Input:

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

Output:

(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*(-(c*d) + b*e)^(7/2)*ArcTan[(Sqrt[c]*Sqr 
t[d + e*x])/Sqrt[-(c*d) + b*e]])/(b*c^(7/2)) - (2*d^(7/2)*ArcTanh[Sqrt[d + 
 e*x]/Sqrt[d]])/b
 

Rubi [A] (verified)

Time = 0.94 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.16, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {1146, 1196, 1196, 1197, 25, 27, 1480, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 1146

\(\displaystyle \frac {\int \frac {(d+e x)^{3/2} \left (c d^2+e (2 c d-b e) x\right )}{c x^2+b x}dx}{c}+\frac {2 e (d+e x)^{5/2}}{5 c}\)

\(\Big \downarrow \) 1196

\(\displaystyle \frac {\frac {\int \frac {\sqrt {d+e x} \left (c^2 d^3+e \left (3 c^2 d^2-3 b c e d+b^2 e^2\right ) x\right )}{c x^2+b x}dx}{c}+\frac {2 e (d+e x)^{3/2} (2 c d-b e)}{3 c}}{c}+\frac {2 e (d+e x)^{5/2}}{5 c}\)

\(\Big \downarrow \) 1196

\(\displaystyle \frac {\frac {\frac {\int \frac {c^3 d^4+e (2 c d-b e) \left (2 c^2 d^2-2 b c e d+b^2 e^2\right ) x}{\sqrt {d+e x} \left (c x^2+b x\right )}dx}{c}+\frac {2 e \sqrt {d+e x} \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )}{c}}{c}+\frac {2 e (d+e x)^{3/2} (2 c d-b e)}{3 c}}{c}+\frac {2 e (d+e x)^{5/2}}{5 c}\)

\(\Big \downarrow \) 1197

\(\displaystyle \frac {\frac {\frac {2 \int -\frac {e \left (d (c d-b e) \left (3 c^2 d^2-3 b c e d+b^2 e^2\right )-(2 c d-b e) \left (2 c^2 d^2-2 b c e d+b^2 e^2\right ) (d+e x)\right )}{c (d+e x)^2-(2 c d-b e) (d+e x)+d (c d-b e)}d\sqrt {d+e x}}{c}+\frac {2 e \sqrt {d+e x} \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )}{c}}{c}+\frac {2 e (d+e x)^{3/2} (2 c d-b e)}{3 c}}{c}+\frac {2 e (d+e x)^{5/2}}{5 c}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {2 e \sqrt {d+e x} \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )}{c}-\frac {2 \int \frac {e \left (d (c d-b e) \left (3 c^2 d^2-3 b c e d+b^2 e^2\right )-(2 c d-b e) \left (2 c^2 d^2-2 b c e d+b^2 e^2\right ) (d+e x)\right )}{c (d+e x)^2-(2 c d-b e) (d+e x)+d (c d-b e)}d\sqrt {d+e x}}{c}}{c}+\frac {2 e (d+e x)^{3/2} (2 c d-b e)}{3 c}}{c}+\frac {2 e (d+e x)^{5/2}}{5 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {2 e \sqrt {d+e x} \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )}{c}-\frac {2 e \int \frac {d (c d-b e) \left (3 c^2 d^2-3 b c e d+b^2 e^2\right )-(2 c d-b e) \left (2 c^2 d^2-2 b c e d+b^2 e^2\right ) (d+e x)}{c (d+e x)^2-(2 c d-b e) (d+e x)+d (c d-b e)}d\sqrt {d+e x}}{c}}{c}+\frac {2 e (d+e x)^{3/2} (2 c d-b e)}{3 c}}{c}+\frac {2 e (d+e x)^{5/2}}{5 c}\)

\(\Big \downarrow \) 1480

\(\displaystyle \frac {\frac {\frac {2 e \sqrt {d+e x} \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )}{c}-\frac {2 e \left (\frac {(c d-b e)^4 \int \frac {1}{-c d+b e+c (d+e x)}d\sqrt {d+e x}}{b e}-\frac {c^4 d^4 \int \frac {1}{c (d+e x)-c d}d\sqrt {d+e x}}{b e}\right )}{c}}{c}+\frac {2 e (d+e x)^{3/2} (2 c d-b e)}{3 c}}{c}+\frac {2 e (d+e x)^{5/2}}{5 c}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\frac {2 e \sqrt {d+e x} \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )}{c}-\frac {2 e \left (\frac {c^3 d^{7/2} \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b e}-\frac {(c d-b e)^{7/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b \sqrt {c} e}\right )}{c}}{c}+\frac {2 e (d+e x)^{3/2} (2 c d-b e)}{3 c}}{c}+\frac {2 e (d+e x)^{5/2}}{5 c}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 1146

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol 
] :> Simp[e*((d + e*x)^(m - 1)/(c*(m - 1))), x] + Simp[1/c   Int[(d + e*x)^ 
(m - 2)*(Simp[c*d^2 - a*e^2 + e*(2*c*d - b*e)*x, x]/(a + b*x + c*x^2)), x], 
 x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[m, 1]
 

rule 1196

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + 
(c_.)*(x_)^2), x_Symbol] :> Simp[g*((d + e*x)^m/(c*m)), x] + Simp[1/c   Int 
[(d + e*x)^(m - 1)*(Simp[c*d*f - a*e*g + (g*c*d - b*e*g + c*e*f)*x, x]/(a + 
 b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && FractionQ[m] & 
& GtQ[m, 0]
 

rule 1197

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)), x_Symbol] :> Simp[2   Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - 
b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; Fr 
eeQ[{a, b, c, d, e, f, g}, x]
 

rule 1480

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : 
> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q))   Int[1/( 
b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q))   Int[1/(b/2 
+ q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] 
 && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
 

Maple [A] (verified)

Time = 1.07 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.90

method	result	size
pseudoelliptic	\(-\frac {2 \left (\left (b e -c d \right )^{4} \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )-\left (-\operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right ) d^{\frac {7}{2}} c^{3}+e \sqrt {e x +d}\, \left (b^{2} e^{2}-\frac {10 e c \left (\frac {e x}{10}+d \right ) b}{3}+\frac {58 c^{2} \left (\frac {3}{58} e^{2} x^{2}+\frac {8}{29} d e x +d^{2}\right )}{15}\right ) b \right ) \sqrt {c \left (b e -c d \right )}\right )}{\sqrt {c \left (b e -c d \right )}\, b \,c^{3}}\)	\(141\)
derivativedivides	\(2 e \left (\frac {\frac {\left (e x +d \right )^{\frac {5}{2}} c^{2}}{5}-\frac {b c e \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {2 c^{2} d \left (e x +d \right )^{\frac {3}{2}}}{3}+b^{2} e^{2} \sqrt {e x +d}-3 b c d e \sqrt {e x +d}+3 c^{2} d^{2} \sqrt {e x +d}}{c^{3}}+\frac {\left (-b^{4} e^{4}+4 d \,e^{3} b^{3} c -6 d^{2} e^{2} b^{2} c^{2}+4 d^{3} e b \,c^{3}-d^{4} c^{4}\right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{c^{3} b e \sqrt {c \left (b e -c d \right )}}-\frac {d^{\frac {7}{2}} \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b e}\right )\)	\(207\)
default	\(2 e \left (\frac {\frac {\left (e x +d \right )^{\frac {5}{2}} c^{2}}{5}-\frac {b c e \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {2 c^{2} d \left (e x +d \right )^{\frac {3}{2}}}{3}+b^{2} e^{2} \sqrt {e x +d}-3 b c d e \sqrt {e x +d}+3 c^{2} d^{2} \sqrt {e x +d}}{c^{3}}+\frac {\left (-b^{4} e^{4}+4 d \,e^{3} b^{3} c -6 d^{2} e^{2} b^{2} c^{2}+4 d^{3} e b \,c^{3}-d^{4} c^{4}\right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{c^{3} b e \sqrt {c \left (b e -c d \right )}}-\frac {d^{\frac {7}{2}} \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b e}\right )\)	\(207\)

Input:

int((e*x+d)^(7/2)/(c*x^2+b*x),x,method=_RETURNVERBOSE)
 

Output:

-2*((b*e-c*d)^4*arctan(c*(e*x+d)^(1/2)/(c*(b*e-c*d))^(1/2))-(-arctanh((e*x 
+d)^(1/2)/d^(1/2))*d^(7/2)*c^3+e*(e*x+d)^(1/2)*(b^2*e^2-10/3*e*c*(1/10*e*x 
+d)*b+58/15*c^2*(3/58*e^2*x^2+8/29*d*e*x+d^2))*b)*(c*(b*e-c*d))^(1/2))/(c* 
(b*e-c*d))^(1/2)/b/c^3
 

Fricas [A] (verification not implemented)

Time = 0.62 (sec) , antiderivative size = 816, normalized size of antiderivative = 5.20 \[ \int \frac {(d+e x)^{7/2}}{b x+c x^2} \, dx =\text {Too large to display} \] Input:

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

Output:

[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)*sqrt(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)*c*s 
qrt(-(c*d - b*e)/c)/(c*d - b*e)) + 2*(3*b*c^2*e^3*x^2 + 58*b*c^2*d^2*e - 5 
0*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(-d)/sqrt(e*x + 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 + (1 
6*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(sqrt(-d)/sqrt(e*x + 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)*arctan(-sqrt(e*x + d)*c*sqrt(-( 
c*d - b*e)/c)/(c*d - b*e)) + (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] (verification not implemented)

Time = 2.62 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.54 \[ \int \frac {(d+e x)^{7/2}}{b x+c x^2} \, dx=\begin {cases} \frac {2 \left (\frac {e^{2} \left (d + e x\right )^{\frac {5}{2}}}{5 c} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (- b e^{3} + 2 c d e^{2}\right )}{3 c^{2}} + \frac {\sqrt {d + e x} \left (b^{2} e^{4} - 3 b c d e^{3} + 3 c^{2} d^{2} e^{2}\right )}{c^{3}} + \frac {d^{4} e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{b \sqrt {- d}} - \frac {e \left (b e - c d\right )^{4} \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {b e - c d}{c}}} \right )}}{b c^{4} \sqrt {\frac {b e - c d}{c}}}\right )}{e} & \text {for}\: e \neq 0 \\d^{\frac {7}{2}} \left (- \frac {2 c \left (\begin {cases} \frac {\frac {b}{2 c} + x}{b} & \text {for}\: c = 0 \\- \frac {\log {\left (b - 2 c \left (\frac {b}{2 c} + x\right ) \right )}}{2 c} & \text {otherwise} \end {cases}\right )}{b} - \frac {2 c \left (\begin {cases} \frac {\frac {b}{2 c} + x}{b} & \text {for}\: c = 0 \\\frac {\log {\left (b + 2 c \left (\frac {b}{2 c} + x\right ) \right )}}{2 c} & \text {otherwise} \end {cases}\right )}{b}\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((2*(e**2*(d + e*x)**(5/2)/(5*c) + (d + e*x)**(3/2)*(-b*e**3 + 2* 
c*d*e**2)/(3*c**2) + sqrt(d + e*x)*(b**2*e**4 - 3*b*c*d*e**3 + 3*c**2*d**2 
*e**2)/c**3 + d**4*e*atan(sqrt(d + e*x)/sqrt(-d))/(b*sqrt(-d)) - e*(b*e - 
c*d)**4*atan(sqrt(d + e*x)/sqrt((b*e - c*d)/c))/(b*c**4*sqrt((b*e - c*d)/c 
)))/e, Ne(e, 0)), (d**(7/2)*(-2*c*Piecewise(((b/(2*c) + x)/b, Eq(c, 0)), ( 
-log(b - 2*c*(b/(2*c) + x))/(2*c), True))/b - 2*c*Piecewise(((b/(2*c) + x) 
/b, Eq(c, 0)), (log(b + 2*c*(b/(2*c) + x))/(2*c), True))/b), True))
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(d+e x)^{7/2}}{b x+c x^2} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(b*e-c*d>0)', see `assume?` for m 
ore detail
 

Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.41 \[ \int \frac {(d+e x)^{7/2}}{b x+c x^2} \, dx=\frac {2 \, d^{4} \arctan \left (\frac {\sqrt {e x + d}}{\sqrt {-d}}\right )}{b \sqrt {-d}} - \frac {2 \, {\left (c^{4} d^{4} - 4 \, b c^{3} d^{3} e + 6 \, b^{2} c^{2} d^{2} e^{2} - 4 \, b^{3} c d e^{3} + b^{4} e^{4}\right )} \arctan \left (\frac {\sqrt {e x + d} c}{\sqrt {-c^{2} d + b c e}}\right )}{\sqrt {-c^{2} d + b c e} b c^{3}} + \frac {2 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} c^{4} e + 10 \, {\left (e x + d\right )}^{\frac {3}{2}} c^{4} d e + 45 \, \sqrt {e x + d} c^{4} d^{2} e - 5 \, {\left (e x + d\right )}^{\frac {3}{2}} b c^{3} e^{2} - 45 \, \sqrt {e x + d} b c^{3} d e^{2} + 15 \, \sqrt {e x + d} b^{2} c^{2} e^{3}\right )}}{15 \, c^{5}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 2482, normalized size of antiderivative = 15.81 \[ \int \frac {(d+e x)^{7/2}}{b x+c x^2} \, dx=\text {Too large to display} \] Input:

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

Output:

((2*e*(b*e - 2*c*d)^2)/c^3 - (2*e*(c*d^2 - b*d*e))/c^2)*(d + e*x)^(1/2) + 
(2*e*(d + e*x)^(5/2))/(5*c) + (atan(((((8*(d + e*x)^(1/2)*(b^8*e^10 + 2*c^ 
8*d^8*e^2 - 8*b*c^7*d^7*e^3 + 28*b^2*c^6*d^6*e^4 - 56*b^3*c^5*d^5*e^5 + 70 
*b^4*c^4*d^4*e^6 - 56*b^5*c^3*d^3*e^7 + 28*b^6*c^2*d^2*e^8 - 8*b^7*c*d*e^9 
))/c^5 + (((8*(b^5*c^4*d*e^6 - 3*b^2*c^7*d^4*e^3 + 6*b^3*c^6*d^3*e^4 - 4*b 
^4*c^5*d^2*e^5))/c^5 + (8*(b^3*c^7*e^3 - 2*b^2*c^8*d*e^2)*(d^7)^(1/2)*(d + 
 e*x)^(1/2))/(b*c^5))*(d^7)^(1/2))/b)*(d^7)^(1/2)*1i)/b + (((8*(d + e*x)^( 
1/2)*(b^8*e^10 + 2*c^8*d^8*e^2 - 8*b*c^7*d^7*e^3 + 28*b^2*c^6*d^6*e^4 - 56 
*b^3*c^5*d^5*e^5 + 70*b^4*c^4*d^4*e^6 - 56*b^5*c^3*d^3*e^7 + 28*b^6*c^2*d^ 
2*e^8 - 8*b^7*c*d*e^9))/c^5 - (((8*(b^5*c^4*d*e^6 - 3*b^2*c^7*d^4*e^3 + 6* 
b^3*c^6*d^3*e^4 - 4*b^4*c^5*d^2*e^5))/c^5 - (8*(b^3*c^7*e^3 - 2*b^2*c^8*d* 
e^2)*(d^7)^(1/2)*(d + e*x)^(1/2))/(b*c^5))*(d^7)^(1/2))/b)*(d^7)^(1/2)*1i) 
/b)/((16*(b^7*d^4*e^10 - 4*c^7*d^11*e^3 + 22*b*c^6*d^10*e^4 - 8*b^6*c*d^5* 
e^9 - 52*b^2*c^5*d^9*e^5 + 69*b^3*c^4*d^8*e^6 - 56*b^4*c^3*d^7*e^7 + 28*b^ 
5*c^2*d^6*e^8))/c^5 + (((8*(d + e*x)^(1/2)*(b^8*e^10 + 2*c^8*d^8*e^2 - 8*b 
*c^7*d^7*e^3 + 28*b^2*c^6*d^6*e^4 - 56*b^3*c^5*d^5*e^5 + 70*b^4*c^4*d^4*e^ 
6 - 56*b^5*c^3*d^3*e^7 + 28*b^6*c^2*d^2*e^8 - 8*b^7*c*d*e^9))/c^5 + (((8*( 
b^5*c^4*d*e^6 - 3*b^2*c^7*d^4*e^3 + 6*b^3*c^6*d^3*e^4 - 4*b^4*c^5*d^2*e^5) 
)/c^5 + (8*(b^3*c^7*e^3 - 2*b^2*c^8*d*e^2)*(d^7)^(1/2)*(d + e*x)^(1/2))/(b 
*c^5))*(d^7)^(1/2))/b)*(d^7)^(1/2))/b - (((8*(d + e*x)^(1/2)*(b^8*e^10 ...
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 329, normalized size of antiderivative = 2.10 \[ \int \frac {(d+e x)^{7/2}}{b x+c x^2} \, dx=\frac {-30 \sqrt {c}\, \sqrt {b e -c d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c}{\sqrt {c}\, \sqrt {b e -c d}}\right ) b^{3} e^{3}+90 \sqrt {c}\, \sqrt {b e -c d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c}{\sqrt {c}\, \sqrt {b e -c d}}\right ) b^{2} c d \,e^{2}-90 \sqrt {c}\, \sqrt {b e -c d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c}{\sqrt {c}\, \sqrt {b e -c d}}\right ) b \,c^{2} d^{2} e +30 \sqrt {c}\, \sqrt {b e -c d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c}{\sqrt {c}\, \sqrt {b e -c d}}\right ) c^{3} d^{3}+30 \sqrt {e x +d}\, b^{3} c \,e^{3}-100 \sqrt {e x +d}\, b^{2} c^{2} d \,e^{2}-10 \sqrt {e x +d}\, b^{2} c^{2} e^{3} x +116 \sqrt {e x +d}\, b \,c^{3} d^{2} e +32 \sqrt {e x +d}\, b \,c^{3} d \,e^{2} x +6 \sqrt {e x +d}\, b \,c^{3} e^{3} x^{2}+15 \sqrt {d}\, \mathrm {log}\left (\sqrt {e x +d}-\sqrt {d}\right ) c^{4} d^{3}-15 \sqrt {d}\, \mathrm {log}\left (\sqrt {e x +d}+\sqrt {d}\right ) c^{4} d^{3}}{15 b \,c^{4}} \] Input:

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

Output:

( - 30*sqrt(c)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - 
c*d)))*b**3*e**3 + 90*sqrt(c)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt 
(c)*sqrt(b*e - c*d)))*b**2*c*d*e**2 - 90*sqrt(c)*sqrt(b*e - c*d)*atan((sqr 
t(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))*b*c**2*d**2*e + 30*sqrt(c)*sqrt(b 
*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))*c**3*d**3 + 30 
*sqrt(d + e*x)*b**3*c*e**3 - 100*sqrt(d + e*x)*b**2*c**2*d*e**2 - 10*sqrt( 
d + e*x)*b**2*c**2*e**3*x + 116*sqrt(d + e*x)*b*c**3*d**2*e + 32*sqrt(d + 
e*x)*b*c**3*d*e**2*x + 6*sqrt(d + e*x)*b*c**3*e**3*x**2 + 15*sqrt(d)*log(s 
qrt(d + e*x) - sqrt(d))*c**4*d**3 - 15*sqrt(d)*log(sqrt(d + e*x) + sqrt(d) 
)*c**4*d**3)/(15*b*c**4)