Integrand size = 24, antiderivative size = 227 \[ \int \frac {(a+b x)^3}{(c+d x) (e+f x)^{7/2}} \, dx=-\frac {2 (b e-a f)^3}{5 f^3 (d e-c f) (e+f x)^{5/2}}+\frac {2 (b e-a f)^2 (2 b d e-3 b c f+a d f)}{3 f^3 (d e-c f)^2 (e+f x)^{3/2}}-\frac {2 (b e-a f) \left (a^2 d^2 f^2+a b d f (d e-3 c f)+b^2 \left (d^2 e^2-3 c d e f+3 c^2 f^2\right )\right )}{f^3 (d e-c f)^3 \sqrt {e+f x}}+\frac {2 (b c-a d)^3 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{\sqrt {d} (d e-c f)^{7/2}} \] Output:
-2/5*(-a*f+b*e)^3/f^3/(-c*f+d*e)/(f*x+e)^(5/2)+2/3*(-a*f+b*e)^2*(a*d*f-3*b *c*f+2*b*d*e)/f^3/(-c*f+d*e)^2/(f*x+e)^(3/2)-2*(-a*f+b*e)*(a^2*d^2*f^2+a*b *d*f*(-3*c*f+d*e)+b^2*(3*c^2*f^2-3*c*d*e*f+d^2*e^2))/f^3/(-c*f+d*e)^3/(f*x +e)^(1/2)+2*(-a*d+b*c)^3*arctanh(d^(1/2)*(f*x+e)^(1/2)/(-c*f+d*e)^(1/2))/d ^(1/2)/(-c*f+d*e)^(7/2)
Time = 0.52 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.31 \[ \int \frac {(a+b x)^3}{(c+d x) (e+f x)^{7/2}} \, dx=\frac {2 (b e-a f) \left (a^2 f^2 \left (3 c^2 f^2-c d f (11 e+5 f x)+d^2 \left (23 e^2+35 e f x+15 f^2 x^2\right )\right )+b^2 \left (d^2 e^2 \left (8 e^2+20 e f x+15 f^2 x^2\right )+3 c^2 f^2 \left (11 e^2+25 e f x+15 f^2 x^2\right )-c d e f \left (26 e^2+65 e f x+45 f^2 x^2\right )\right )+a b f \left (3 c^2 f^2 (3 e+5 f x)+d^2 e \left (14 e^2+35 e f x+15 f^2 x^2\right )-c d f \left (53 e^2+110 e f x+45 f^2 x^2\right )\right )\right )}{15 f^3 (-d e+c f)^3 (e+f x)^{5/2}}+\frac {2 (b c-a d)^3 \arctan \left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {-d e+c f}}\right )}{\sqrt {d} (-d e+c f)^{7/2}} \] Input:
Integrate[(a + b*x)^3/((c + d*x)*(e + f*x)^(7/2)),x]
Output:
(2*(b*e - a*f)*(a^2*f^2*(3*c^2*f^2 - c*d*f*(11*e + 5*f*x) + d^2*(23*e^2 + 35*e*f*x + 15*f^2*x^2)) + b^2*(d^2*e^2*(8*e^2 + 20*e*f*x + 15*f^2*x^2) + 3 *c^2*f^2*(11*e^2 + 25*e*f*x + 15*f^2*x^2) - c*d*e*f*(26*e^2 + 65*e*f*x + 4 5*f^2*x^2)) + a*b*f*(3*c^2*f^2*(3*e + 5*f*x) + d^2*e*(14*e^2 + 35*e*f*x + 15*f^2*x^2) - c*d*f*(53*e^2 + 110*e*f*x + 45*f^2*x^2))))/(15*f^3*(-(d*e) + c*f)^3*(e + f*x)^(5/2)) + (2*(b*c - a*d)^3*ArcTan[(Sqrt[d]*Sqrt[e + f*x]) /Sqrt[-(d*e) + c*f]])/(Sqrt[d]*(-(d*e) + c*f)^(7/2))
Time = 0.45 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {98, 2009}
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.
| \(\displaystyle \int \frac {(a+b x)^3}{(c+d x) (e+f x)^{7/2}} \, dx\) |
| \(\Big \downarrow \) 98 |
| \(\displaystyle \int \left (\frac {-a^3 d^2 f^3+3 a^2 b c d f^3-3 a b^2 c^2 f^3+b^3 e \left (3 c^2 f^2-3 c d e f+d^2 e^2\right )}{f^2 (e+f x)^{3/2} (d e-c f)^3}+\frac {(a f-b e)^2 (-a d f+3 b c f-2 b d e)}{f^2 (e+f x)^{5/2} (c f-d e)^2}+\frac {(a f-b e)^3}{f^2 (e+f x)^{7/2} (c f-d e)}-\frac {(b c-a d)^3}{(c+d x) \sqrt {e+f x} (d e-c f)^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 (b e-a f) \left (a^2 d^2 f^2+a b d f (d e-3 c f)+b^2 \left (3 c^2 f^2-3 c d e f+d^2 e^2\right )\right )}{f^3 \sqrt {e+f x} (d e-c f)^3}+\frac {2 (b c-a d)^3 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{\sqrt {d} (d e-c f)^{7/2}}+\frac {2 (b e-a f)^2 (a d f-3 b c f+2 b d e)}{3 f^3 (e+f x)^{3/2} (d e-c f)^2}-\frac {2 (b e-a f)^3}{5 f^3 (e+f x)^{5/2} (d e-c f)}\) |
Input:
Int[(a + b*x)^3/((c + d*x)*(e + f*x)^(7/2)),x]
Output:
(-2*(b*e - a*f)^3)/(5*f^3*(d*e - c*f)*(e + f*x)^(5/2)) + (2*(b*e - a*f)^2* (2*b*d*e - 3*b*c*f + a*d*f))/(3*f^3*(d*e - c*f)^2*(e + f*x)^(3/2)) - (2*(b *e - a*f)*(a^2*d^2*f^2 + a*b*d*f*(d*e - 3*c*f) + b^2*(d^2*e^2 - 3*c*d*e*f + 3*c^2*f^2)))/(f^3*(d*e - c*f)^3*Sqrt[e + f*x]) + (2*(b*c - a*d)^3*ArcTan h[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/(Sqrt[d]*(d*e - c*f)^(7/2))
Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x _)), x_] :> Int[ExpandIntegrand[(e + f*x)^FractionalPart[p], (c + d*x)^n*(( e + f*x)^IntegerPart[p]/(a + b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]
Time = 0.57 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.27
| method | result | size |
| pseudoelliptic | \(-\frac {2 \left (5 f^{3} \left (f x +e \right )^{\frac {5}{2}} \left (a d -b c \right )^{3} \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )+\left (a f -b e \right ) \left (\left (15 b^{2} c^{2} x^{2}+5 a c x \left (-3 x d +c \right ) b +a^{2} \left (5 d^{2} x^{2}-\frac {5}{3} c d x +c^{2}\right )\right ) f^{4}-\frac {11 \left (-\frac {75 \left (-\frac {3 x d}{5}+c \right ) c x \,b^{2}}{11}-\frac {9 a \left (\frac {5}{3} d^{2} x^{2}-\frac {110}{9} c d x +c^{2}\right ) b}{11}+a^{2} d \left (-\frac {35 x d}{11}+c \right )\right ) e \,f^{3}}{3}+\frac {23 \left (\frac {\left (15 d^{2} x^{2}-65 c d x +33 c^{2}\right ) b^{2}}{23}-\frac {53 a d \left (-\frac {35 x d}{53}+c \right ) b}{23}+a^{2} d^{2}\right ) e^{2} f^{2}}{3}+\frac {14 d \left (\frac {\left (10 x d -13 c \right ) b}{7}+a d \right ) b \,e^{3} f}{3}+\frac {8 b^{2} d^{2} e^{4}}{3}\right ) \sqrt {\left (c f -d e \right ) d}\right )}{5 \left (f x +e \right )^{\frac {5}{2}} \sqrt {\left (c f -d e \right ) d}\, f^{3} \left (c f -d e \right )^{3}}\) | \(288\) |
| derivativedivides | \(\frac {-\frac {2 f^{3} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{3} \sqrt {\left (c f -d e \right ) d}}-\frac {2 \left (a^{3} f^{3}-3 a^{2} b e \,f^{2}+3 a \,b^{2} e^{2} f -b^{3} e^{3}\right )}{5 \left (c f -d e \right ) \left (f x +e \right )^{\frac {5}{2}}}-\frac {2 \left (-a^{3} d \,f^{3}+3 a^{2} b c \,f^{3}-6 a \,b^{2} c e \,f^{2}+3 a \,b^{2} d \,e^{2} f +3 b^{3} c \,e^{2} f -2 b^{3} d \,e^{3}\right )}{3 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {3}{2}}}-\frac {2 \left (a^{3} d^{2} f^{3}-3 a^{2} b c d \,f^{3}+3 a \,b^{2} c^{2} f^{3}-3 b^{3} c^{2} e \,f^{2}+3 b^{3} c d \,e^{2} f -b^{3} d^{2} e^{3}\right )}{\left (c f -d e \right )^{3} \sqrt {f x +e}}}{f^{3}}\) | \(314\) |
| default | \(\frac {-\frac {2 f^{3} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{3} \sqrt {\left (c f -d e \right ) d}}-\frac {2 \left (a^{3} f^{3}-3 a^{2} b e \,f^{2}+3 a \,b^{2} e^{2} f -b^{3} e^{3}\right )}{5 \left (c f -d e \right ) \left (f x +e \right )^{\frac {5}{2}}}-\frac {2 \left (-a^{3} d \,f^{3}+3 a^{2} b c \,f^{3}-6 a \,b^{2} c e \,f^{2}+3 a \,b^{2} d \,e^{2} f +3 b^{3} c \,e^{2} f -2 b^{3} d \,e^{3}\right )}{3 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {3}{2}}}-\frac {2 \left (a^{3} d^{2} f^{3}-3 a^{2} b c d \,f^{3}+3 a \,b^{2} c^{2} f^{3}-3 b^{3} c^{2} e \,f^{2}+3 b^{3} c d \,e^{2} f -b^{3} d^{2} e^{3}\right )}{\left (c f -d e \right )^{3} \sqrt {f x +e}}}{f^{3}}\) | \(314\) |
Input:
int((b*x+a)^3/(d*x+c)/(f*x+e)^(7/2),x,method=_RETURNVERBOSE)
Output:
-2/5/(f*x+e)^(5/2)/((c*f-d*e)*d)^(1/2)*(5*f^3*(f*x+e)^(5/2)*(a*d-b*c)^3*ar ctan(d*(f*x+e)^(1/2)/((c*f-d*e)*d)^(1/2))+(a*f-b*e)*((15*b^2*c^2*x^2+5*a*c *x*(-3*d*x+c)*b+a^2*(5*d^2*x^2-5/3*c*d*x+c^2))*f^4-11/3*(-75/11*(-3/5*x*d+ c)*c*x*b^2-9/11*a*(5/3*d^2*x^2-110/9*c*d*x+c^2)*b+a^2*d*(-35/11*x*d+c))*e* f^3+23/3*(1/23*(15*d^2*x^2-65*c*d*x+33*c^2)*b^2-53/23*a*d*(-35/53*x*d+c)*b +a^2*d^2)*e^2*f^2+14/3*d*(1/7*(10*d*x-13*c)*b+a*d)*b*e^3*f+8/3*b^2*d^2*e^4 )*((c*f-d*e)*d)^(1/2))/f^3/(c*f-d*e)^3
Leaf count of result is larger than twice the leaf count of optimal. 1009 vs. \(2 (207) = 414\).
Time = 0.26 (sec) , antiderivative size = 2032, normalized size of antiderivative = 8.95 \[ \int \frac {(a+b x)^3}{(c+d x) (e+f x)^{7/2}} \, dx=\text {Too large to display} \] Input:
integrate((b*x+a)^3/(d*x+c)/(f*x+e)^(7/2),x, algorithm="fricas")
Output:
[1/15*(15*((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*f^6*x^3 + 3 *(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*e*f^5*x^2 + 3*(b^3*c^ 3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*e^2*f^4*x + (b^3*c^3 - 3*a*b^ 2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*e^3*f^3)*sqrt(d^2*e - c*d*f)*log((d*f*x + 2*d*e - c*f + 2*sqrt(d^2*e - c*d*f)*sqrt(f*x + e))/(d*x + c)) - 2*(8*b^ 3*d^4*e^6 + 3*a^3*c^3*d*f^6 - 2*(17*b^3*c*d^3 - 3*a*b^2*d^4)*e^5*f + (59*b ^3*c^2*d^2 - 33*a*b^2*c*d^3 + 9*a^2*b*d^4)*e^4*f^2 - (33*b^3*c^3*d - 3*a*b ^2*c^2*d^2 - 33*a^2*b*c*d^3 + 23*a^3*d^4)*e^3*f^3 + 2*(12*a*b^2*c^3*d - 24 *a^2*b*c^2*d^2 + 17*a^3*c*d^3)*e^2*f^4 + 2*(3*a^2*b*c^3*d - 7*a^3*c^2*d^2) *e*f^5 + 15*(b^3*d^4*e^4*f^2 - 4*b^3*c*d^3*e^3*f^3 + 6*b^3*c^2*d^2*e^2*f^4 - (3*b^3*c^3*d + 3*a*b^2*c^2*d^2 - 3*a^2*b*c*d^3 + a^3*d^4)*e*f^5 + (3*a* b^2*c^3*d - 3*a^2*b*c^2*d^2 + a^3*c*d^3)*f^6)*x^2 + 5*(4*b^3*d^4*e^5*f - ( 17*b^3*c*d^3 - 3*a*b^2*d^4)*e^4*f^2 + 4*(7*b^3*c^2*d^2 - 3*a*b^2*c*d^3)*e^ 3*f^3 - (15*b^3*c^3*d + 3*a*b^2*c^2*d^2 - 21*a^2*b*c*d^3 + 7*a^3*d^4)*e^2* f^4 + 4*(3*a*b^2*c^3*d - 6*a^2*b*c^2*d^2 + 2*a^3*c*d^3)*e*f^5 + (3*a^2*b*c ^3*d - a^3*c^2*d^2)*f^6)*x)*sqrt(f*x + e))/(d^5*e^7*f^3 - 4*c*d^4*e^6*f^4 + 6*c^2*d^3*e^5*f^5 - 4*c^3*d^2*e^4*f^6 + c^4*d*e^3*f^7 + (d^5*e^4*f^6 - 4 *c*d^4*e^3*f^7 + 6*c^2*d^3*e^2*f^8 - 4*c^3*d^2*e*f^9 + c^4*d*f^10)*x^3 + 3 *(d^5*e^5*f^5 - 4*c*d^4*e^4*f^6 + 6*c^2*d^3*e^3*f^7 - 4*c^3*d^2*e^2*f^8 + c^4*d*e*f^9)*x^2 + 3*(d^5*e^6*f^4 - 4*c*d^4*e^5*f^5 + 6*c^2*d^3*e^4*f^6...
Time = 16.25 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.44 \[ \int \frac {(a+b x)^3}{(c+d x) (e+f x)^{7/2}} \, dx=\begin {cases} \frac {2 \left (- \frac {\left (a f - b e\right ) \left (a^{2} d^{2} f^{2} - 3 a b c d f^{2} + a b d^{2} e f + 3 b^{2} c^{2} f^{2} - 3 b^{2} c d e f + b^{2} d^{2} e^{2}\right )}{f^{2} \sqrt {e + f x} \left (c f - d e\right )^{3}} + \frac {\left (a f - b e\right )^{2} \left (a d f - 3 b c f + 2 b d e\right )}{3 f^{2} \left (e + f x\right )^{\frac {3}{2}} \left (c f - d e\right )^{2}} - \frac {\left (a f - b e\right )^{3}}{5 f^{2} \left (e + f x\right )^{\frac {5}{2}} \left (c f - d e\right )} - \frac {f \left (a d - b c\right )^{3} \operatorname {atan}{\left (\frac {\sqrt {e + f x}}{\sqrt {\frac {c f - d e}{d}}} \right )}}{d \sqrt {\frac {c f - d e}{d}} \left (c f - d e\right )^{3}}\right )}{f} & \text {for}\: f \neq 0 \\\frac {\frac {b^{3} x^{3}}{3 d} + \frac {x^{2} \cdot \left (3 a b^{2} d - b^{3} c\right )}{2 d^{2}} + \frac {x \left (3 a^{2} b d^{2} - 3 a b^{2} c d + b^{3} c^{2}\right )}{d^{3}} + \frac {\left (a d - b c\right )^{3} \left (\begin {cases} \frac {x}{c} & \text {for}\: d = 0 \\\frac {\log {\left (c + d x \right )}}{d} & \text {otherwise} \end {cases}\right )}{d^{3}}}{e^{\frac {7}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate((b*x+a)**3/(d*x+c)/(f*x+e)**(7/2),x)
Output:
Piecewise((2*(-(a*f - b*e)*(a**2*d**2*f**2 - 3*a*b*c*d*f**2 + a*b*d**2*e*f + 3*b**2*c**2*f**2 - 3*b**2*c*d*e*f + b**2*d**2*e**2)/(f**2*sqrt(e + f*x) *(c*f - d*e)**3) + (a*f - b*e)**2*(a*d*f - 3*b*c*f + 2*b*d*e)/(3*f**2*(e + f*x)**(3/2)*(c*f - d*e)**2) - (a*f - b*e)**3/(5*f**2*(e + f*x)**(5/2)*(c* f - d*e)) - f*(a*d - b*c)**3*atan(sqrt(e + f*x)/sqrt((c*f - d*e)/d))/(d*sq rt((c*f - d*e)/d)*(c*f - d*e)**3))/f, Ne(f, 0)), ((b**3*x**3/(3*d) + x**2* (3*a*b**2*d - b**3*c)/(2*d**2) + x*(3*a**2*b*d**2 - 3*a*b**2*c*d + b**3*c* *2)/d**3 + (a*d - b*c)**3*Piecewise((x/c, Eq(d, 0)), (log(c + d*x)/d, True ))/d**3)/e**(7/2), True))
Exception generated. \[ \int \frac {(a+b x)^3}{(c+d x) (e+f x)^{7/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((b*x+a)^3/(d*x+c)/(f*x+e)^(7/2),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(c*f-d*e>0)', see `assume?` for m ore detail
Leaf count of result is larger than twice the leaf count of optimal. 604 vs. \(2 (207) = 414\).
Time = 0.14 (sec) , antiderivative size = 604, normalized size of antiderivative = 2.66 \[ \int \frac {(a+b x)^3}{(c+d x) (e+f x)^{7/2}} \, dx=-\frac {2 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {-d^{2} e + c d f}}\right )}{{\left (d^{3} e^{3} - 3 \, c d^{2} e^{2} f + 3 \, c^{2} d e f^{2} - c^{3} f^{3}\right )} \sqrt {-d^{2} e + c d f}} - \frac {2 \, {\left (15 \, {\left (f x + e\right )}^{2} b^{3} d^{2} e^{3} - 10 \, {\left (f x + e\right )} b^{3} d^{2} e^{4} + 3 \, b^{3} d^{2} e^{5} - 45 \, {\left (f x + e\right )}^{2} b^{3} c d e^{2} f + 25 \, {\left (f x + e\right )} b^{3} c d e^{3} f + 15 \, {\left (f x + e\right )} a b^{2} d^{2} e^{3} f - 6 \, b^{3} c d e^{4} f - 9 \, a b^{2} d^{2} e^{4} f + 45 \, {\left (f x + e\right )}^{2} b^{3} c^{2} e f^{2} - 15 \, {\left (f x + e\right )} b^{3} c^{2} e^{2} f^{2} - 45 \, {\left (f x + e\right )} a b^{2} c d e^{2} f^{2} + 3 \, b^{3} c^{2} e^{3} f^{2} + 18 \, a b^{2} c d e^{3} f^{2} + 9 \, a^{2} b d^{2} e^{3} f^{2} - 45 \, {\left (f x + e\right )}^{2} a b^{2} c^{2} f^{3} + 45 \, {\left (f x + e\right )}^{2} a^{2} b c d f^{3} - 15 \, {\left (f x + e\right )}^{2} a^{3} d^{2} f^{3} + 30 \, {\left (f x + e\right )} a b^{2} c^{2} e f^{3} + 15 \, {\left (f x + e\right )} a^{2} b c d e f^{3} - 5 \, {\left (f x + e\right )} a^{3} d^{2} e f^{3} - 9 \, a b^{2} c^{2} e^{2} f^{3} - 18 \, a^{2} b c d e^{2} f^{3} - 3 \, a^{3} d^{2} e^{2} f^{3} - 15 \, {\left (f x + e\right )} a^{2} b c^{2} f^{4} + 5 \, {\left (f x + e\right )} a^{3} c d f^{4} + 9 \, a^{2} b c^{2} e f^{4} + 6 \, a^{3} c d e f^{4} - 3 \, a^{3} c^{2} f^{5}\right )}}{15 \, {\left (d^{3} e^{3} f^{3} - 3 \, c d^{2} e^{2} f^{4} + 3 \, c^{2} d e f^{5} - c^{3} f^{6}\right )} {\left (f x + e\right )}^{\frac {5}{2}}} \] Input:
integrate((b*x+a)^3/(d*x+c)/(f*x+e)^(7/2),x, algorithm="giac")
Output:
-2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*arctan(sqrt(f*x + e )*d/sqrt(-d^2*e + c*d*f))/((d^3*e^3 - 3*c*d^2*e^2*f + 3*c^2*d*e*f^2 - c^3* f^3)*sqrt(-d^2*e + c*d*f)) - 2/15*(15*(f*x + e)^2*b^3*d^2*e^3 - 10*(f*x + e)*b^3*d^2*e^4 + 3*b^3*d^2*e^5 - 45*(f*x + e)^2*b^3*c*d*e^2*f + 25*(f*x + e)*b^3*c*d*e^3*f + 15*(f*x + e)*a*b^2*d^2*e^3*f - 6*b^3*c*d*e^4*f - 9*a*b^ 2*d^2*e^4*f + 45*(f*x + e)^2*b^3*c^2*e*f^2 - 15*(f*x + e)*b^3*c^2*e^2*f^2 - 45*(f*x + e)*a*b^2*c*d*e^2*f^2 + 3*b^3*c^2*e^3*f^2 + 18*a*b^2*c*d*e^3*f^ 2 + 9*a^2*b*d^2*e^3*f^2 - 45*(f*x + e)^2*a*b^2*c^2*f^3 + 45*(f*x + e)^2*a^ 2*b*c*d*f^3 - 15*(f*x + e)^2*a^3*d^2*f^3 + 30*(f*x + e)*a*b^2*c^2*e*f^3 + 15*(f*x + e)*a^2*b*c*d*e*f^3 - 5*(f*x + e)*a^3*d^2*e*f^3 - 9*a*b^2*c^2*e^2 *f^3 - 18*a^2*b*c*d*e^2*f^3 - 3*a^3*d^2*e^2*f^3 - 15*(f*x + e)*a^2*b*c^2*f ^4 + 5*(f*x + e)*a^3*c*d*f^4 + 9*a^2*b*c^2*e*f^4 + 6*a^3*c*d*e*f^4 - 3*a^3 *c^2*f^5)/((d^3*e^3*f^3 - 3*c*d^2*e^2*f^4 + 3*c^2*d*e*f^5 - c^3*f^6)*(f*x + e)^(5/2))
Time = 0.21 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.59 \[ \int \frac {(a+b x)^3}{(c+d x) (e+f x)^{7/2}} \, dx=-\frac {\frac {2\,\left (a^3\,f^3-3\,a^2\,b\,e\,f^2+3\,a\,b^2\,e^2\,f-b^3\,e^3\right )}{5\,\left (c\,f-d\,e\right )}+\frac {2\,{\left (e+f\,x\right )}^2\,\left (a^3\,d^2\,f^3-3\,a^2\,b\,c\,d\,f^3+3\,a\,b^2\,c^2\,f^3-3\,b^3\,c^2\,e\,f^2+3\,b^3\,c\,d\,e^2\,f-b^3\,d^2\,e^3\right )}{{\left (c\,f-d\,e\right )}^3}-\frac {2\,\left (e+f\,x\right )\,\left (d\,a^3\,f^3-3\,c\,a^2\,b\,f^3-3\,d\,a\,b^2\,e^2\,f+6\,c\,a\,b^2\,e\,f^2+2\,d\,b^3\,e^3-3\,c\,b^3\,e^2\,f\right )}{3\,{\left (c\,f-d\,e\right )}^2}}{f^3\,{\left (e+f\,x\right )}^{5/2}}-\frac {2\,\mathrm {atan}\left (\frac {2\,\sqrt {d}\,\sqrt {e+f\,x}\,{\left (a\,d-b\,c\right )}^3\,\left (c^3\,f^3-3\,c^2\,d\,e\,f^2+3\,c\,d^2\,e^2\,f-d^3\,e^3\right )}{{\left (c\,f-d\,e\right )}^{7/2}\,\left (2\,a^3\,d^3-6\,a^2\,b\,c\,d^2+6\,a\,b^2\,c^2\,d-2\,b^3\,c^3\right )}\right )\,{\left (a\,d-b\,c\right )}^3}{\sqrt {d}\,{\left (c\,f-d\,e\right )}^{7/2}} \] Input:
int((a + b*x)^3/((e + f*x)^(7/2)*(c + d*x)),x)
Output:
- ((2*(a^3*f^3 - b^3*e^3 + 3*a*b^2*e^2*f - 3*a^2*b*e*f^2))/(5*(c*f - d*e)) + (2*(e + f*x)^2*(a^3*d^2*f^3 - b^3*d^2*e^3 + 3*a*b^2*c^2*f^3 - 3*b^3*c^2 *e*f^2 - 3*a^2*b*c*d*f^3 + 3*b^3*c*d*e^2*f))/(c*f - d*e)^3 - (2*(e + f*x)* (a^3*d*f^3 + 2*b^3*d*e^3 - 3*a^2*b*c*f^3 - 3*b^3*c*e^2*f + 6*a*b^2*c*e*f^2 - 3*a*b^2*d*e^2*f))/(3*(c*f - d*e)^2))/(f^3*(e + f*x)^(5/2)) - (2*atan((2 *d^(1/2)*(e + f*x)^(1/2)*(a*d - b*c)^3*(c^3*f^3 - d^3*e^3 + 3*c*d^2*e^2*f - 3*c^2*d*e*f^2))/((c*f - d*e)^(7/2)*(2*a^3*d^3 - 2*b^3*c^3 + 6*a*b^2*c^2* d - 6*a^2*b*c*d^2)))*(a*d - b*c)^3)/(d^(1/2)*(c*f - d*e)^(7/2))
Time = 0.16 (sec) , antiderivative size = 1501, normalized size of antiderivative = 6.61 \[ \int \frac {(a+b x)^3}{(c+d x) (e+f x)^{7/2}} \, dx =\text {Too large to display} \] Input:
int((b*x+a)^3/(d*x+c)/(f*x+e)^(7/2),x)
Output:
(2*( - 15*sqrt(d)*sqrt(e + f*x)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sq rt(d)*sqrt(c*f - d*e)))*a**3*d**3*e**2*f**3 - 30*sqrt(d)*sqrt(e + f*x)*sqr t(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*a**3*d**3*e *f**4*x - 15*sqrt(d)*sqrt(e + f*x)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/ (sqrt(d)*sqrt(c*f - d*e)))*a**3*d**3*f**5*x**2 + 45*sqrt(d)*sqrt(e + f*x)* sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*a**2*b*c *d**2*e**2*f**3 + 90*sqrt(d)*sqrt(e + f*x)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*a**2*b*c*d**2*e*f**4*x + 45*sqrt(d)*sqr t(e + f*x)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e) ))*a**2*b*c*d**2*f**5*x**2 - 45*sqrt(d)*sqrt(e + f*x)*sqrt(c*f - d*e)*atan ((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*a*b**2*c**2*d*e**2*f**3 - 90 *sqrt(d)*sqrt(e + f*x)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqr t(c*f - d*e)))*a*b**2*c**2*d*e*f**4*x - 45*sqrt(d)*sqrt(e + f*x)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*a*b**2*c**2*d*f** 5*x**2 + 15*sqrt(d)*sqrt(e + f*x)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/( sqrt(d)*sqrt(c*f - d*e)))*b**3*c**3*e**2*f**3 + 30*sqrt(d)*sqrt(e + f*x)*s qrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*b**3*c**3 *e*f**4*x + 15*sqrt(d)*sqrt(e + f*x)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d )/(sqrt(d)*sqrt(c*f - d*e)))*b**3*c**3*f**5*x**2 - 3*a**3*c**3*d*f**6 + 14 *a**3*c**2*d**2*e*f**5 + 5*a**3*c**2*d**2*f**6*x - 34*a**3*c*d**3*e**2*...