Integrand size = 18, antiderivative size = 216 \[ \int \frac {\cos ^2(a+b x)}{(c+d x)^{7/2}} \, dx=-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}-\frac {2 \cos ^2(a+b x)}{5 d (c+d x)^{5/2}}+\frac {32 b^2 \cos ^2(a+b x)}{15 d^3 \sqrt {c+d x}}+\frac {32 b^{5/2} \sqrt {\pi } \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{15 d^{7/2}}+\frac {32 b^{5/2} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{15 d^{7/2}}+\frac {8 b \cos (a+b x) \sin (a+b x)}{15 d^2 (c+d x)^{3/2}} \] Output:
-16/15*b^2/d^3/(d*x+c)^(1/2)-2/5*cos(b*x+a)^2/d/(d*x+c)^(5/2)+32/15*b^2*co s(b*x+a)^2/d^3/(d*x+c)^(1/2)+32/15*b^(5/2)*Pi^(1/2)*cos(2*a-2*b*c/d)*Fresn elS(2*b^(1/2)*(d*x+c)^(1/2)/d^(1/2)/Pi^(1/2))/d^(7/2)+32/15*b^(5/2)*Pi^(1/ 2)*FresnelC(2*b^(1/2)*(d*x+c)^(1/2)/d^(1/2)/Pi^(1/2))*sin(2*a-2*b*c/d)/d^( 7/2)+8/15*b*cos(b*x+a)*sin(b*x+a)/d^2/(d*x+c)^(3/2)
Result contains complex when optimal does not.
Time = 0.90 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.10 \[ \int \frac {\cos ^2(a+b x)}{(c+d x)^{7/2}} \, dx=\frac {-6 d^2+e^{2 i a} \left (-3 d^2 e^{2 i b x}+4 b e^{-\frac {2 i b c}{d}} (c+d x) \left (e^{\frac {2 i b (c+d x)}{d}} (-i d+4 b (c+d x))-4 i \sqrt {2} d \left (-\frac {i b (c+d x)}{d}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {2 i b (c+d x)}{d}\right )\right )\right )+e^{-2 i (a+b x)} \left (-3 d^2-2 i b (c+d x) \left (-2 d+8 i b (c+d x)-8 \sqrt {2} d e^{\frac {2 i b (c+d x)}{d}} \left (\frac {i b (c+d x)}{d}\right )^{3/2} \Gamma \left (\frac {1}{2},\frac {2 i b (c+d x)}{d}\right )\right )\right )}{30 d^3 (c+d x)^{5/2}} \] Input:
Integrate[Cos[a + b*x]^2/(c + d*x)^(7/2),x]
Output:
(-6*d^2 + E^((2*I)*a)*(-3*d^2*E^((2*I)*b*x) + (4*b*(c + d*x)*(E^(((2*I)*b* (c + d*x))/d)*((-I)*d + 4*b*(c + d*x)) - (4*I)*Sqrt[2]*d*(((-I)*b*(c + d*x ))/d)^(3/2)*Gamma[1/2, ((-2*I)*b*(c + d*x))/d]))/E^(((2*I)*b*c)/d)) + (-3* d^2 - (2*I)*b*(c + d*x)*(-2*d + (8*I)*b*(c + d*x) - 8*Sqrt[2]*d*E^(((2*I)* b*(c + d*x))/d)*((I*b*(c + d*x))/d)^(3/2)*Gamma[1/2, ((2*I)*b*(c + d*x))/d ]))/E^((2*I)*(a + b*x)))/(30*d^3*(c + d*x)^(5/2))
Time = 0.99 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.03, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.722, Rules used = {3042, 3795, 17, 3042, 3794, 27, 3042, 3787, 3042, 3785, 3786, 3832, 3833}
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.
| \(\displaystyle \int \frac {\cos ^2(a+b x)}{(c+d x)^{7/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (a+b x+\frac {\pi }{2}\right )^2}{(c+d x)^{7/2}}dx\) |
| \(\Big \downarrow \) 3795 |
| \(\displaystyle -\frac {16 b^2 \int \frac {\cos ^2(a+b x)}{(c+d x)^{3/2}}dx}{15 d^2}+\frac {8 b^2 \int \frac {1}{(c+d x)^{3/2}}dx}{15 d^2}+\frac {8 b \sin (a+b x) \cos (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \cos ^2(a+b x)}{5 d (c+d x)^{5/2}}\) |
| \(\Big \downarrow \) 17 |
| \(\displaystyle -\frac {16 b^2 \int \frac {\cos ^2(a+b x)}{(c+d x)^{3/2}}dx}{15 d^2}+\frac {8 b \sin (a+b x) \cos (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \cos ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {16 b^2 \int \frac {\sin \left (a+b x+\frac {\pi }{2}\right )^2}{(c+d x)^{3/2}}dx}{15 d^2}+\frac {8 b \sin (a+b x) \cos (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \cos ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}\) |
| \(\Big \downarrow \) 3794 |
| \(\displaystyle -\frac {16 b^2 \left (\frac {4 b \int -\frac {\sin (2 a+2 b x)}{2 \sqrt {c+d x}}dx}{d}-\frac {2 \cos ^2(a+b x)}{d \sqrt {c+d x}}\right )}{15 d^2}+\frac {8 b \sin (a+b x) \cos (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \cos ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {16 b^2 \left (-\frac {2 b \int \frac {\sin (2 a+2 b x)}{\sqrt {c+d x}}dx}{d}-\frac {2 \cos ^2(a+b x)}{d \sqrt {c+d x}}\right )}{15 d^2}+\frac {8 b \sin (a+b x) \cos (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \cos ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {16 b^2 \left (-\frac {2 b \int \frac {\sin (2 a+2 b x)}{\sqrt {c+d x}}dx}{d}-\frac {2 \cos ^2(a+b x)}{d \sqrt {c+d x}}\right )}{15 d^2}+\frac {8 b \sin (a+b x) \cos (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \cos ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}\) |
| \(\Big \downarrow \) 3787 |
| \(\displaystyle -\frac {16 b^2 \left (-\frac {2 b \left (\sin \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\cos \left (\frac {2 b c}{d}+2 b x\right )}{\sqrt {c+d x}}dx+\cos \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{\sqrt {c+d x}}dx\right )}{d}-\frac {2 \cos ^2(a+b x)}{d \sqrt {c+d x}}\right )}{15 d^2}+\frac {8 b \sin (a+b x) \cos (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \cos ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {16 b^2 \left (-\frac {2 b \left (\sin \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x+\frac {\pi }{2}\right )}{\sqrt {c+d x}}dx+\cos \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{\sqrt {c+d x}}dx\right )}{d}-\frac {2 \cos ^2(a+b x)}{d \sqrt {c+d x}}\right )}{15 d^2}+\frac {8 b \sin (a+b x) \cos (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \cos ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}\) |
| \(\Big \downarrow \) 3785 |
| \(\displaystyle -\frac {16 b^2 \left (-\frac {2 b \left (\frac {2 \sin \left (2 a-\frac {2 b c}{d}\right ) \int \cos \left (\frac {2 b (c+d x)}{d}\right )d\sqrt {c+d x}}{d}+\cos \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{\sqrt {c+d x}}dx\right )}{d}-\frac {2 \cos ^2(a+b x)}{d \sqrt {c+d x}}\right )}{15 d^2}+\frac {8 b \sin (a+b x) \cos (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \cos ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}\) |
| \(\Big \downarrow \) 3786 |
| \(\displaystyle -\frac {16 b^2 \left (-\frac {2 b \left (\frac {2 \sin \left (2 a-\frac {2 b c}{d}\right ) \int \cos \left (\frac {2 b (c+d x)}{d}\right )d\sqrt {c+d x}}{d}+\frac {2 \cos \left (2 a-\frac {2 b c}{d}\right ) \int \sin \left (\frac {2 b (c+d x)}{d}\right )d\sqrt {c+d x}}{d}\right )}{d}-\frac {2 \cos ^2(a+b x)}{d \sqrt {c+d x}}\right )}{15 d^2}+\frac {8 b \sin (a+b x) \cos (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \cos ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}\) |
| \(\Big \downarrow \) 3832 |
| \(\displaystyle -\frac {16 b^2 \left (-\frac {2 b \left (\frac {2 \sin \left (2 a-\frac {2 b c}{d}\right ) \int \cos \left (\frac {2 b (c+d x)}{d}\right )d\sqrt {c+d x}}{d}+\frac {\sqrt {\pi } \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{\sqrt {b} \sqrt {d}}\right )}{d}-\frac {2 \cos ^2(a+b x)}{d \sqrt {c+d x}}\right )}{15 d^2}+\frac {8 b \sin (a+b x) \cos (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \cos ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}\) |
| \(\Big \downarrow \) 3833 |
| \(\displaystyle -\frac {16 b^2 \left (-\frac {2 b \left (\frac {\sqrt {\pi } \sin \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{\sqrt {b} \sqrt {d}}+\frac {\sqrt {\pi } \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{\sqrt {b} \sqrt {d}}\right )}{d}-\frac {2 \cos ^2(a+b x)}{d \sqrt {c+d x}}\right )}{15 d^2}+\frac {8 b \sin (a+b x) \cos (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \cos ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}\) |
Input:
Int[Cos[a + b*x]^2/(c + d*x)^(7/2),x]
Output:
(-16*b^2)/(15*d^3*Sqrt[c + d*x]) - (2*Cos[a + b*x]^2)/(5*d*(c + d*x)^(5/2) ) - (16*b^2*((-2*Cos[a + b*x]^2)/(d*Sqrt[c + d*x]) - (2*b*((Sqrt[Pi]*Cos[2 *a - (2*b*c)/d]*FresnelS[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])])/(S qrt[b]*Sqrt[d]) + (Sqrt[Pi]*FresnelC[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sq rt[Pi])]*Sin[2*a - (2*b*c)/d])/(Sqrt[b]*Sqrt[d])))/d))/(15*d^2) + (8*b*Cos [a + b*x]*Sin[a + b*x])/(15*d^2*(c + d*x)^(3/2))
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> S imp[2/d Subst[Int[Cos[f*(x^2/d)], x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]
Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[2/d Subst[Int[Sin[f*(x^2/d)], x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f }, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]
Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Cos [(d*e - c*f)/d] Int[Sin[c*(f/d) + f*x]/Sqrt[c + d*x], x], x] + Simp[Sin[( d*e - c*f)/d] Int[Cos[c*(f/d) + f*x]/Sqrt[c + d*x], x], x] /; FreeQ[{c, d , e, f}, x] && ComplexFreeQ[f] && NeQ[d*e - c*f, 0]
Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Si mp[(c + d*x)^(m + 1)*(Sin[e + f*x]^n/(d*(m + 1))), x] - Simp[f*(n/(d*(m + 1 ))) Int[ExpandTrigReduce[(c + d*x)^(m + 1), Cos[e + f*x]*Sin[e + f*x]^(n - 1), x], x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && GeQ[m, -2] & & LtQ[m, -1]
Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbo l] :> Simp[(c + d*x)^(m + 1)*((b*Sin[e + f*x])^n/(d*(m + 1))), x] + (-Simp[ b*f*n*(c + d*x)^(m + 2)*Cos[e + f*x]*((b*Sin[e + f*x])^(n - 1)/(d^2*(m + 1) *(m + 2))), x] + Simp[b^2*f^2*n*((n - 1)/(d^2*(m + 1)*(m + 2))) Int[(c + d*x)^(m + 2)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[f^2*(n^2/(d^2*(m + 1)* (m + 2))) Int[(c + d*x)^(m + 2)*(b*Sin[e + f*x])^n, x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && LtQ[m, -2]
Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
Time = 1.76 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.06
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{5 \left (d x +c \right )^{\frac {5}{2}}}-\frac {\cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 b c}{d}\right )}{5 \left (d x +c \right )^{\frac {5}{2}}}-\frac {4 b \left (-\frac {\sin \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 b c}{d}\right )}{3 \left (d x +c \right )^{\frac {3}{2}}}+\frac {4 b \left (-\frac {\cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 b c}{d}\right )}{\sqrt {d x +c}}-\frac {2 b \sqrt {\pi }\, \left (\cos \left (\frac {2 a d -2 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {2 a d -2 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{d \sqrt {\frac {b}{d}}}\right )}{3 d}\right )}{5 d}}{d}\) | \(230\) |
| default | \(\frac {-\frac {1}{5 \left (d x +c \right )^{\frac {5}{2}}}-\frac {\cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 b c}{d}\right )}{5 \left (d x +c \right )^{\frac {5}{2}}}-\frac {4 b \left (-\frac {\sin \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 b c}{d}\right )}{3 \left (d x +c \right )^{\frac {3}{2}}}+\frac {4 b \left (-\frac {\cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 b c}{d}\right )}{\sqrt {d x +c}}-\frac {2 b \sqrt {\pi }\, \left (\cos \left (\frac {2 a d -2 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {2 a d -2 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{d \sqrt {\frac {b}{d}}}\right )}{3 d}\right )}{5 d}}{d}\) | \(230\) |
Input:
int(cos(b*x+a)^2/(d*x+c)^(7/2),x,method=_RETURNVERBOSE)
Output:
2/d*(-1/10/(d*x+c)^(5/2)-1/10/(d*x+c)^(5/2)*cos(2*b*(d*x+c)/d+2*(a*d-b*c)/ d)-2/5*b/d*(-1/3/(d*x+c)^(3/2)*sin(2*b*(d*x+c)/d+2*(a*d-b*c)/d)+4/3*b/d*(- 1/(d*x+c)^(1/2)*cos(2*b*(d*x+c)/d+2*(a*d-b*c)/d)-2*b/d*Pi^(1/2)/(b/d)^(1/2 )*(cos(2*(a*d-b*c)/d)*FresnelS(2/Pi^(1/2)/(b/d)^(1/2)*b*(d*x+c)^(1/2)/d)+s in(2*(a*d-b*c)/d)*FresnelC(2/Pi^(1/2)/(b/d)^(1/2)*b*(d*x+c)^(1/2)/d)))))
Time = 0.11 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.50 \[ \int \frac {\cos ^2(a+b x)}{(c+d x)^{7/2}} \, dx=\frac {2 \, {\left (16 \, {\left (\pi b^{2} d^{3} x^{3} + 3 \, \pi b^{2} c d^{2} x^{2} + 3 \, \pi b^{2} c^{2} d x + \pi b^{2} c^{3}\right )} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {S}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) + 16 \, {\left (\pi b^{2} d^{3} x^{3} + 3 \, \pi b^{2} c d^{2} x^{2} + 3 \, \pi b^{2} c^{2} d x + \pi b^{2} c^{3}\right )} \sqrt {\frac {b}{\pi d}} \operatorname {C}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - {\left (8 \, b^{2} d^{2} x^{2} + 16 \, b^{2} c d x + 8 \, b^{2} c^{2} - {\left (16 \, b^{2} d^{2} x^{2} + 32 \, b^{2} c d x + 16 \, b^{2} c^{2} - 3 \, d^{2}\right )} \cos \left (b x + a\right )^{2} - 4 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right )\right )} \sqrt {d x + c}\right )}}{15 \, {\left (d^{6} x^{3} + 3 \, c d^{5} x^{2} + 3 \, c^{2} d^{4} x + c^{3} d^{3}\right )}} \] Input:
integrate(cos(b*x+a)^2/(d*x+c)^(7/2),x, algorithm="fricas")
Output:
2/15*(16*(pi*b^2*d^3*x^3 + 3*pi*b^2*c*d^2*x^2 + 3*pi*b^2*c^2*d*x + pi*b^2* c^3)*sqrt(b/(pi*d))*cos(-2*(b*c - a*d)/d)*fresnel_sin(2*sqrt(d*x + c)*sqrt (b/(pi*d))) + 16*(pi*b^2*d^3*x^3 + 3*pi*b^2*c*d^2*x^2 + 3*pi*b^2*c^2*d*x + pi*b^2*c^3)*sqrt(b/(pi*d))*fresnel_cos(2*sqrt(d*x + c)*sqrt(b/(pi*d)))*si n(-2*(b*c - a*d)/d) - (8*b^2*d^2*x^2 + 16*b^2*c*d*x + 8*b^2*c^2 - (16*b^2* d^2*x^2 + 32*b^2*c*d*x + 16*b^2*c^2 - 3*d^2)*cos(b*x + a)^2 - 4*(b*d^2*x + b*c*d)*cos(b*x + a)*sin(b*x + a))*sqrt(d*x + c))/(d^6*x^3 + 3*c*d^5*x^2 + 3*c^2*d^4*x + c^3*d^3)
\[ \int \frac {\cos ^2(a+b x)}{(c+d x)^{7/2}} \, dx=\int \frac {\cos ^{2}{\left (a + b x \right )}}{\left (c + d x\right )^{\frac {7}{2}}}\, dx \] Input:
integrate(cos(b*x+a)**2/(d*x+c)**(7/2),x)
Output:
Integral(cos(a + b*x)**2/(c + d*x)**(7/2), x)
Result contains complex when optimal does not.
Time = 0.22 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.63 \[ \int \frac {\cos ^2(a+b x)}{(c+d x)^{7/2}} \, dx=\frac {5 \, \sqrt {2} {\left ({\left (\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {5}{2}, \frac {2 i \, {\left (d x + c\right )} b}{d}\right ) - \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {5}{2}, -\frac {2 i \, {\left (d x + c\right )} b}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + {\left (-\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {5}{2}, \frac {2 i \, {\left (d x + c\right )} b}{d}\right ) + \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {5}{2}, -\frac {2 i \, {\left (d x + c\right )} b}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \left (\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {5}{2}} - 2}{10 \, {\left (d x + c\right )}^{\frac {5}{2}} d} \] Input:
integrate(cos(b*x+a)^2/(d*x+c)^(7/2),x, algorithm="maxima")
Output:
1/10*(5*sqrt(2)*(((I + 1)*sqrt(2)*gamma(-5/2, 2*I*(d*x + c)*b/d) - (I - 1) *sqrt(2)*gamma(-5/2, -2*I*(d*x + c)*b/d))*cos(-2*(b*c - a*d)/d) + (-(I - 1 )*sqrt(2)*gamma(-5/2, 2*I*(d*x + c)*b/d) + (I + 1)*sqrt(2)*gamma(-5/2, -2* I*(d*x + c)*b/d))*sin(-2*(b*c - a*d)/d))*((d*x + c)*b/d)^(5/2) - 2)/((d*x + c)^(5/2)*d)
\[ \int \frac {\cos ^2(a+b x)}{(c+d x)^{7/2}} \, dx=\int { \frac {\cos \left (b x + a\right )^{2}}{{\left (d x + c\right )}^{\frac {7}{2}}} \,d x } \] Input:
integrate(cos(b*x+a)^2/(d*x+c)^(7/2),x, algorithm="giac")
Output:
integrate(cos(b*x + a)^2/(d*x + c)^(7/2), x)
Timed out. \[ \int \frac {\cos ^2(a+b x)}{(c+d x)^{7/2}} \, dx=\int \frac {{\cos \left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^{7/2}} \,d x \] Input:
int(cos(a + b*x)^2/(c + d*x)^(7/2),x)
Output:
int(cos(a + b*x)^2/(c + d*x)^(7/2), x)
\[ \int \frac {\cos ^2(a+b x)}{(c+d x)^{7/2}} \, dx=\int \frac {\cos \left (b x +a \right )^{2}}{\sqrt {d x +c}\, c^{3}+3 \sqrt {d x +c}\, c^{2} d x +3 \sqrt {d x +c}\, c \,d^{2} x^{2}+\sqrt {d x +c}\, d^{3} x^{3}}d x \] Input:
int(cos(b*x+a)^2/(d*x+c)^(7/2),x)
Output:
int(cos(a + b*x)**2/(sqrt(c + d*x)*c**3 + 3*sqrt(c + d*x)*c**2*d*x + 3*sqr t(c + d*x)*c*d**2*x**2 + sqrt(c + d*x)*d**3*x**3),x)