\(\int \frac {(a+b x)^{7/2}}{(c+d x)^{5/2}} \, dx\) [363]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [F(-2)]
Giac [B] (verification not implemented)
Mupad [F(-1)]
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 19, antiderivative size = 170 \[ \int \frac {(a+b x)^{7/2}}{(c+d x)^{5/2}} \, dx=-\frac {2 (a+b x)^{7/2}}{3 d (c+d x)^{3/2}}-\frac {14 b (a+b x)^{5/2}}{3 d^2 \sqrt {c+d x}}-\frac {35 b^2 (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d^4}+\frac {35 b^2 (a+b x)^{3/2} \sqrt {c+d x}}{6 d^3}+\frac {35 b^{3/2} (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 d^{9/2}} \] Output:

-2/3*(b*x+a)^(7/2)/d/(d*x+c)^(3/2)-14/3*b*(b*x+a)^(5/2)/d^2/(d*x+c)^(1/2)- 
35/4*b^2*(-a*d+b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/d^4+35/6*b^2*(b*x+a)^(3/2) 
*(d*x+c)^(1/2)/d^3+35/4*b^(3/2)*(-a*d+b*c)^2*arctanh(d^(1/2)*(b*x+a)^(1/2) 
/b^(1/2)/(d*x+c)^(1/2))/d^(9/2)
 

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.98 \[ \int \frac {(a+b x)^{7/2}}{(c+d x)^{5/2}} \, dx=-\frac {\sqrt {a+b x} \left (8 a^3 d^3+8 a^2 b d^2 (7 c+10 d x)-a b^2 d \left (175 c^2+238 c d x+39 d^2 x^2\right )+b^3 \left (105 c^3+140 c^2 d x+21 c d^2 x^2-6 d^3 x^3\right )\right )}{12 d^4 (c+d x)^{3/2}}+\frac {35 b^{3/2} (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{4 d^{9/2}} \] Input:

Integrate[(a + b*x)^(7/2)/(c + d*x)^(5/2),x]
 

Output:

-1/12*(Sqrt[a + b*x]*(8*a^3*d^3 + 8*a^2*b*d^2*(7*c + 10*d*x) - a*b^2*d*(17 
5*c^2 + 238*c*d*x + 39*d^2*x^2) + b^3*(105*c^3 + 140*c^2*d*x + 21*c*d^2*x^ 
2 - 6*d^3*x^3)))/(d^4*(c + d*x)^(3/2)) + (35*b^(3/2)*(b*c - a*d)^2*ArcTanh 
[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(4*d^(9/2))
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {57, 57, 60, 60, 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.

\(\displaystyle \int \frac {(a+b x)^{7/2}}{(c+d x)^{5/2}} \, dx\)

\(\Big \downarrow \) 57

\(\displaystyle \frac {7 b \int \frac {(a+b x)^{5/2}}{(c+d x)^{3/2}}dx}{3 d}-\frac {2 (a+b x)^{7/2}}{3 d (c+d x)^{3/2}}\)

\(\Big \downarrow \) 57

\(\displaystyle \frac {7 b \left (\frac {5 b \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}}dx}{d}-\frac {2 (a+b x)^{5/2}}{d \sqrt {c+d x}}\right )}{3 d}-\frac {2 (a+b x)^{7/2}}{3 d (c+d x)^{3/2}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {7 b \left (\frac {5 b \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}}dx}{4 d}\right )}{d}-\frac {2 (a+b x)^{5/2}}{d \sqrt {c+d x}}\right )}{3 d}-\frac {2 (a+b x)^{7/2}}{3 d (c+d x)^{3/2}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {7 b \left (\frac {5 b \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 d}\right )}{4 d}\right )}{d}-\frac {2 (a+b x)^{5/2}}{d \sqrt {c+d x}}\right )}{3 d}-\frac {2 (a+b x)^{7/2}}{3 d (c+d x)^{3/2}}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {7 b \left (\frac {5 b \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{d}\right )}{4 d}\right )}{d}-\frac {2 (a+b x)^{5/2}}{d \sqrt {c+d x}}\right )}{3 d}-\frac {2 (a+b x)^{7/2}}{3 d (c+d x)^{3/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {7 b \left (\frac {5 b \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} d^{3/2}}\right )}{4 d}\right )}{d}-\frac {2 (a+b x)^{5/2}}{d \sqrt {c+d x}}\right )}{3 d}-\frac {2 (a+b x)^{7/2}}{3 d (c+d x)^{3/2}}\)

Input:

Int[(a + b*x)^(7/2)/(c + d*x)^(5/2),x]
 

Output:

(-2*(a + b*x)^(7/2))/(3*d*(c + d*x)^(3/2)) + (7*b*((-2*(a + b*x)^(5/2))/(d 
*Sqrt[c + d*x]) + (5*b*(((a + b*x)^(3/2)*Sqrt[c + d*x])/(2*d) - (3*(b*c - 
a*d)*((Sqrt[a + b*x]*Sqrt[c + d*x])/d - ((b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt 
[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(Sqrt[b]*d^(3/2))))/(4*d)))/d))/(3*d)
 

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 60
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[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 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 [F]

\[\int \frac {\left (b x +a \right )^{\frac {7}{2}}}{\left (x d +c \right )^{\frac {5}{2}}}d x\]

Input:

int((b*x+a)^(7/2)/(d*x+c)^(5/2),x)
 

Output:

int((b*x+a)^(7/2)/(d*x+c)^(5/2),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 316 vs. \(2 (132) = 264\).

Time = 0.35 (sec) , antiderivative size = 657, normalized size of antiderivative = 3.86 \[ \int \frac {(a+b x)^{7/2}}{(c+d x)^{5/2}} \, dx=\left [\frac {105 \, {\left (b^{3} c^{4} - 2 \, a b^{2} c^{3} d + a^{2} b c^{2} d^{2} + {\left (b^{3} c^{2} d^{2} - 2 \, a b^{2} c d^{3} + a^{2} b d^{4}\right )} x^{2} + 2 \, {\left (b^{3} c^{3} d - 2 \, a b^{2} c^{2} d^{2} + a^{2} b c d^{3}\right )} x\right )} \sqrt {\frac {b}{d}} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d^{2} x + b c d + a d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {b}{d}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (6 \, b^{3} d^{3} x^{3} - 105 \, b^{3} c^{3} + 175 \, a b^{2} c^{2} d - 56 \, a^{2} b c d^{2} - 8 \, a^{3} d^{3} - 3 \, {\left (7 \, b^{3} c d^{2} - 13 \, a b^{2} d^{3}\right )} x^{2} - 2 \, {\left (70 \, b^{3} c^{2} d - 119 \, a b^{2} c d^{2} + 40 \, a^{2} b d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, {\left (d^{6} x^{2} + 2 \, c d^{5} x + c^{2} d^{4}\right )}}, -\frac {105 \, {\left (b^{3} c^{4} - 2 \, a b^{2} c^{3} d + a^{2} b c^{2} d^{2} + {\left (b^{3} c^{2} d^{2} - 2 \, a b^{2} c d^{3} + a^{2} b d^{4}\right )} x^{2} + 2 \, {\left (b^{3} c^{3} d - 2 \, a b^{2} c^{2} d^{2} + a^{2} b c d^{3}\right )} x\right )} \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {b}{d}}}{2 \, {\left (b^{2} d x^{2} + a b c + {\left (b^{2} c + a b d\right )} x\right )}}\right ) - 2 \, {\left (6 \, b^{3} d^{3} x^{3} - 105 \, b^{3} c^{3} + 175 \, a b^{2} c^{2} d - 56 \, a^{2} b c d^{2} - 8 \, a^{3} d^{3} - 3 \, {\left (7 \, b^{3} c d^{2} - 13 \, a b^{2} d^{3}\right )} x^{2} - 2 \, {\left (70 \, b^{3} c^{2} d - 119 \, a b^{2} c d^{2} + 40 \, a^{2} b d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{24 \, {\left (d^{6} x^{2} + 2 \, c d^{5} x + c^{2} d^{4}\right )}}\right ] \] Input:

integrate((b*x+a)^(7/2)/(d*x+c)^(5/2),x, algorithm="fricas")
 

Output:

[1/48*(105*(b^3*c^4 - 2*a*b^2*c^3*d + a^2*b*c^2*d^2 + (b^3*c^2*d^2 - 2*a*b 
^2*c*d^3 + a^2*b*d^4)*x^2 + 2*(b^3*c^3*d - 2*a*b^2*c^2*d^2 + a^2*b*c*d^3)* 
x)*sqrt(b/d)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b*d^ 
2*x + b*c*d + a*d^2)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(b/d) + 8*(b^2*c*d + 
a*b*d^2)*x) + 4*(6*b^3*d^3*x^3 - 105*b^3*c^3 + 175*a*b^2*c^2*d - 56*a^2*b* 
c*d^2 - 8*a^3*d^3 - 3*(7*b^3*c*d^2 - 13*a*b^2*d^3)*x^2 - 2*(70*b^3*c^2*d - 
 119*a*b^2*c*d^2 + 40*a^2*b*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(d^6*x^2 
+ 2*c*d^5*x + c^2*d^4), -1/24*(105*(b^3*c^4 - 2*a*b^2*c^3*d + a^2*b*c^2*d^ 
2 + (b^3*c^2*d^2 - 2*a*b^2*c*d^3 + a^2*b*d^4)*x^2 + 2*(b^3*c^3*d - 2*a*b^2 
*c^2*d^2 + a^2*b*c*d^3)*x)*sqrt(-b/d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqr 
t(b*x + a)*sqrt(d*x + c)*sqrt(-b/d)/(b^2*d*x^2 + a*b*c + (b^2*c + a*b*d)*x 
)) - 2*(6*b^3*d^3*x^3 - 105*b^3*c^3 + 175*a*b^2*c^2*d - 56*a^2*b*c*d^2 - 8 
*a^3*d^3 - 3*(7*b^3*c*d^2 - 13*a*b^2*d^3)*x^2 - 2*(70*b^3*c^2*d - 119*a*b^ 
2*c*d^2 + 40*a^2*b*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(d^6*x^2 + 2*c*d^5 
*x + c^2*d^4)]
 

Sympy [F]

\[ \int \frac {(a+b x)^{7/2}}{(c+d x)^{5/2}} \, dx=\int \frac {\left (a + b x\right )^{\frac {7}{2}}}{\left (c + d x\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((b*x+a)**(7/2)/(d*x+c)**(5/2),x)
 

Output:

Integral((a + b*x)**(7/2)/(c + d*x)**(5/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+b x)^{7/2}}{(c+d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((b*x+a)^(7/2)/(d*x+c)^(5/2),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. 380 vs. \(2 (132) = 264\).

Time = 0.22 (sec) , antiderivative size = 380, normalized size of antiderivative = 2.24 \[ \int \frac {(a+b x)^{7/2}}{(c+d x)^{5/2}} \, dx=\frac {{\left ({\left (3 \, {\left (b x + a\right )} {\left (\frac {2 \, {\left (b^{6} c d^{6} - a b^{5} d^{7}\right )} {\left (b x + a\right )}}{b^{2} c d^{7} {\left | b \right |} - a b d^{8} {\left | b \right |}} - \frac {7 \, {\left (b^{7} c^{2} d^{5} - 2 \, a b^{6} c d^{6} + a^{2} b^{5} d^{7}\right )}}{b^{2} c d^{7} {\left | b \right |} - a b d^{8} {\left | b \right |}}\right )} - \frac {140 \, {\left (b^{8} c^{3} d^{4} - 3 \, a b^{7} c^{2} d^{5} + 3 \, a^{2} b^{6} c d^{6} - a^{3} b^{5} d^{7}\right )}}{b^{2} c d^{7} {\left | b \right |} - a b d^{8} {\left | b \right |}}\right )} {\left (b x + a\right )} - \frac {105 \, {\left (b^{9} c^{4} d^{3} - 4 \, a b^{8} c^{3} d^{4} + 6 \, a^{2} b^{7} c^{2} d^{5} - 4 \, a^{3} b^{6} c d^{6} + a^{4} b^{5} d^{7}\right )}}{b^{2} c d^{7} {\left | b \right |} - a b d^{8} {\left | b \right |}}\right )} \sqrt {b x + a}}{12 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} - \frac {35 \, {\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{4 \, \sqrt {b d} d^{4} {\left | b \right |}} \] Input:

integrate((b*x+a)^(7/2)/(d*x+c)^(5/2),x, algorithm="giac")
 

Output:

1/12*((3*(b*x + a)*(2*(b^6*c*d^6 - a*b^5*d^7)*(b*x + a)/(b^2*c*d^7*abs(b) 
- a*b*d^8*abs(b)) - 7*(b^7*c^2*d^5 - 2*a*b^6*c*d^6 + a^2*b^5*d^7)/(b^2*c*d 
^7*abs(b) - a*b*d^8*abs(b))) - 140*(b^8*c^3*d^4 - 3*a*b^7*c^2*d^5 + 3*a^2* 
b^6*c*d^6 - a^3*b^5*d^7)/(b^2*c*d^7*abs(b) - a*b*d^8*abs(b)))*(b*x + a) - 
105*(b^9*c^4*d^3 - 4*a*b^8*c^3*d^4 + 6*a^2*b^7*c^2*d^5 - 4*a^3*b^6*c*d^6 + 
 a^4*b^5*d^7)/(b^2*c*d^7*abs(b) - a*b*d^8*abs(b)))*sqrt(b*x + a)/(b^2*c + 
(b*x + a)*b*d - a*b*d)^(3/2) - 35/4*(b^5*c^2 - 2*a*b^4*c*d + a^2*b^3*d^2)* 
log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/( 
sqrt(b*d)*d^4*abs(b))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^{7/2}}{(c+d x)^{5/2}} \, dx=\int \frac {{\left (a+b\,x\right )}^{7/2}}{{\left (c+d\,x\right )}^{5/2}} \,d x \] Input:

int((a + b*x)^(7/2)/(c + d*x)^(5/2),x)
 

Output:

int((a + b*x)^(7/2)/(c + d*x)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 817, normalized size of antiderivative = 4.81 \[ \int \frac {(a+b x)^{7/2}}{(c+d x)^{5/2}} \, dx =\text {Too large to display} \] Input:

int((b*x+a)^(7/2)/(d*x+c)^(5/2),x)
 

Output:

( - 64*sqrt(c + d*x)*sqrt(a + b*x)*a**3*d**4 - 448*sqrt(c + d*x)*sqrt(a + 
b*x)*a**2*b*c*d**3 - 640*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b*d**4*x + 1400* 
sqrt(c + d*x)*sqrt(a + b*x)*a*b**2*c**2*d**2 + 1904*sqrt(c + d*x)*sqrt(a + 
 b*x)*a*b**2*c*d**3*x + 312*sqrt(c + d*x)*sqrt(a + b*x)*a*b**2*d**4*x**2 - 
 840*sqrt(c + d*x)*sqrt(a + b*x)*b**3*c**3*d - 1120*sqrt(c + d*x)*sqrt(a + 
 b*x)*b**3*c**2*d**2*x - 168*sqrt(c + d*x)*sqrt(a + b*x)*b**3*c*d**3*x**2 
+ 48*sqrt(c + d*x)*sqrt(a + b*x)*b**3*d**4*x**3 + 840*sqrt(d)*sqrt(b)*log( 
(sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**2*b*c* 
*2*d**2 + 1680*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c 
 + d*x))/sqrt(a*d - b*c))*a**2*b*c*d**3*x + 840*sqrt(d)*sqrt(b)*log((sqrt( 
d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**2*b*d**4*x** 
2 - 1680*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x 
))/sqrt(a*d - b*c))*a*b**2*c**3*d - 3360*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt 
(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a*b**2*c**2*d**2*x - 1 
680*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sq 
rt(a*d - b*c))*a*b**2*c*d**3*x**2 + 840*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt( 
a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*b**3*c**4 + 1680*sqrt(d 
)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b 
*c))*b**3*c**3*d*x + 840*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt 
(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*b**3*c**2*d**2*x**2 + 175*sqrt(d)*s...