\(\int \frac {(c+d x)^{3/2} (e+f x)}{(a+b x)^{7/2}} \, dx\) [1185]

Optimal result

Integrand size = 24, antiderivative size = 138 \[ \int \frac {(c+d x)^{3/2} (e+f x)}{(a+b x)^{7/2}} \, dx=-\frac {2 d f \sqrt {c+d x}}{b^3 \sqrt {a+b x}}-\frac {2 f (c+d x)^{3/2}}{3 b^2 (a+b x)^{3/2}}-\frac {2 (b e-a f) (c+d x)^{5/2}}{5 b (b c-a d) (a+b x)^{5/2}}+\frac {2 d^{3/2} f \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{7/2}} \] Output:

-2*d*f*(d*x+c)^(1/2)/b^3/(b*x+a)^(1/2)-2/3*f*(d*x+c)^(3/2)/b^2/(b*x+a)^(3/ 
2)-2/5*(-a*f+b*e)*(d*x+c)^(5/2)/b/(-a*d+b*c)/(b*x+a)^(5/2)+2*d^(3/2)*f*arc 
tanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(7/2)
 

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.24 \[ \int \frac {(c+d x)^{3/2} (e+f x)}{(a+b x)^{7/2}} \, dx=-\frac {2 (c+d x)^{5/2} \left (3 b^3 e-3 a b^2 f+\frac {15 b c d f (a+b x)^2}{(c+d x)^2}-\frac {15 a d^2 f (a+b x)^2}{(c+d x)^2}+\frac {5 b^2 c f (a+b x)}{c+d x}-\frac {5 a b d f (a+b x)}{c+d x}\right )}{15 b^3 (b c-a d) (a+b x)^{5/2}}+\frac {2 d^{3/2} f \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{7/2}} \] Input:

Integrate[((c + d*x)^(3/2)*(e + f*x))/(a + b*x)^(7/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {87, 57, 57, 66, 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.

\(\displaystyle \int \frac {(c+d x)^{3/2} (e+f x)}{(a+b x)^{7/2}} \, dx\)

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

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

Input:

Int[((c + d*x)^(3/2)*(e + f*x))/(a + b*x)^(7/2),x]
 

Output:

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

Defintions of rubi rules used

rule 57

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & 
& GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m 
+ n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c 
, d, m, n, x]
 

rule 66

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] &&  !GtQ[c - a*(d/b), 0]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(779\) vs. \(2(110)=220\).

Time = 0.33 (sec) , antiderivative size = 780, normalized size of antiderivative = 5.65

method	result	size
default	\(\frac {\sqrt {x d +c}\, \left (15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a \,b^{3} d^{3} f \,x^{3}-15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) b^{4} c \,d^{2} f \,x^{3}+45 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a^{2} b^{2} d^{3} f \,x^{2}-45 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a \,b^{3} c \,d^{2} f \,x^{2}+45 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a^{3} b \,d^{3} f x -45 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a^{2} b^{2} c \,d^{2} f x -46 a \,b^{2} d^{2} f \,x^{2} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+40 b^{3} c d f \,x^{2} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+6 b^{3} d^{2} e \,x^{2} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a^{4} d^{3} f -15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a^{3} b c \,d^{2} f -70 a^{2} b \,d^{2} f x \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+48 a \,b^{2} c d f x \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+10 b^{3} c^{2} f x \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+12 b^{3} c d e x \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}-30 a^{3} d^{2} f \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+20 a^{2} b c d f \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+4 a \,b^{2} c^{2} f \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+6 b^{3} c^{2} e \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\right )}{15 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \left (a d -b c \right ) \sqrt {d b}\, \left (b x +a \right )^{\frac {5}{2}} b^{3}}\)	\(780\)

Input:

int((d*x+c)^(3/2)*(f*x+e)/(b*x+a)^(7/2),x,method=_RETURNVERBOSE)
 

Output:

1/15*(d*x+c)^(1/2)*(15*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/ 
2)+a*d+b*c)/(d*b)^(1/2))*a*b^3*d^3*f*x^3-15*ln(1/2*(2*b*d*x+2*((b*x+a)*(d* 
x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*b^4*c*d^2*f*x^3+45*ln(1/2*(2 
*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^2*b^2 
*d^3*f*x^2-45*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b* 
c)/(d*b)^(1/2))*a*b^3*c*d^2*f*x^2+45*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^( 
1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^3*b*d^3*f*x-45*ln(1/2*(2*b*d*x+2* 
((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^2*b^2*c*d^2*f* 
x-46*a*b^2*d^2*f*x^2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+40*b^3*c*d*f*x^2* 
((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+6*b^3*d^2*e*x^2*((b*x+a)*(d*x+c))^(1/2 
)*(d*b)^(1/2)+15*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d 
+b*c)/(d*b)^(1/2))*a^4*d^3*f-15*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)* 
(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^3*b*c*d^2*f-70*a^2*b*d^2*f*x*((b*x+a)* 
(d*x+c))^(1/2)*(d*b)^(1/2)+48*a*b^2*c*d*f*x*((b*x+a)*(d*x+c))^(1/2)*(d*b)^ 
(1/2)+10*b^3*c^2*f*x*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+12*b^3*c*d*e*x*(( 
b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)-30*a^3*d^2*f*((b*x+a)*(d*x+c))^(1/2)*(d* 
b)^(1/2)+20*a^2*b*c*d*f*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+4*a*b^2*c^2*f* 
((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+6*b^3*c^2*e*((b*x+a)*(d*x+c))^(1/2)*(d 
*b)^(1/2))/((b*x+a)*(d*x+c))^(1/2)/(a*d-b*c)/(d*b)^(1/2)/(b*x+a)^(5/2)/b^3
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 370 vs. \(2 (110) = 220\).

Time = 2.34 (sec) , antiderivative size = 765, normalized size of antiderivative = 5.54 \[ \int \frac {(c+d x)^{3/2} (e+f x)}{(a+b x)^{7/2}} \, dx =\text {Too large to display} \] Input:

integrate((d*x+c)^(3/2)*(f*x+e)/(b*x+a)^(7/2),x, algorithm="fricas")
 

Output:

[1/30*(15*((b^4*c*d - a*b^3*d^2)*f*x^3 + 3*(a*b^3*c*d - a^2*b^2*d^2)*f*x^2 
 + 3*(a^2*b^2*c*d - a^3*b*d^2)*f*x + (a^3*b*c*d - a^4*d^2)*f)*sqrt(d/b)*lo 
g(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b^2*d*x + b^2*c + a 
*b*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(d/b) + 8*(b^2*c*d + a*b*d^2)*x) - 4 
*(3*b^3*c^2*e + (3*b^3*d^2*e + (20*b^3*c*d - 23*a*b^2*d^2)*f)*x^2 + (2*a*b 
^2*c^2 + 10*a^2*b*c*d - 15*a^3*d^2)*f + (6*b^3*c*d*e + (5*b^3*c^2 + 24*a*b 
^2*c*d - 35*a^2*b*d^2)*f)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^3*b^4*c - a^4 
*b^3*d + (b^7*c - a*b^6*d)*x^3 + 3*(a*b^6*c - a^2*b^5*d)*x^2 + 3*(a^2*b^5* 
c - a^3*b^4*d)*x), -1/15*(15*((b^4*c*d - a*b^3*d^2)*f*x^3 + 3*(a*b^3*c*d - 
 a^2*b^2*d^2)*f*x^2 + 3*(a^2*b^2*c*d - a^3*b*d^2)*f*x + (a^3*b*c*d - a^4*d 
^2)*f)*sqrt(-d/b)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x 
+ c)*sqrt(-d/b)/(b*d^2*x^2 + a*c*d + (b*c*d + a*d^2)*x)) + 2*(3*b^3*c^2*e 
+ (3*b^3*d^2*e + (20*b^3*c*d - 23*a*b^2*d^2)*f)*x^2 + (2*a*b^2*c^2 + 10*a^ 
2*b*c*d - 15*a^3*d^2)*f + (6*b^3*c*d*e + (5*b^3*c^2 + 24*a*b^2*c*d - 35*a^ 
2*b*d^2)*f)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^3*b^4*c - a^4*b^3*d + (b^7* 
c - a*b^6*d)*x^3 + 3*(a*b^6*c - a^2*b^5*d)*x^2 + 3*(a^2*b^5*c - a^3*b^4*d) 
*x)]
 

Sympy [F]

\[ \int \frac {(c+d x)^{3/2} (e+f x)}{(a+b x)^{7/2}} \, dx=\int \frac {\left (c + d x\right )^{\frac {3}{2}} \left (e + f x\right )}{\left (a + b x\right )^{\frac {7}{2}}}\, dx \] Input:

integrate((d*x+c)**(3/2)*(f*x+e)/(b*x+a)**(7/2),x)
 

Output:

Integral((c + d*x)**(3/2)*(e + f*x)/(a + b*x)**(7/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d x)^{3/2} (e+f x)}{(a+b x)^{7/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((d*x+c)^(3/2)*(f*x+e)/(b*x+a)^(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(a*d-b*c>0)', see `assume?` for m 
ore detail
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1360 vs. \(2 (110) = 220\).

Time = 0.44 (sec) , antiderivative size = 1360, normalized size of antiderivative = 9.86 \[ \int \frac {(c+d x)^{3/2} (e+f x)}{(a+b x)^{7/2}} \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)^(3/2)*(f*x+e)/(b*x+a)^(7/2),x, algorithm="giac")
 

Output:

-sqrt(b*d)*d*f*abs(b)*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a 
)*b*d - a*b*d))^2)/b^5 - 4/15*(3*sqrt(b*d)*b^9*c^4*d^2*e*abs(b) - 12*sqrt( 
b*d)*a*b^8*c^3*d^3*e*abs(b) + 18*sqrt(b*d)*a^2*b^7*c^2*d^4*e*abs(b) - 12*s 
qrt(b*d)*a^3*b^6*c*d^5*e*abs(b) + 3*sqrt(b*d)*a^4*b^5*d^6*e*abs(b) + 20*sq 
rt(b*d)*b^9*c^5*d*f*abs(b) - 103*sqrt(b*d)*a*b^8*c^4*d^2*f*abs(b) + 212*sq 
rt(b*d)*a^2*b^7*c^3*d^3*f*abs(b) - 218*sqrt(b*d)*a^3*b^6*c^2*d^4*f*abs(b) 
+ 112*sqrt(b*d)*a^4*b^5*c*d^5*f*abs(b) - 23*sqrt(b*d)*a^5*b^4*d^6*f*abs(b) 
 - 70*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a* 
b*d))^2*b^7*c^4*d*f*abs(b) + 280*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt 
(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^6*c^3*d^2*f*abs(b) - 420*sqrt(b*d)* 
(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^5* 
c^2*d^3*f*abs(b) + 280*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + ( 
b*x + a)*b*d - a*b*d))^2*a^3*b^4*c*d^4*f*abs(b) - 70*sqrt(b*d)*(sqrt(b*d)* 
sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^4*b^3*d^5*f*abs(b 
) + 30*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a 
*b*d))^4*b^5*c^2*d^2*e*abs(b) - 60*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sq 
rt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a*b^4*c*d^3*e*abs(b) + 30*sqrt(b*d)*( 
sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^2*b^3*d 
^4*e*abs(b) + 110*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + 
 a)*b*d - a*b*d))^4*b^5*c^3*d*f*abs(b) - 360*sqrt(b*d)*(sqrt(b*d)*sqrt(...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{3/2} (e+f x)}{(a+b x)^{7/2}} \, dx=\int \frac {\left (e+f\,x\right )\,{\left (c+d\,x\right )}^{3/2}}{{\left (a+b\,x\right )}^{7/2}} \,d x \] Input:

int(((e + f*x)*(c + d*x)^(3/2))/(a + b*x)^(7/2),x)
 

Output:

int(((e + f*x)*(c + d*x)^(3/2))/(a + b*x)^(7/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 748, normalized size of antiderivative = 5.42 \[ \int \frac {(c+d x)^{3/2} (e+f x)}{(a+b x)^{7/2}} \, dx =\text {Too large to display} \] Input:

int((d*x+c)^(3/2)*(f*x+e)/(b*x+a)^(7/2),x)
 

Output:

(2*(15*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)* 
sqrt(c + d*x))/sqrt(a*d - b*c))*a**3*d**2*f - 15*sqrt(d)*sqrt(b)*sqrt(a + 
b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))* 
a**2*b*c*d*f + 30*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) 
 + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**2*b*d**2*f*x - 30*sqrt(d)*sq 
rt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sq 
rt(a*d - b*c))*a*b**2*c*d*f*x + 15*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt 
(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a*b**2*d**2*f* 
x**2 - 15*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt( 
b)*sqrt(c + d*x))/sqrt(a*d - b*c))*b**3*c*d*f*x**2 + 5*sqrt(d)*sqrt(b)*sqr 
t(a + b*x)*a**3*d**2*f - 8*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**2*b*c*d*f + 3* 
sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**2*b*d**2*e + 10*sqrt(d)*sqrt(b)*sqrt(a + 
b*x)*a**2*b*d**2*f*x - 16*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a*b**2*c*d*f*x + 6 
*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a*b**2*d**2*e*x + 5*sqrt(d)*sqrt(b)*sqrt(a 
+ b*x)*a*b**2*d**2*f*x**2 - 8*sqrt(d)*sqrt(b)*sqrt(a + b*x)*b**3*c*d*f*x** 
2 + 3*sqrt(d)*sqrt(b)*sqrt(a + b*x)*b**3*d**2*e*x**2 - 15*sqrt(c + d*x)*a* 
*3*b*d**2*f + 10*sqrt(c + d*x)*a**2*b**2*c*d*f - 35*sqrt(c + d*x)*a**2*b** 
2*d**2*f*x + 2*sqrt(c + d*x)*a*b**3*c**2*f + 24*sqrt(c + d*x)*a*b**3*c*d*f 
*x - 23*sqrt(c + d*x)*a*b**3*d**2*f*x**2 + 3*sqrt(c + d*x)*b**4*c**2*e + 5 
*sqrt(c + d*x)*b**4*c**2*f*x + 6*sqrt(c + d*x)*b**4*c*d*e*x + 20*sqrt(c...