Integrand size = 35, antiderivative size = 172 \[ \int \frac {(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 a^{5/2} B \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d}+\frac {2 a^3 (32 A+35 B) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (8 A+5 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)} \]
2*a^(5/2)*B*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/d+2/5*a*A*(a +a*cos(d*x+c))^(3/2)*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2/15*a^3*(32*A+35*B)*si n(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2)+2/15*a^2*(8*A+5*B)*sin( d*x+c)*(a+a*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(3/2)
Time = 0.52 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.76 \[ \int \frac {(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (30 \sqrt {2} B \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {5}{2}}(c+d x)+2 (49 A+40 B+2 (14 A+5 B) \cos (c+d x)+(43 A+40 B) \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{30 d \cos ^{\frac {5}{2}}(c+d x)} \]
(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(30*Sqrt[2]*B*ArcSin[Sqrt [2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^(5/2) + 2*(49*A + 40*B + 2*(14*A + 5*B) *Cos[c + d*x] + (43*A + 40*B)*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(30*d*C os[c + d*x]^(5/2))
Time = 0.97 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {3042, 3454, 27, 3042, 3454, 27, 3042, 3459, 3042, 3253, 223}
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.
| \(\displaystyle \int \frac {(a \cos (c+d x)+a)^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {2}{5} \int \frac {(\cos (c+d x) a+a)^{3/2} (a (8 A+5 B)+5 a B \cos (c+d x))}{2 \cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {(\cos (c+d x) a+a)^{3/2} (a (8 A+5 B)+5 a B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (8 A+5 B)+5 a B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {\sqrt {\cos (c+d x) a+a} \left ((32 A+35 B) a^2+15 B \cos (c+d x) a^2\right )}{2 \cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a^2 (8 A+5 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {\sqrt {\cos (c+d x) a+a} \left ((32 A+35 B) a^2+15 B \cos (c+d x) a^2\right )}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a^2 (8 A+5 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((32 A+35 B) a^2+15 B \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a^2 (8 A+5 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3459 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (15 a^2 B \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx+\frac {2 a^3 (32 A+35 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 (8 A+5 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (15 a^2 B \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^3 (32 A+35 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 (8 A+5 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3253 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {2 a^3 (32 A+35 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {30 a^2 B \int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {2 a^2 (8 A+5 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \frac {1}{5} \left (\frac {2 a^2 (8 A+5 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (\frac {30 a^{5/2} B \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {2 a^3 (32 A+35 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
(2*a*A*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + ((2*a^2*(8*A + 5*B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d*Cos[c + d *x]^(3/2)) + ((30*a^(5/2)*B*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (2*a^3*(32*A + 35*B)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*Sq rt[a + a*Cos[c + d*x]]))/3)/5
3.2.87.3.1 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[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.) *(x_)]], x_Symbol] :> Simp[-2/f Subst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Co s[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x] && E qQ[a^2 - b^2, 0] && EqQ[d, a/b]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-b^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp [a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B *(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 ])
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp [(-b^2)*(B*c - A*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1) *(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b* c - 2*a*d*(n + 1)))/(2*d*(n + 1)*(b*c + a*d)) Int[Sqrt[a + b*Sin[e + f*x] ]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x ] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1]
Time = 7.94 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.32
| method | result | size |
| default | \(\frac {2 a^{2} \left (15 B \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \left (\cos ^{3}\left (d x +c \right )\right )+15 B \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \left (\cos ^{2}\left (d x +c \right )\right )+43 A \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+40 B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+14 A \sin \left (d x +c \right ) \cos \left (d x +c \right )+5 B \sin \left (d x +c \right ) \cos \left (d x +c \right )+3 A \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{15 d \left (1+\cos \left (d x +c \right )\right ) \cos \left (d x +c \right )^{\frac {5}{2}}}\) | \(227\) |
| parts | \(\frac {2 A \sin \left (d x +c \right ) \left (43 \left (\cos ^{2}\left (d x +c \right )\right )+14 \cos \left (d x +c \right )+3\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, a^{2}}{15 d \left (1+\cos \left (d x +c \right )\right ) \cos \left (d x +c \right )^{\frac {5}{2}}}+\frac {2 B \left (3 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+3 \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+8 \cos \left (d x +c \right ) \sin \left (d x +c \right )+\sin \left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, a^{2}}{3 d \left (1+\cos \left (d x +c \right )\right ) \cos \left (d x +c \right )^{\frac {3}{2}}}\) | \(237\) |
2/15*a^2/d*(15*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(tan(d*x+c)*(cos( d*x+c)/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^3+15*B*(cos(d*x+c)/(1+cos(d*x+c)) )^(1/2)*arctan(tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^2+ 43*A*sin(d*x+c)*cos(d*x+c)^2+40*B*sin(d*x+c)*cos(d*x+c)^2+14*A*sin(d*x+c)* cos(d*x+c)+5*B*sin(d*x+c)*cos(d*x+c)+3*A*sin(d*x+c))*(a*(1+cos(d*x+c)))^(1 /2)/(1+cos(d*x+c))/cos(d*x+c)^(5/2)
Time = 0.33 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.94 \[ \int \frac {(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \, {\left ({\left ({\left (43 \, A + 40 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (14 \, A + 5 \, B\right )} a^{2} \cos \left (d x + c\right ) + 3 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 15 \, {\left (B a^{2} \cos \left (d x + c\right )^{4} + B a^{2} \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )\right )}}{15 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \]
2/15*(((43*A + 40*B)*a^2*cos(d*x + c)^2 + (14*A + 5*B)*a^2*cos(d*x + c) + 3*A*a^2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c) - 15*(B* a^2*cos(d*x + c)^4 + B*a^2*cos(d*x + c)^3)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))))/(d*cos(d*x + c)^4 + d *cos(d*x + c)^3)
Timed out. \[ \int \frac {(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 1548 vs. \(2 (148) = 296\).
Time = 0.47 (sec) , antiderivative size = 1548, normalized size of antiderivative = 9.00 \[ \int \frac {(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\text {Too large to display} \]
1/30*(5*(30*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(3/4)*a^(5/2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1) ) - 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^( 1/4)*((12*a^2*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(2*d *x + 2*c) - 3*a^2*sin(2*d*x + 2*c) - 4*(3*a^2*cos(2*d*x + 2*c) + 4*a^2)*si n(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(3/2*arctan2(sin(2* d*x + 2*c), cos(2*d*x + 2*c) + 1)) + (12*a^2*sin(2*d*x + 2*c)*sin(3/2*arct an2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 3*a^2*cos(2*d*x + 2*c) - a^2 + 4*(3*a^2*cos(2*d*x + 2*c) + 4*a^2)*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2 *d*x + 2*c))))*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*s qrt(a) + 3*((a^2*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*x + 2*c)^2 + 2*a^2*cos(2 *d*x + 2*c) + a^2)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*co s(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2 *c)))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - cos(1/2*a rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2* cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2 *arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d* x + 2*c), cos(2*d*x + 2*c)))) + 1) - (a^2*cos(2*d*x + 2*c)^2 + a^2*sin(...
Timed out. \[ \int \frac {(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^{7/2}} \,d x \]