Integrand size = 35, antiderivative size = 79 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\frac {A \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {(2 A+3 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}} \]
1/3*A*sin(d*x+c)/d/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(1/2)+1/3*(2*A+3*C)*sin (d*x+c)/d/cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(1/2)
Time = 0.08 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.65 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\frac {\sin (c+d x) \left (3 (A+C)+A \tan ^2(c+d x)\right )}{3 d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}} \]
Time = 0.33 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.84, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {2032, 3042, 3491, 3042, 4254, 24}
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.
| \(\displaystyle \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 2032 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \left (C \cos ^2(c+d x)+A\right ) \sec ^4(c+d x)dx}{\sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx}{\sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3491 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{3} (2 A+3 C) \int \sec ^2(c+d x)dx+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{\sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{3} (2 A+3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{\sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}-\frac {(2 A+3 C) \int 1d(-\tan (c+d x))}{3 d}\right )}{\sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {(2 A+3 C) \tan (c+d x)}{3 d}+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{\sqrt {b \cos (c+d x)}}\) |
(Sqrt[Cos[c + d*x]]*(((2*A + 3*C)*Tan[c + d*x])/(3*d) + (A*Sec[c + d*x]^2* Tan[c + d*x])/(3*d)))/Sqrt[b*Cos[c + d*x]]
3.2.22.3.1 Defintions of rubi rules used
Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[a^(m - 1/ 2)*b^(n + 1/2)*(Sqrt[a*v]/Sqrt[b*v]) Int[v^(m + n)*Fx, x], x] /; FreeQ[{a , b, m}, x] && !IntegerQ[m] && ILtQ[n - 1/2, 0] && IntegerQ[m + n]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x _)]^2), x_Symbol] :> Simp[A*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Simp[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1)) Int[(b*Sin[e + f* x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Time = 7.68 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.68
| method | result | size |
| default | \(\frac {\left (2 A \left (\cos ^{2}\left (d x +c \right )\right )+3 C \left (\cos ^{2}\left (d x +c \right )\right )+A \right ) \sin \left (d x +c \right )}{3 d \sqrt {\cos \left (d x +c \right ) b}\, \cos \left (d x +c \right )^{\frac {5}{2}}}\) | \(54\) |
| parts | \(\frac {A \left (2 \left (\cos ^{2}\left (d x +c \right )\right )+1\right ) \sin \left (d x +c \right )}{3 d \sqrt {\cos \left (d x +c \right ) b}\, \cos \left (d x +c \right )^{\frac {5}{2}}}+\frac {C \sin \left (d x +c \right )}{d \sqrt {\cos \left (d x +c \right ) b}\, \sqrt {\cos \left (d x +c \right )}}\) | \(73\) |
| risch | \(\frac {i \left (3 C \,{\mathrm e}^{3 i \left (d x +c \right )}+\left (8 A +9 C \right ) \cos \left (d x +c \right )+i \left (4 A +3 C \right ) \sin \left (d x +c \right )\right )}{3 \sqrt {\cos \left (d x +c \right ) b}\, \sqrt {\cos \left (d x +c \right )}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} d}\) | \(81\) |
1/3/d*(2*A*cos(d*x+c)^2+3*C*cos(d*x+c)^2+A)*sin(d*x+c)/(cos(d*x+c)*b)^(1/2 )/cos(d*x+c)^(5/2)
Time = 0.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.63 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\frac {{\left ({\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + A\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, b d \cos \left (d x + c\right )^{\frac {7}{2}}} \]
1/3*((2*A + 3*C)*cos(d*x + c)^2 + A)*sqrt(b*cos(d*x + c))*sin(d*x + c)/(b* d*cos(d*x + c)^(7/2))
Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 355 vs. \(2 (67) = 134\).
Time = 0.60 (sec) , antiderivative size = 355, normalized size of antiderivative = 4.49 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\frac {2 \, {\left (\frac {3 \, C \sqrt {b} \sin \left (2 \, d x + 2 \, c\right )}{b \cos \left (2 \, d x + 2 \, c\right )^{2} + b \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, b \cos \left (2 \, d x + 2 \, c\right ) + b} + \frac {2 \, {\left ({\left (3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (6 \, d x + 6 \, c\right ) + 3 \, {\left (3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (4 \, d x + 4 \, c\right ) - 3 \, \cos \left (6 \, d x + 6 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) - 9 \, \cos \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right )\right )} A}{{\left (2 \, {\left (3 \, \cos \left (4 \, d x + 4 \, c\right ) + 3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (6 \, d x + 6 \, c\right ) + \cos \left (6 \, d x + 6 \, c\right )^{2} + 6 \, {\left (3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (4 \, d x + 4 \, c\right ) + 9 \, \cos \left (4 \, d x + 4 \, c\right )^{2} + 9 \, \cos \left (2 \, d x + 2 \, c\right )^{2} + 6 \, {\left (\sin \left (4 \, d x + 4 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right )\right )} \sin \left (6 \, d x + 6 \, c\right ) + \sin \left (6 \, d x + 6 \, c\right )^{2} + 9 \, \sin \left (4 \, d x + 4 \, c\right )^{2} + 18 \, \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 9 \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 6 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sqrt {b}}\right )}}{3 \, d} \]
2/3*(3*C*sqrt(b)*sin(2*d*x + 2*c)/(b*cos(2*d*x + 2*c)^2 + b*sin(2*d*x + 2* c)^2 + 2*b*cos(2*d*x + 2*c) + b) + 2*((3*cos(2*d*x + 2*c) + 1)*sin(6*d*x + 6*c) + 3*(3*cos(2*d*x + 2*c) + 1)*sin(4*d*x + 4*c) - 3*cos(6*d*x + 6*c)*s in(2*d*x + 2*c) - 9*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*A/((2*(3*cos(4*d*x + 4*c) + 3*cos(2*d*x + 2*c) + 1)*cos(6*d*x + 6*c) + cos(6*d*x + 6*c)^2 + 6 *(3*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 9*cos(4*d*x + 4*c)^2 + 9*cos( 2*d*x + 2*c)^2 + 6*(sin(4*d*x + 4*c) + sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + sin(6*d*x + 6*c)^2 + 9*sin(4*d*x + 4*c)^2 + 18*sin(4*d*x + 4*c)*sin(2*d* x + 2*c) + 9*sin(2*d*x + 2*c)^2 + 6*cos(2*d*x + 2*c) + 1)*sqrt(b)))/d
\[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{\sqrt {b \cos \left (d x + c\right )} \cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]
Time = 2.90 (sec) , antiderivative size = 220, normalized size of antiderivative = 2.78 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\frac {\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (18\,A\,\sin \left (2\,c+2\,d\,x\right )+12\,A\,\sin \left (4\,c+4\,d\,x\right )+2\,A\,\sin \left (6\,c+6\,d\,x\right )+15\,C\,\sin \left (2\,c+2\,d\,x\right )+12\,C\,\sin \left (4\,c+4\,d\,x\right )+3\,C\,\sin \left (6\,c+6\,d\,x\right )+A\,20{}\mathrm {i}+C\,30{}\mathrm {i}+A\,\cos \left (2\,c+2\,d\,x\right )\,30{}\mathrm {i}+A\,\cos \left (4\,c+4\,d\,x\right )\,12{}\mathrm {i}+A\,\cos \left (6\,c+6\,d\,x\right )\,2{}\mathrm {i}+C\,\cos \left (2\,c+2\,d\,x\right )\,45{}\mathrm {i}+C\,\cos \left (4\,c+4\,d\,x\right )\,18{}\mathrm {i}+C\,\cos \left (6\,c+6\,d\,x\right )\,3{}\mathrm {i}\right )}{3\,b\,d\,\sqrt {\cos \left (c+d\,x\right )}\,\left (15\,\cos \left (2\,c+2\,d\,x\right )+6\,\cos \left (4\,c+4\,d\,x\right )+\cos \left (6\,c+6\,d\,x\right )+10\right )} \]
((b*cos(c + d*x))^(1/2)*(A*20i + C*30i + A*cos(2*c + 2*d*x)*30i + A*cos(4* c + 4*d*x)*12i + A*cos(6*c + 6*d*x)*2i + C*cos(2*c + 2*d*x)*45i + C*cos(4* c + 4*d*x)*18i + C*cos(6*c + 6*d*x)*3i + 18*A*sin(2*c + 2*d*x) + 12*A*sin( 4*c + 4*d*x) + 2*A*sin(6*c + 6*d*x) + 15*C*sin(2*c + 2*d*x) + 12*C*sin(4*c + 4*d*x) + 3*C*sin(6*c + 6*d*x)))/(3*b*d*cos(c + d*x)^(1/2)*(15*cos(2*c + 2*d*x) + 6*cos(4*c + 4*d*x) + cos(6*c + 6*d*x) + 10))