\(\int \frac {(f+g x) (a+b \log (c (d+e x^2)^p))}{(h x)^{7/2}} \, dx\) [609]

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

Optimal result

Integrand size = 29, antiderivative size = 460 \[ \int \frac {(f+g x) \left (a+b \log \left (c \left (d+e x^2\right )^p\right )\right )}{(h x)^{7/2}} \, dx=-\frac {8 b e f p}{5 d h^3 \sqrt {h x}}+\frac {2 \sqrt {2} b e^{5/4} f p \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{e} \sqrt {h x}}{\sqrt [4]{d} \sqrt {h}}\right )}{5 d^{5/4} h^{7/2}}-\frac {2 \sqrt {2} b e^{3/4} g p \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{e} \sqrt {h x}}{\sqrt [4]{d} \sqrt {h}}\right )}{3 d^{3/4} h^{7/2}}-\frac {2 \sqrt {2} b e^{5/4} f p \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{e} \sqrt {h x}}{\sqrt [4]{d} \sqrt {h}}\right )}{5 d^{5/4} h^{7/2}}+\frac {2 \sqrt {2} b e^{3/4} g p \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{e} \sqrt {h x}}{\sqrt [4]{d} \sqrt {h}}\right )}{3 d^{3/4} h^{7/2}}+\frac {2 \sqrt {2} b e^{5/4} f p \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt [4]{e} \sqrt {h x}}{\sqrt {h} \left (\sqrt {d}+\sqrt {e} x\right )}\right )}{5 d^{5/4} h^{7/2}}+\frac {2 \sqrt {2} b e^{3/4} g p \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt [4]{e} \sqrt {h x}}{\sqrt {h} \left (\sqrt {d}+\sqrt {e} x\right )}\right )}{3 d^{3/4} h^{7/2}}-\frac {2 f \left (a+b \log \left (c \left (d+e x^2\right )^p\right )\right )}{5 h (h x)^{5/2}}-\frac {2 g \left (a+b \log \left (c \left (d+e x^2\right )^p\right )\right )}{3 h^2 (h x)^{3/2}} \] Output:

-8/5*b*e*f*p/d/h^3/(h*x)^(1/2)+2/5*2^(1/2)*b*e^(5/4)*f*p*arctan(1-2^(1/2)* 
e^(1/4)*(h*x)^(1/2)/d^(1/4)/h^(1/2))/d^(5/4)/h^(7/2)-2/3*2^(1/2)*b*e^(3/4) 
*g*p*arctan(1-2^(1/2)*e^(1/4)*(h*x)^(1/2)/d^(1/4)/h^(1/2))/d^(3/4)/h^(7/2) 
-2/5*2^(1/2)*b*e^(5/4)*f*p*arctan(1+2^(1/2)*e^(1/4)*(h*x)^(1/2)/d^(1/4)/h^ 
(1/2))/d^(5/4)/h^(7/2)+2/3*2^(1/2)*b*e^(3/4)*g*p*arctan(1+2^(1/2)*e^(1/4)* 
(h*x)^(1/2)/d^(1/4)/h^(1/2))/d^(3/4)/h^(7/2)+2/5*2^(1/2)*b*e^(5/4)*f*p*arc 
tanh(2^(1/2)*d^(1/4)*e^(1/4)*(h*x)^(1/2)/h^(1/2)/(d^(1/2)+e^(1/2)*x))/d^(5 
/4)/h^(7/2)+2/3*2^(1/2)*b*e^(3/4)*g*p*arctanh(2^(1/2)*d^(1/4)*e^(1/4)*(h*x 
)^(1/2)/h^(1/2)/(d^(1/2)+e^(1/2)*x))/d^(3/4)/h^(7/2)-2/5*f*(a+b*ln(c*(e*x^ 
2+d)^p))/h/(h*x)^(5/2)-2/3*g*(a+b*ln(c*(e*x^2+d)^p))/h^2/(h*x)^(3/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.26 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.51 \[ \int \frac {(f+g x) \left (a+b \log \left (c \left (d+e x^2\right )^p\right )\right )}{(h x)^{7/2}} \, dx=\frac {x \left (-24 b e f p x^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},1,\frac {3}{4},-\frac {e x^2}{d}\right )-5 \sqrt {2} b \sqrt [4]{d} e^{3/4} g p x^{5/2} \left (2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )-2 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )+\log \left (\sqrt {d}-\sqrt {2} \sqrt [4]{d} \sqrt [4]{e} \sqrt {x}+\sqrt {e} x\right )-\log \left (\sqrt {d}+\sqrt {2} \sqrt [4]{d} \sqrt [4]{e} \sqrt {x}+\sqrt {e} x\right )\right )-6 d f \left (a+b \log \left (c \left (d+e x^2\right )^p\right )\right )-10 d g x \left (a+b \log \left (c \left (d+e x^2\right )^p\right )\right )\right )}{15 d (h x)^{7/2}} \] Input:

Integrate[((f + g*x)*(a + b*Log[c*(d + e*x^2)^p]))/(h*x)^(7/2),x]
 

Output:

(x*(-24*b*e*f*p*x^2*Hypergeometric2F1[-1/4, 1, 3/4, -((e*x^2)/d)] - 5*Sqrt 
[2]*b*d^(1/4)*e^(3/4)*g*p*x^(5/2)*(2*ArcTan[1 - (Sqrt[2]*e^(1/4)*Sqrt[x])/ 
d^(1/4)] - 2*ArcTan[1 + (Sqrt[2]*e^(1/4)*Sqrt[x])/d^(1/4)] + Log[Sqrt[d] - 
 Sqrt[2]*d^(1/4)*e^(1/4)*Sqrt[x] + Sqrt[e]*x] - Log[Sqrt[d] + Sqrt[2]*d^(1 
/4)*e^(1/4)*Sqrt[x] + Sqrt[e]*x]) - 6*d*f*(a + b*Log[c*(d + e*x^2)^p]) - 1 
0*d*g*x*(a + b*Log[c*(d + e*x^2)^p])))/(15*d*(h*x)^(7/2))
 

Rubi [A] (verified)

Time = 1.46 (sec) , antiderivative size = 608, normalized size of antiderivative = 1.32, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2917, 27, 2926, 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.

\(\displaystyle \int \frac {(f+g x) \left (a+b \log \left (c \left (d+e x^2\right )^p\right )\right )}{(h x)^{7/2}} \, dx\)

\(\Big \downarrow \) 2917

\(\displaystyle \frac {2 \int \frac {(f h+g x h) \left (a+b \log \left (c \left (e x^2+d\right )^p\right )\right )}{h^4 x^3}d\sqrt {h x}}{h}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \int \frac {(f h+g x h) \left (a+b \log \left (c \left (e x^2+d\right )^p\right )\right )}{h^3 x^3}d\sqrt {h x}}{h^2}\)

\(\Big \downarrow \) 2926

\(\displaystyle \frac {2 \int \left (\frac {g \left (a+b \log \left (c \left (e x^2+d\right )^p\right )\right )}{h^2 x^2}+\frac {f \left (a+b \log \left (c \left (e x^2+d\right )^p\right )\right )}{h^2 x^3}\right )d\sqrt {h x}}{h^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 \left (-\frac {f h \left (a+b \log \left (c \left (d+e x^2\right )^p\right )\right )}{5 (h x)^{5/2}}-\frac {g \left (a+b \log \left (c \left (d+e x^2\right )^p\right )\right )}{3 (h x)^{3/2}}+\frac {\sqrt {2} b e^{5/4} f p \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{e} \sqrt {h x}}{\sqrt [4]{d} \sqrt {h}}\right )}{5 d^{5/4} h^{3/2}}-\frac {\sqrt {2} b e^{5/4} f p \arctan \left (\frac {\sqrt {2} \sqrt [4]{e} \sqrt {h x}}{\sqrt [4]{d} \sqrt {h}}+1\right )}{5 d^{5/4} h^{3/2}}-\frac {\sqrt {2} b e^{3/4} g p \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{e} \sqrt {h x}}{\sqrt [4]{d} \sqrt {h}}\right )}{3 d^{3/4} h^{3/2}}+\frac {\sqrt {2} b e^{3/4} g p \arctan \left (\frac {\sqrt {2} \sqrt [4]{e} \sqrt {h x}}{\sqrt [4]{d} \sqrt {h}}+1\right )}{3 d^{3/4} h^{3/2}}-\frac {b e^{5/4} f p \log \left (-\sqrt {2} \sqrt [4]{d} \sqrt [4]{e} \sqrt {h} \sqrt {h x}+\sqrt {d} h+\sqrt {e} h x\right )}{5 \sqrt {2} d^{5/4} h^{3/2}}+\frac {b e^{5/4} f p \log \left (\sqrt {2} \sqrt [4]{d} \sqrt [4]{e} \sqrt {h} \sqrt {h x}+\sqrt {d} h+\sqrt {e} h x\right )}{5 \sqrt {2} d^{5/4} h^{3/2}}-\frac {b e^{3/4} g p \log \left (-\sqrt {2} \sqrt [4]{d} \sqrt [4]{e} \sqrt {h} \sqrt {h x}+\sqrt {d} h+\sqrt {e} h x\right )}{3 \sqrt {2} d^{3/4} h^{3/2}}+\frac {b e^{3/4} g p \log \left (\sqrt {2} \sqrt [4]{d} \sqrt [4]{e} \sqrt {h} \sqrt {h x}+\sqrt {d} h+\sqrt {e} h x\right )}{3 \sqrt {2} d^{3/4} h^{3/2}}-\frac {4 b e f p}{5 d h \sqrt {h x}}\right )}{h^2}\)

Input:

Int[((f + g*x)*(a + b*Log[c*(d + e*x^2)^p]))/(h*x)^(7/2),x]
 

Output:

(2*((-4*b*e*f*p)/(5*d*h*Sqrt[h*x]) + (Sqrt[2]*b*e^(5/4)*f*p*ArcTan[1 - (Sq 
rt[2]*e^(1/4)*Sqrt[h*x])/(d^(1/4)*Sqrt[h])])/(5*d^(5/4)*h^(3/2)) - (Sqrt[2 
]*b*e^(3/4)*g*p*ArcTan[1 - (Sqrt[2]*e^(1/4)*Sqrt[h*x])/(d^(1/4)*Sqrt[h])]) 
/(3*d^(3/4)*h^(3/2)) - (Sqrt[2]*b*e^(5/4)*f*p*ArcTan[1 + (Sqrt[2]*e^(1/4)* 
Sqrt[h*x])/(d^(1/4)*Sqrt[h])])/(5*d^(5/4)*h^(3/2)) + (Sqrt[2]*b*e^(3/4)*g* 
p*ArcTan[1 + (Sqrt[2]*e^(1/4)*Sqrt[h*x])/(d^(1/4)*Sqrt[h])])/(3*d^(3/4)*h^ 
(3/2)) - (f*h*(a + b*Log[c*(d + e*x^2)^p]))/(5*(h*x)^(5/2)) - (g*(a + b*Lo 
g[c*(d + e*x^2)^p]))/(3*(h*x)^(3/2)) - (b*e^(5/4)*f*p*Log[Sqrt[d]*h + Sqrt 
[e]*h*x - Sqrt[2]*d^(1/4)*e^(1/4)*Sqrt[h]*Sqrt[h*x]])/(5*Sqrt[2]*d^(5/4)*h 
^(3/2)) - (b*e^(3/4)*g*p*Log[Sqrt[d]*h + Sqrt[e]*h*x - Sqrt[2]*d^(1/4)*e^( 
1/4)*Sqrt[h]*Sqrt[h*x]])/(3*Sqrt[2]*d^(3/4)*h^(3/2)) + (b*e^(5/4)*f*p*Log[ 
Sqrt[d]*h + Sqrt[e]*h*x + Sqrt[2]*d^(1/4)*e^(1/4)*Sqrt[h]*Sqrt[h*x]])/(5*S 
qrt[2]*d^(5/4)*h^(3/2)) + (b*e^(3/4)*g*p*Log[Sqrt[d]*h + Sqrt[e]*h*x + Sqr 
t[2]*d^(1/4)*e^(1/4)*Sqrt[h]*Sqrt[h*x]])/(3*Sqrt[2]*d^(3/4)*h^(3/2))))/h^2
 

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 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2917
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))^(p_.)]*(b_.))^(q_.)*((h_.) 
*(x_))^(m_)*((f_.) + (g_.)*(x_))^(r_.), x_Symbol] :> With[{k = Denominator[ 
m]}, Simp[k/h   Subst[Int[x^(k*(m + 1) - 1)*(f + g*(x^k/h))^r*(a + b*Log[c* 
(d + e*(x^(k*n)/h^n))^p])^q, x], x, (h*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, 
 e, f, g, h, p, r}, x] && FractionQ[m] && IntegerQ[n] && IntegerQ[r]
 

rule 2926
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m 
_.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol] :> Int[ExpandIntegrand[(a + b 
*Log[c*(d + e*x^n)^p])^q, x^m*(f + g*x^s)^r, x], x] /; FreeQ[{a, b, c, d, e 
, f, g, m, n, p, q, r, s}, x] && IGtQ[q, 0] && IntegerQ[m] && IntegerQ[r] & 
& IntegerQ[s]
 
Maple [F]

\[\int \frac {\left (g x +f \right ) \left (a +b \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )\right )}{\left (h x \right )^{\frac {7}{2}}}d x\]

Input:

int((g*x+f)*(a+b*ln(c*(e*x^2+d)^p))/(h*x)^(7/2),x)
 

Output:

int((g*x+f)*(a+b*ln(c*(e*x^2+d)^p))/(h*x)^(7/2),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1348 vs. \(2 (320) = 640\).

Time = 0.16 (sec) , antiderivative size = 1348, normalized size of antiderivative = 2.93 \[ \int \frac {(f+g x) \left (a+b \log \left (c \left (d+e x^2\right )^p\right )\right )}{(h x)^{7/2}} \, dx=\text {Too large to display} \] Input:

integrate((g*x+f)*(a+b*log(c*(e*x^2+d)^p))/(h*x)^(7/2),x, algorithm="frica 
s")
 

Output:

2/15*(d*h^4*x^3*sqrt((d^2*h^7*sqrt(-(81*b^4*e^5*f^4 - 450*b^4*d*e^4*f^2*g^ 
2 + 625*b^4*d^2*e^3*g^4)*p^4/(d^5*h^14)) + 30*b^2*e^2*f*g*p^2)/(d^2*h^7))* 
log(-32*(81*b^3*e^4*f^4 - 625*b^3*d^2*e^2*g^4)*sqrt(h*x)*p^3 + 32*(3*d^4*f 
*h^11*sqrt(-(81*b^4*e^5*f^4 - 450*b^4*d*e^4*f^2*g^2 + 625*b^4*d^2*e^3*g^4) 
*p^4/(d^5*h^14)) - 5*(9*b^2*d^2*e^2*f^2*g - 25*b^2*d^3*e*g^3)*h^4*p^2)*sqr 
t((d^2*h^7*sqrt(-(81*b^4*e^5*f^4 - 450*b^4*d*e^4*f^2*g^2 + 625*b^4*d^2*e^3 
*g^4)*p^4/(d^5*h^14)) + 30*b^2*e^2*f*g*p^2)/(d^2*h^7))) - d*h^4*x^3*sqrt(( 
d^2*h^7*sqrt(-(81*b^4*e^5*f^4 - 450*b^4*d*e^4*f^2*g^2 + 625*b^4*d^2*e^3*g^ 
4)*p^4/(d^5*h^14)) + 30*b^2*e^2*f*g*p^2)/(d^2*h^7))*log(-32*(81*b^3*e^4*f^ 
4 - 625*b^3*d^2*e^2*g^4)*sqrt(h*x)*p^3 - 32*(3*d^4*f*h^11*sqrt(-(81*b^4*e^ 
5*f^4 - 450*b^4*d*e^4*f^2*g^2 + 625*b^4*d^2*e^3*g^4)*p^4/(d^5*h^14)) - 5*( 
9*b^2*d^2*e^2*f^2*g - 25*b^2*d^3*e*g^3)*h^4*p^2)*sqrt((d^2*h^7*sqrt(-(81*b 
^4*e^5*f^4 - 450*b^4*d*e^4*f^2*g^2 + 625*b^4*d^2*e^3*g^4)*p^4/(d^5*h^14)) 
+ 30*b^2*e^2*f*g*p^2)/(d^2*h^7))) - d*h^4*x^3*sqrt(-(d^2*h^7*sqrt(-(81*b^4 
*e^5*f^4 - 450*b^4*d*e^4*f^2*g^2 + 625*b^4*d^2*e^3*g^4)*p^4/(d^5*h^14)) - 
30*b^2*e^2*f*g*p^2)/(d^2*h^7))*log(-32*(81*b^3*e^4*f^4 - 625*b^3*d^2*e^2*g 
^4)*sqrt(h*x)*p^3 + 32*(3*d^4*f*h^11*sqrt(-(81*b^4*e^5*f^4 - 450*b^4*d*e^4 
*f^2*g^2 + 625*b^4*d^2*e^3*g^4)*p^4/(d^5*h^14)) + 5*(9*b^2*d^2*e^2*f^2*g - 
 25*b^2*d^3*e*g^3)*h^4*p^2)*sqrt(-(d^2*h^7*sqrt(-(81*b^4*e^5*f^4 - 450*b^4 
*d*e^4*f^2*g^2 + 625*b^4*d^2*e^3*g^4)*p^4/(d^5*h^14)) - 30*b^2*e^2*f*g*...
 

Sympy [F(-1)]

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

integrate((g*x+f)*(a+b*ln(c*(e*x**2+d)**p))/(h*x)**(7/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 723 vs. \(2 (320) = 640\).

Time = 0.14 (sec) , antiderivative size = 723, normalized size of antiderivative = 1.57 \[ \int \frac {(f+g x) \left (a+b \log \left (c \left (d+e x^2\right )^p\right )\right )}{(h x)^{7/2}} \, dx =\text {Too large to display} \] Input:

integrate((g*x+f)*(a+b*log(c*(e*x^2+d)^p))/(h*x)^(7/2),x, algorithm="maxim 
a")
 

Output:

1/5*b*e*f*p*(e*(sqrt(2)*log(sqrt(e)*h*x + sqrt(2)*(d*h^2)^(1/4)*sqrt(h*x)* 
e^(1/4) + sqrt(d)*h)/((d*h^2)^(1/4)*e^(3/4)) - sqrt(2)*log(sqrt(e)*h*x - s 
qrt(2)*(d*h^2)^(1/4)*sqrt(h*x)*e^(1/4) + sqrt(d)*h)/((d*h^2)^(1/4)*e^(3/4) 
) - sqrt(2)*log(-(sqrt(2)*sqrt(-sqrt(d)*sqrt(e)*h) + sqrt(2)*(d*h^2)^(1/4) 
*e^(1/4) - 2*sqrt(h*x)*sqrt(e))/(sqrt(2)*sqrt(-sqrt(d)*sqrt(e)*h) - sqrt(2 
)*(d*h^2)^(1/4)*e^(1/4) + 2*sqrt(h*x)*sqrt(e)))/(sqrt(-sqrt(d)*sqrt(e)*h)* 
sqrt(e)) - sqrt(2)*log(-(sqrt(2)*sqrt(-sqrt(d)*sqrt(e)*h) - sqrt(2)*(d*h^2 
)^(1/4)*e^(1/4) - 2*sqrt(h*x)*sqrt(e))/(sqrt(2)*sqrt(-sqrt(d)*sqrt(e)*h) + 
 sqrt(2)*(d*h^2)^(1/4)*e^(1/4) + 2*sqrt(h*x)*sqrt(e)))/(sqrt(-sqrt(d)*sqrt 
(e)*h)*sqrt(e)))/d - 8/(sqrt(h*x)*d))/h^3 - 2/3*b*g*x^2*log((e*x^2 + d)^p* 
c)/(h*x)^(7/2) + 1/3*(sqrt(2)*h^2*log(sqrt(e)*h*x + sqrt(2)*(d*h^2)^(1/4)* 
sqrt(h*x)*e^(1/4) + sqrt(d)*h)/((d*h^2)^(3/4)*e^(1/4)) - sqrt(2)*h^2*log(s 
qrt(e)*h*x - sqrt(2)*(d*h^2)^(1/4)*sqrt(h*x)*e^(1/4) + sqrt(d)*h)/((d*h^2) 
^(3/4)*e^(1/4)) + sqrt(2)*h*log(-(sqrt(2)*sqrt(-sqrt(d)*sqrt(e)*h) + sqrt( 
2)*(d*h^2)^(1/4)*e^(1/4) - 2*sqrt(h*x)*sqrt(e))/(sqrt(2)*sqrt(-sqrt(d)*sqr 
t(e)*h) - sqrt(2)*(d*h^2)^(1/4)*e^(1/4) + 2*sqrt(h*x)*sqrt(e)))/(sqrt(-sqr 
t(d)*sqrt(e)*h)*sqrt(d)) + sqrt(2)*h*log(-(sqrt(2)*sqrt(-sqrt(d)*sqrt(e)*h 
) - sqrt(2)*(d*h^2)^(1/4)*e^(1/4) - 2*sqrt(h*x)*sqrt(e))/(sqrt(2)*sqrt(-sq 
rt(d)*sqrt(e)*h) + sqrt(2)*(d*h^2)^(1/4)*e^(1/4) + 2*sqrt(h*x)*sqrt(e)))/( 
sqrt(-sqrt(d)*sqrt(e)*h)*sqrt(d)))*b*e*g*p/h^4 - 2/3*a*g*x^2/(h*x)^(7/2...
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 494, normalized size of antiderivative = 1.07 \[ \int \frac {(f+g x) \left (a+b \log \left (c \left (d+e x^2\right )^p\right )\right )}{(h x)^{7/2}} \, dx=\frac {\frac {2 \, {\left (5 \, \sqrt {2} \left (d e^{3} h^{2}\right )^{\frac {1}{4}} b d e g h p - 3 \, \sqrt {2} \left (d e^{3} h^{2}\right )^{\frac {3}{4}} b f p\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {d h^{2}}{e}\right )^{\frac {1}{4}} + 2 \, \sqrt {h x}\right )}}{2 \, \left (\frac {d h^{2}}{e}\right )^{\frac {1}{4}}}\right )}{d^{2} e h} + \frac {2 \, {\left (5 \, \sqrt {2} \left (d e^{3} h^{2}\right )^{\frac {1}{4}} b d e g h p - 3 \, \sqrt {2} \left (d e^{3} h^{2}\right )^{\frac {3}{4}} b f p\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {d h^{2}}{e}\right )^{\frac {1}{4}} - 2 \, \sqrt {h x}\right )}}{2 \, \left (\frac {d h^{2}}{e}\right )^{\frac {1}{4}}}\right )}{d^{2} e h} + \frac {{\left (5 \, \sqrt {2} \left (d e^{3} h^{2}\right )^{\frac {1}{4}} b d e g h p + 3 \, \sqrt {2} \left (d e^{3} h^{2}\right )^{\frac {3}{4}} b f p\right )} \log \left (h x + \sqrt {2} \left (\frac {d h^{2}}{e}\right )^{\frac {1}{4}} \sqrt {h x} + \sqrt {\frac {d h^{2}}{e}}\right )}{d^{2} e h} - \frac {{\left (5 \, \sqrt {2} \left (d e^{3} h^{2}\right )^{\frac {1}{4}} b d e g h p + 3 \, \sqrt {2} \left (d e^{3} h^{2}\right )^{\frac {3}{4}} b f p\right )} \log \left (h x - \sqrt {2} \left (\frac {d h^{2}}{e}\right )^{\frac {1}{4}} \sqrt {h x} + \sqrt {\frac {d h^{2}}{e}}\right )}{d^{2} e h} - \frac {2 \, {\left (5 \, b g h^{3} p x + 3 \, b f h^{3} p\right )} \log \left (e h^{2} x^{2} + d h^{2}\right )}{\sqrt {h x} h^{2} x^{2}} - \frac {2 \, {\left (12 \, b e f h^{3} p x^{2} - 5 \, b d g h^{3} p x \log \left (h^{2}\right ) - 3 \, b d f h^{3} p \log \left (h^{2}\right ) + 5 \, b d g h^{3} x \log \left (c\right ) + 5 \, a d g h^{3} x + 3 \, b d f h^{3} \log \left (c\right ) + 3 \, a d f h^{3}\right )}}{\sqrt {h x} d h^{2} x^{2}}}{15 \, h^{4}} \] Input:

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

Output:

1/15*(2*(5*sqrt(2)*(d*e^3*h^2)^(1/4)*b*d*e*g*h*p - 3*sqrt(2)*(d*e^3*h^2)^( 
3/4)*b*f*p)*arctan(1/2*sqrt(2)*(sqrt(2)*(d*h^2/e)^(1/4) + 2*sqrt(h*x))/(d* 
h^2/e)^(1/4))/(d^2*e*h) + 2*(5*sqrt(2)*(d*e^3*h^2)^(1/4)*b*d*e*g*h*p - 3*s 
qrt(2)*(d*e^3*h^2)^(3/4)*b*f*p)*arctan(-1/2*sqrt(2)*(sqrt(2)*(d*h^2/e)^(1/ 
4) - 2*sqrt(h*x))/(d*h^2/e)^(1/4))/(d^2*e*h) + (5*sqrt(2)*(d*e^3*h^2)^(1/4 
)*b*d*e*g*h*p + 3*sqrt(2)*(d*e^3*h^2)^(3/4)*b*f*p)*log(h*x + sqrt(2)*(d*h^ 
2/e)^(1/4)*sqrt(h*x) + sqrt(d*h^2/e))/(d^2*e*h) - (5*sqrt(2)*(d*e^3*h^2)^( 
1/4)*b*d*e*g*h*p + 3*sqrt(2)*(d*e^3*h^2)^(3/4)*b*f*p)*log(h*x - sqrt(2)*(d 
*h^2/e)^(1/4)*sqrt(h*x) + sqrt(d*h^2/e))/(d^2*e*h) - 2*(5*b*g*h^3*p*x + 3* 
b*f*h^3*p)*log(e*h^2*x^2 + d*h^2)/(sqrt(h*x)*h^2*x^2) - 2*(12*b*e*f*h^3*p* 
x^2 - 5*b*d*g*h^3*p*x*log(h^2) - 3*b*d*f*h^3*p*log(h^2) + 5*b*d*g*h^3*x*lo 
g(c) + 5*a*d*g*h^3*x + 3*b*d*f*h^3*log(c) + 3*a*d*f*h^3)/(sqrt(h*x)*d*h^2* 
x^2))/h^4
 

Mupad [F(-1)]

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

int(((f + g*x)*(a + b*log(c*(d + e*x^2)^p)))/(h*x)^(7/2),x)
 

Output:

int(((f + g*x)*(a + b*log(c*(d + e*x^2)^p)))/(h*x)^(7/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 400, normalized size of antiderivative = 0.87 \[ \int \frac {(f+g x) \left (a+b \log \left (c \left (d+e x^2\right )^p\right )\right )}{(h x)^{7/2}} \, dx=\frac {\sqrt {h}\, \left (6 \sqrt {x}\, e^{\frac {5}{4}} d^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {e^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {2}-2 \sqrt {x}\, \sqrt {e}}{e^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {2}}\right ) b f p \,x^{2}-10 \sqrt {x}\, e^{\frac {3}{4}} d^{\frac {5}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {e^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {2}-2 \sqrt {x}\, \sqrt {e}}{e^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {2}}\right ) b g p \,x^{2}-6 \sqrt {x}\, e^{\frac {5}{4}} d^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {e^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {2}+2 \sqrt {x}\, \sqrt {e}}{e^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {2}}\right ) b f p \,x^{2}+10 \sqrt {x}\, e^{\frac {3}{4}} d^{\frac {5}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {e^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {2}+2 \sqrt {x}\, \sqrt {e}}{e^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {2}}\right ) b g p \,x^{2}-6 \sqrt {x}\, e^{\frac {5}{4}} d^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, e^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {2}+\sqrt {d}+\sqrt {e}\, x \right ) b f p \,x^{2}+3 \sqrt {x}\, e^{\frac {5}{4}} d^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) b f \,x^{2}-10 \sqrt {x}\, e^{\frac {3}{4}} d^{\frac {5}{4}} \sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, e^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {2}+\sqrt {d}+\sqrt {e}\, x \right ) b g p \,x^{2}+5 \sqrt {x}\, e^{\frac {3}{4}} d^{\frac {5}{4}} \sqrt {2}\, \mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) b g \,x^{2}-6 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) b \,d^{2} f -10 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) b \,d^{2} g x -6 a \,d^{2} f -10 a \,d^{2} g x -24 b d e f p \,x^{2}\right )}{15 \sqrt {x}\, d^{2} h^{4} x^{2}} \] Input:

int((g*x+f)*(a+b*log(c*(e*x^2+d)^p))/(h*x)^(7/2),x)
 

Output:

(sqrt(h)*(6*sqrt(x)*e**(1/4)*d**(3/4)*sqrt(2)*atan((e**(1/4)*d**(1/4)*sqrt 
(2) - 2*sqrt(x)*sqrt(e))/(e**(1/4)*d**(1/4)*sqrt(2)))*b*e*f*p*x**2 - 10*sq 
rt(x)*e**(3/4)*d**(1/4)*sqrt(2)*atan((e**(1/4)*d**(1/4)*sqrt(2) - 2*sqrt(x 
)*sqrt(e))/(e**(1/4)*d**(1/4)*sqrt(2)))*b*d*g*p*x**2 - 6*sqrt(x)*e**(1/4)* 
d**(3/4)*sqrt(2)*atan((e**(1/4)*d**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(e))/(e** 
(1/4)*d**(1/4)*sqrt(2)))*b*e*f*p*x**2 + 10*sqrt(x)*e**(3/4)*d**(1/4)*sqrt( 
2)*atan((e**(1/4)*d**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(e))/(e**(1/4)*d**(1/4) 
*sqrt(2)))*b*d*g*p*x**2 - 6*sqrt(x)*e**(1/4)*d**(3/4)*sqrt(2)*log( - sqrt( 
x)*e**(1/4)*d**(1/4)*sqrt(2) + sqrt(d) + sqrt(e)*x)*b*e*f*p*x**2 + 3*sqrt( 
x)*e**(1/4)*d**(3/4)*sqrt(2)*log((d + e*x**2)**p*c)*b*e*f*x**2 - 10*sqrt(x 
)*e**(3/4)*d**(1/4)*sqrt(2)*log( - sqrt(x)*e**(1/4)*d**(1/4)*sqrt(2) + sqr 
t(d) + sqrt(e)*x)*b*d*g*p*x**2 + 5*sqrt(x)*e**(3/4)*d**(1/4)*sqrt(2)*log(( 
d + e*x**2)**p*c)*b*d*g*x**2 - 6*log((d + e*x**2)**p*c)*b*d**2*f - 10*log( 
(d + e*x**2)**p*c)*b*d**2*g*x - 6*a*d**2*f - 10*a*d**2*g*x - 24*b*d*e*f*p* 
x**2))/(15*sqrt(x)*d**2*h**4*x**2)