Integrand size = 33, antiderivative size = 378 \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=-\frac {4 a \left (2 A b^2-a^2 C+3 b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^3 (a+b)^{3/2} d}+\frac {2 \left (2 a^2 C+3 a b (A+C)-b^2 (A+3 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^2 \sqrt {a+b} \left (a^2-b^2\right ) d}-\frac {2 \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {4 a \left (2 A b^2-a^2 C+3 b^2 C\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \]
-4/3*a*(2*A*b^2-C*a^2+3*C*b^2)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2) /(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+se c(d*x+c))/(a-b))^(1/2)/(a-b)/b^3/(a+b)^(3/2)/d+2/3*(2*C*a^2+3*a*b*(A+C)-b^ 2*(A+3*C))*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/ (a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/ 2)/b^2/(a^2-b^2)/d/(a+b)^(1/2)-2/3*(A*b^2+C*a^2)*tan(d*x+c)/b/(a^2-b^2)/d/ (a+b*sec(d*x+c))^(3/2)-4/3*a*(2*A*b^2-C*a^2+3*C*b^2)*tan(d*x+c)/b/(a^2-b^2 )^2/d/(a+b*sec(d*x+c))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(3369\) vs. \(2(378)=756\).
Time = 25.23 (sec) , antiderivative size = 3369, normalized size of antiderivative = 8.91 \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Result too large to show} \]
((b + a*Cos[c + d*x])^3*Sec[c + d*x]*(A + C*Sec[c + d*x]^2)*((-8*a*(-2*A*b ^2 + a^2*C - 3*b^2*C)*Sin[c + d*x])/(3*b^2*(a^2 - b^2)^2) + (4*(A*b^2*Sin[ c + d*x] + a^2*C*Sin[c + d*x]))/(3*a*(a^2 - b^2)*(b + a*Cos[c + d*x])^2) + (4*(-5*a^2*A*b^2*Sin[c + d*x] + A*b^4*Sin[c + d*x] + a^4*C*Sin[c + d*x] - 5*a^2*b^2*C*Sin[c + d*x]))/(3*a*b*(-a^2 + b^2)^2*(b + a*Cos[c + d*x]))))/ (d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^(5/2)) + (8*(b + a* Cos[c + d*x])^2*((-8*a*A*b)/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqr t[Sec[c + d*x]]) + (4*a^3*C)/(3*b*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]* Sqrt[Sec[c + d*x]]) - (4*a*b*C)/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*S qrt[Sec[c + d*x]]) - (2*a^2*A*Sqrt[Sec[c + d*x]])/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (2*A*b^2*Sqrt[Sec[c + d*x]])/(3*(-a^2 + b^2)^2*Sqrt[ b + a*Cos[c + d*x]]) - (10*a^2*C*Sqrt[Sec[c + d*x]])/(3*(-a^2 + b^2)^2*Sqr t[b + a*Cos[c + d*x]]) + (4*a^4*C*Sqrt[Sec[c + d*x]])/(3*b^2*(-a^2 + b^2)^ 2*Sqrt[b + a*Cos[c + d*x]]) + (2*b^2*C*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)^2 *Sqrt[b + a*Cos[c + d*x]]) - (8*a^2*A*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]]) /(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (4*a^2*C*Cos[2*(c + d*x)]*S qrt[Sec[c + d*x]])/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (4*a^4*C*Co s[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(3*b^2*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]))*Sqrt[Sec[c + d*x]]*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(A + C* Sec[c + d*x]^2)*(2*a*(a + b)*(-2*A*b^2 + (a^2 - 3*b^2)*C)*Sqrt[Cos[c + ...
Time = 1.41 (sec) , antiderivative size = 393, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4569, 27, 3042, 4491, 27, 3042, 4493, 3042, 4319, 4492}
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.
| \(\displaystyle \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4569 |
| \(\displaystyle -\frac {2 \int -\frac {\sec (c+d x) \left (3 a b (A+C)-\left (-2 C a^2+A b^2+3 b^2 C\right ) \sec (c+d x)\right )}{2 (a+b \sec (c+d x))^{3/2}}dx}{3 b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sec (c+d x) \left (3 a b (A+C)-\left (-2 C a^2+A b^2+3 b^2 C\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^{3/2}}dx}{3 b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 a b (A+C)+\left (2 C a^2-A b^2-3 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{3 b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4491 |
| \(\displaystyle \frac {-\frac {2 \int -\frac {\sec (c+d x) \left (b \left ((3 A+C) a^2+b^2 (A+3 C)\right )+2 a \left (-C a^2+2 A b^2+3 b^2 C\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{a^2-b^2}-\frac {4 a \left (2 A b^2-C \left (a^2-3 b^2\right )\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\sec (c+d x) \left (b \left ((3 A+C) a^2+b^2 (A+3 C)\right )+2 a \left (-C a^2+2 A b^2+3 b^2 C\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{a^2-b^2}-\frac {4 a \left (2 A b^2-C \left (a^2-3 b^2\right )\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left ((3 A+C) a^2+b^2 (A+3 C)\right )+2 a \left (-C a^2+2 A b^2+3 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2-b^2}-\frac {4 a \left (2 A b^2-C \left (a^2-3 b^2\right )\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4493 |
| \(\displaystyle \frac {\frac {(a-b) \left (2 a^2 C+3 a b (A+C)-b^2 (A+3 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+2 a \left (a^2 (-C)+2 A b^2+3 b^2 C\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx}{a^2-b^2}-\frac {4 a \left (2 A b^2-C \left (a^2-3 b^2\right )\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {(a-b) \left (2 a^2 C+3 a b (A+C)-b^2 (A+3 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+2 a \left (a^2 (-C)+2 A b^2+3 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2-b^2}-\frac {4 a \left (2 A b^2-C \left (a^2-3 b^2\right )\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {\frac {2 a \left (a^2 (-C)+2 A b^2+3 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (a-b) \sqrt {a+b} \left (2 a^2 C+3 a b (A+C)-b^2 (A+3 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{a^2-b^2}-\frac {4 a \left (2 A b^2-C \left (a^2-3 b^2\right )\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {\frac {\frac {2 (a-b) \sqrt {a+b} \left (2 a^2 C+3 a b (A+C)-b^2 (A+3 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {4 a (a-b) \sqrt {a+b} \left (a^2 (-C)+2 A b^2+3 b^2 C\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}}{a^2-b^2}-\frac {4 a \left (2 A b^2-C \left (a^2-3 b^2\right )\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
(-2*(A*b^2 + a^2*C)*Tan[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^ (3/2)) + (((-4*a*(a - b)*Sqrt[a + b]*(2*A*b^2 - a^2*C + 3*b^2*C)*Cot[c + d *x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b )]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b^2*d) + (2*(a - b)*Sqrt[a + b]*(2*a^2*C + 3*a*b*(A + C) - b^2*(A + 3*C))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b] ], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Se c[c + d*x]))/(a - b))])/(b*d))/(a^2 - b^2) - (4*a*(2*A*b^2 - (a^2 - 3*b^2) *C)*Tan[c + d*x])/((a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]))/(3*b*(a^2 - b^ 2))
3.8.50.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f* x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin[Sqrt [a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1 /((m + 1)*(a^2 - b^2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp [(a*A - b*B)*(m + 1) - (A*b - a*B)*(m + 2)*Csc[e + f*x], x], x], x] /; Free Q[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[m , -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(A - B) Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Simp[B Int[Csc[e + f*x]*((1 + Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A, B} , x] && NeQ[a^2 - b^2, 0] && NeQ[A^2 - B^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[ (e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-(A*b^2 + a^2*C) )*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^( m + 1)*Simp[a*b*(A + C)*(m + 1) - (A*b^2 + a^2*C + b*(A*b + b*C)*(m + 1))*C sc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C}, x] && LtQ[m, -1] && Ne Q[a^2 - b^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(5563\) vs. \(2(348)=696\).
Time = 17.29 (sec) , antiderivative size = 5564, normalized size of antiderivative = 14.72
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(5564\) |
| default | \(\text {Expression too large to display}\) | \(5624\) |
\[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
integral((C*sec(d*x + c)^3 + A*sec(d*x + c))*sqrt(b*sec(d*x + c) + a)/(b^3 *sec(d*x + c)^3 + 3*a*b^2*sec(d*x + c)^2 + 3*a^2*b*sec(d*x + c) + a^3), x)
\[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
Timed out. \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{\cos \left (c+d\,x\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]