\(\int \frac {1}{(d+e x)^{7/2} (b x+c x^2)} \, dx\) [112]

Optimal result

Integrand size = 21, antiderivative size = 187 \[ \int \frac {1}{(d+e x)^{7/2} \left (b x+c x^2\right )} \, dx=-\frac {2 e}{5 d (c d-b e) (d+e x)^{5/2}}-\frac {2 e (2 c d-b e)}{3 d^2 (c d-b e)^2 (d+e x)^{3/2}}-\frac {2 e \left (3 c^2 d^2-3 b c d e+b^2 e^2\right )}{d^3 (c d-b e)^3 \sqrt {d+e x}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b d^{7/2}}+\frac {2 c^{7/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b (c d-b e)^{7/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.97 \[ \int \frac {1}{(d+e x)^{7/2} \left (b x+c x^2\right )} \, dx=\frac {2 e \left (-3 b c d e \left (22 d^2+35 d e x+15 e^2 x^2\right )+b^2 e^2 \left (23 d^2+35 d e x+15 e^2 x^2\right )+c^2 d^2 \left (58 d^2+100 d e x+45 e^2 x^2\right )\right )}{15 d^3 (-c d+b e)^3 (d+e x)^{5/2}}+\frac {2 c^{7/2} \arctan \left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {-c d+b e}}\right )}{b (-c d+b e)^{7/2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b d^{7/2}} \] Input:

Integrate[1/((d + e*x)^(7/2)*(b*x + c*x^2)),x]
 

Output:

(2*e*(-3*b*c*d*e*(22*d^2 + 35*d*e*x + 15*e^2*x^2) + b^2*e^2*(23*d^2 + 35*d 
*e*x + 15*e^2*x^2) + c^2*d^2*(58*d^2 + 100*d*e*x + 45*e^2*x^2)))/(15*d^3*( 
-(c*d) + b*e)^3*(d + e*x)^(5/2)) + (2*c^(7/2)*ArcTan[(Sqrt[c]*Sqrt[d + e*x 
])/Sqrt[-(c*d) + b*e]])/(b*(-(c*d) + b*e)^(7/2)) - (2*ArcTanh[Sqrt[d + e*x 
]/Sqrt[d]])/(b*d^(7/2))
 

Rubi [A] (verified)

Time = 0.99 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.35, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1147, 1198, 1198, 1197, 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.

\(\displaystyle \int \frac {1}{\left (b x+c x^2\right ) (d+e x)^{7/2}} \, dx\)

\(\Big \downarrow \) 1147

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

\(\Big \downarrow \) 1198

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

\(\Big \downarrow \) 1198

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

\(\Big \downarrow \) 1197

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1480

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

\(\Big \downarrow \) 221

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

Input:

Int[1/((d + e*x)^(7/2)*(b*x + c*x^2)),x]
 

Output:

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

Defintions of rubi rules used

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 1147

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

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 1198

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

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 = 0.74 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.95

method	result	size
pseudoelliptic	\(2 e \left (-\frac {\operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{e b \,d^{\frac {7}{2}}}+\frac {b e -2 c d}{3 d^{2} \left (b e -c d \right )^{2} \left (e x +d \right )^{\frac {3}{2}}}+\frac {b^{2} e^{2}-3 b c d e +3 c^{2} d^{2}}{d^{3} \left (b e -c d \right )^{3} \sqrt {e x +d}}+\frac {1}{5 d \left (b e -c d \right ) \left (e x +d \right )^{\frac {5}{2}}}+\frac {c^{4} \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{\left (b e -c d \right )^{3} b e \sqrt {c \left (b e -c d \right )}}\right )\)	\(177\)
derivativedivides	\(2 e \left (\frac {c^{4} \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{\left (b e -c d \right )^{3} b e \sqrt {c \left (b e -c d \right )}}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{e b \,d^{\frac {7}{2}}}-\frac {-b e +2 c d}{3 d^{2} \left (b e -c d \right )^{2} \left (e x +d \right )^{\frac {3}{2}}}-\frac {-b^{2} e^{2}+3 b c d e -3 c^{2} d^{2}}{d^{3} \left (b e -c d \right )^{3} \sqrt {e x +d}}+\frac {1}{5 d \left (b e -c d \right ) \left (e x +d \right )^{\frac {5}{2}}}\right )\)	\(180\)
default	\(2 e \left (\frac {c^{4} \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{\left (b e -c d \right )^{3} b e \sqrt {c \left (b e -c d \right )}}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{e b \,d^{\frac {7}{2}}}-\frac {-b e +2 c d}{3 d^{2} \left (b e -c d \right )^{2} \left (e x +d \right )^{\frac {3}{2}}}-\frac {-b^{2} e^{2}+3 b c d e -3 c^{2} d^{2}}{d^{3} \left (b e -c d \right )^{3} \sqrt {e x +d}}+\frac {1}{5 d \left (b e -c d \right ) \left (e x +d \right )^{\frac {5}{2}}}\right )\)	\(180\)

Input:

int(1/(e*x+d)^(7/2)/(c*x^2+b*x),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 614 vs. \(2 (161) = 322\).

Time = 0.59 (sec) , antiderivative size = 2532, normalized size of antiderivative = 13.54 \[ \int \frac {1}{(d+e x)^{7/2} \left (b x+c x^2\right )} \, dx=\text {Too large to display} \] Input:

integrate(1/(e*x+d)^(7/2)/(c*x^2+b*x),x, algorithm="fricas")
 

Output:

[-1/15*(15*(c^3*d^4*e^3*x^3 + 3*c^3*d^5*e^2*x^2 + 3*c^3*d^6*e*x + c^3*d^7) 
*sqrt(c/(c*d - b*e))*log((c*e*x + 2*c*d - b*e - 2*(c*d - b*e)*sqrt(e*x + d 
)*sqrt(c/(c*d - b*e)))/(c*x + b)) - 15*(c^3*d^6 - 3*b*c^2*d^5*e + 3*b^2*c* 
d^4*e^2 - b^3*d^3*e^3 + (c^3*d^3*e^3 - 3*b*c^2*d^2*e^4 + 3*b^2*c*d*e^5 - b 
^3*e^6)*x^3 + 3*(c^3*d^4*e^2 - 3*b*c^2*d^3*e^3 + 3*b^2*c*d^2*e^4 - b^3*d*e 
^5)*x^2 + 3*(c^3*d^5*e - 3*b*c^2*d^4*e^2 + 3*b^2*c*d^3*e^3 - b^3*d^2*e^4)* 
x)*sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 2*(58*b*c^2*d^5* 
e - 66*b^2*c*d^4*e^2 + 23*b^3*d^3*e^3 + 15*(3*b*c^2*d^3*e^3 - 3*b^2*c*d^2* 
e^4 + b^3*d*e^5)*x^2 + 5*(20*b*c^2*d^4*e^2 - 21*b^2*c*d^3*e^3 + 7*b^3*d^2* 
e^4)*x)*sqrt(e*x + d))/(b*c^3*d^10 - 3*b^2*c^2*d^9*e + 3*b^3*c*d^8*e^2 - b 
^4*d^7*e^3 + (b*c^3*d^7*e^3 - 3*b^2*c^2*d^6*e^4 + 3*b^3*c*d^5*e^5 - b^4*d^ 
4*e^6)*x^3 + 3*(b*c^3*d^8*e^2 - 3*b^2*c^2*d^7*e^3 + 3*b^3*c*d^6*e^4 - b^4* 
d^5*e^5)*x^2 + 3*(b*c^3*d^9*e - 3*b^2*c^2*d^8*e^2 + 3*b^3*c*d^7*e^3 - b^4* 
d^6*e^4)*x), -1/15*(30*(c^3*d^4*e^3*x^3 + 3*c^3*d^5*e^2*x^2 + 3*c^3*d^6*e* 
x + c^3*d^7)*sqrt(-c/(c*d - b*e))*arctan(sqrt(e*x + d)*sqrt(-c/(c*d - b*e) 
)) - 15*(c^3*d^6 - 3*b*c^2*d^5*e + 3*b^2*c*d^4*e^2 - b^3*d^3*e^3 + (c^3*d^ 
3*e^3 - 3*b*c^2*d^2*e^4 + 3*b^2*c*d*e^5 - b^3*e^6)*x^3 + 3*(c^3*d^4*e^2 - 
3*b*c^2*d^3*e^3 + 3*b^2*c*d^2*e^4 - b^3*d*e^5)*x^2 + 3*(c^3*d^5*e - 3*b*c^ 
2*d^4*e^2 + 3*b^2*c*d^3*e^3 - b^3*d^2*e^4)*x)*sqrt(d)*log((e*x - 2*sqrt(e* 
x + d)*sqrt(d) + 2*d)/x) + 2*(58*b*c^2*d^5*e - 66*b^2*c*d^4*e^2 + 23*b^...
 

Sympy [A] (verification not implemented)

Time = 2.67 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.40 \[ \int \frac {1}{(d+e x)^{7/2} \left (b x+c x^2\right )} \, dx=\begin {cases} \frac {2 \left (\frac {e^{2}}{5 d \left (d + e x\right )^{\frac {5}{2}} \left (b e - c d\right )} + \frac {e^{2} \left (b e - 2 c d\right )}{3 d^{2} \left (d + e x\right )^{\frac {3}{2}} \left (b e - c d\right )^{2}} + \frac {e^{2} \left (b^{2} e^{2} - 3 b c d e + 3 c^{2} d^{2}\right )}{d^{3} \sqrt {d + e x} \left (b e - c d\right )^{3}} + \frac {c^{3} e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {b e - c d}{c}}} \right )}}{b \sqrt {\frac {b e - c d}{c}} \left (b e - c d\right )^{3}} + \frac {e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{b d^{3} \sqrt {- d}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {- \frac {2 c \left (\begin {cases} \frac {\frac {b}{2 c} + x}{b} & \text {for}\: c = 0 \\- \frac {\log {\left (b - 2 c \left (\frac {b}{2 c} + x\right ) \right )}}{2 c} & \text {otherwise} \end {cases}\right )}{b} - \frac {2 c \left (\begin {cases} \frac {\frac {b}{2 c} + x}{b} & \text {for}\: c = 0 \\\frac {\log {\left (b + 2 c \left (\frac {b}{2 c} + x\right ) \right )}}{2 c} & \text {otherwise} \end {cases}\right )}{b}}{d^{\frac {7}{2}}} & \text {otherwise} \end {cases} \] Input:

integrate(1/(e*x+d)**(7/2)/(c*x**2+b*x),x)
 

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(d+e x)^{7/2} \left (b x+c x^2\right )} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(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 = 0.29 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.51 \[ \int \frac {1}{(d+e x)^{7/2} \left (b x+c x^2\right )} \, dx=-\frac {2 \, c^{4} \arctan \left (\frac {\sqrt {e x + d} c}{\sqrt {-c^{2} d + b c e}}\right )}{{\left (b c^{3} d^{3} - 3 \, b^{2} c^{2} d^{2} e + 3 \, b^{3} c d e^{2} - b^{4} e^{3}\right )} \sqrt {-c^{2} d + b c e}} - \frac {2 \, {\left (45 \, {\left (e x + d\right )}^{2} c^{2} d^{2} e + 10 \, {\left (e x + d\right )} c^{2} d^{3} e + 3 \, c^{2} d^{4} e - 45 \, {\left (e x + d\right )}^{2} b c d e^{2} - 15 \, {\left (e x + d\right )} b c d^{2} e^{2} - 6 \, b c d^{3} e^{2} + 15 \, {\left (e x + d\right )}^{2} b^{2} e^{3} + 5 \, {\left (e x + d\right )} b^{2} d e^{3} + 3 \, b^{2} d^{2} e^{3}\right )}}{15 \, {\left (c^{3} d^{6} - 3 \, b c^{2} d^{5} e + 3 \, b^{2} c d^{4} e^{2} - b^{3} d^{3} e^{3}\right )} {\left (e x + d\right )}^{\frac {5}{2}}} + \frac {2 \, \arctan \left (\frac {\sqrt {e x + d}}{\sqrt {-d}}\right )}{b \sqrt {-d} d^{3}} \] Input:

integrate(1/(e*x+d)^(7/2)/(c*x^2+b*x),x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 6.48 (sec) , antiderivative size = 4068, normalized size of antiderivative = 21.75 \[ \int \frac {1}{(d+e x)^{7/2} \left (b x+c x^2\right )} \, dx=\text {Too large to display} \] Input:

int(1/((b*x + c*x^2)*(d + e*x)^(7/2)),x)
 

Output:

(atan((((-c^7*(b*e - c*d)^7)^(1/2)*((d + e*x)^(1/2)*(16*c^18*d^24*e^2 - 19 
2*b*c^17*d^23*e^3 + 1128*b^2*c^16*d^22*e^4 - 4312*b^3*c^15*d^21*e^5 + 1192 
8*b^4*c^14*d^20*e^6 - 25032*b^5*c^13*d^19*e^7 + 40712*b^6*c^12*d^18*e^8 - 
51768*b^7*c^11*d^17*e^9 + 51552*b^8*c^10*d^16*e^10 - 40048*b^9*c^9*d^15*e^ 
11 + 24024*b^10*c^8*d^14*e^12 - 10920*b^11*c^7*d^13*e^13 + 3640*b^12*c^6*d 
^12*e^14 - 840*b^13*c^5*d^11*e^15 + 120*b^14*c^4*d^10*e^16 - 8*b^15*c^3*d^ 
9*e^17) - ((-c^7*(b*e - c*d)^7)^(1/2)*(432*b^3*c^16*d^26*e^4 - 32*b^2*c^17 
*d^27*e^3 - 2720*b^4*c^15*d^25*e^5 + 10600*b^5*c^14*d^24*e^6 - 28608*b^6*c 
^13*d^23*e^7 + 56672*b^7*c^12*d^22*e^8 - 85184*b^8*c^11*d^21*e^9 + 99000*b 
^9*c^10*d^20*e^10 - 89760*b^10*c^9*d^19*e^11 + 63536*b^11*c^8*d^18*e^12 - 
34848*b^12*c^7*d^17*e^13 + 14552*b^13*c^6*d^16*e^14 - 4480*b^14*c^5*d^15*e 
^15 + 960*b^15*c^4*d^14*e^16 - 128*b^16*c^3*d^13*e^17 + 8*b^17*c^2*d^12*e^ 
18 + ((-c^7*(b*e - c*d)^7)^(1/2)*(d + e*x)^(1/2)*(16*b^2*c^18*d^31*e^2 - 2 
48*b^3*c^17*d^30*e^3 + 1800*b^4*c^16*d^29*e^4 - 8120*b^5*c^15*d^28*e^5 + 2 
5480*b^6*c^14*d^27*e^6 - 58968*b^7*c^13*d^26*e^7 + 104104*b^8*c^12*d^25*e^ 
8 - 143000*b^9*c^11*d^24*e^9 + 154440*b^10*c^10*d^23*e^10 - 131560*b^11*c^ 
9*d^22*e^11 + 88088*b^12*c^8*d^21*e^12 - 45864*b^13*c^7*d^20*e^13 + 18200* 
b^14*c^6*d^19*e^14 - 5320*b^15*c^5*d^18*e^15 + 1080*b^16*c^4*d^17*e^16 - 1 
36*b^17*c^3*d^16*e^17 + 8*b^18*c^2*d^15*e^18))/(b*(b*e - c*d)^7)))/(b*(b*e 
 - c*d)^7))*1i)/(b*(b*e - c*d)^7) + ((-c^7*(b*e - c*d)^7)^(1/2)*((d + e...
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 1458, normalized size of antiderivative = 7.80 \[ \int \frac {1}{(d+e x)^{7/2} \left (b x+c x^2\right )} \, dx =\text {Too large to display} \] Input:

int(1/(e*x+d)^(7/2)/(c*x^2+b*x),x)
 

Output:

(30*sqrt(c)*sqrt(d + e*x)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)* 
sqrt(b*e - c*d)))*c**3*d**6 + 60*sqrt(c)*sqrt(d + e*x)*sqrt(b*e - c*d)*ata 
n((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))*c**3*d**5*e*x + 30*sqrt(c)* 
sqrt(d + e*x)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c 
*d)))*c**3*d**4*e**2*x**2 + 15*sqrt(d)*sqrt(d + e*x)*log(sqrt(d + e*x) - s 
qrt(d))*b**4*d**2*e**4 + 30*sqrt(d)*sqrt(d + e*x)*log(sqrt(d + e*x) - sqrt 
(d))*b**4*d*e**5*x + 15*sqrt(d)*sqrt(d + e*x)*log(sqrt(d + e*x) - sqrt(d)) 
*b**4*e**6*x**2 - 60*sqrt(d)*sqrt(d + e*x)*log(sqrt(d + e*x) - sqrt(d))*b* 
*3*c*d**3*e**3 - 120*sqrt(d)*sqrt(d + e*x)*log(sqrt(d + e*x) - sqrt(d))*b* 
*3*c*d**2*e**4*x - 60*sqrt(d)*sqrt(d + e*x)*log(sqrt(d + e*x) - sqrt(d))*b 
**3*c*d*e**5*x**2 + 90*sqrt(d)*sqrt(d + e*x)*log(sqrt(d + e*x) - sqrt(d))* 
b**2*c**2*d**4*e**2 + 180*sqrt(d)*sqrt(d + e*x)*log(sqrt(d + e*x) - sqrt(d 
))*b**2*c**2*d**3*e**3*x + 90*sqrt(d)*sqrt(d + e*x)*log(sqrt(d + e*x) - sq 
rt(d))*b**2*c**2*d**2*e**4*x**2 - 60*sqrt(d)*sqrt(d + e*x)*log(sqrt(d + e* 
x) - sqrt(d))*b*c**3*d**5*e - 120*sqrt(d)*sqrt(d + e*x)*log(sqrt(d + e*x) 
- sqrt(d))*b*c**3*d**4*e**2*x - 60*sqrt(d)*sqrt(d + e*x)*log(sqrt(d + e*x) 
 - sqrt(d))*b*c**3*d**3*e**3*x**2 + 15*sqrt(d)*sqrt(d + e*x)*log(sqrt(d + 
e*x) - sqrt(d))*c**4*d**6 + 30*sqrt(d)*sqrt(d + e*x)*log(sqrt(d + e*x) - s 
qrt(d))*c**4*d**5*e*x + 15*sqrt(d)*sqrt(d + e*x)*log(sqrt(d + e*x) - sqrt( 
d))*c**4*d**4*e**2*x**2 - 15*sqrt(d)*sqrt(d + e*x)*log(sqrt(d + e*x) + ...