Integrand size = 35, antiderivative size = 79 \[ \int \frac {\sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {A \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {(2 A+3 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)} \] Output:
1/3*A*(b*cos(d*x+c))^(1/2)*sin(d*x+c)/d/cos(d*x+c)^(7/2)+1/3*(2*A+3*C)*(b* cos(d*x+c))^(1/2)*sin(d*x+c)/d/cos(d*x+c)^(3/2)
Time = 0.23 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.65 \[ \int \frac {\sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {\sqrt {b \cos (c+d x)} \sin (c+d x) \left (3 (A+C)+A \tan ^2(c+d x)\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)} \] Input:
Integrate[(Sqrt[b*Cos[c + d*x]]*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(9/2) ,x]
Output:
(Sqrt[b*Cos[c + d*x]]*Sin[c + d*x]*(3*(A + C) + A*Tan[c + d*x]^2))/(3*d*Co s[c + d*x]^(3/2))
Time = 0.34 (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 = {2031, 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.
\(\displaystyle \int \frac {\sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 2031 |
\(\displaystyle \frac {\sqrt {b \cos (c+d x)} \int \left (C \cos ^2(c+d x)+A\right ) \sec ^4(c+d x)dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {b \cos (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3491 |
\(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (\frac {1}{3} (2 A+3 C) \int \sec ^2(c+d x)dx+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (\frac {1}{3} (2 A+3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}-\frac {(2 A+3 C) \int 1d(-\tan (c+d x))}{3 d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (\frac {(2 A+3 C) \tan (c+d x)}{3 d}+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{\sqrt {\cos (c+d x)}}\) |
Input:
Int[(Sqrt[b*Cos[c + d*x]]*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(9/2),x]
Output:
(Sqrt[b*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[Cos[c + d*x]]
Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[a^(m + 1/ 2)*b^(n - 1/2)*(Sqrt[b*v]/Sqrt[a*v]) Int[v^(m + n)*Fx, x], x] /; FreeQ[{a , b, m}, x] && !IntegerQ[m] && IGtQ[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 = 0.38 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.68
method | result | size |
default | \(\frac {\sin \left (d x +c \right ) \left (2 A \cos \left (d x +c \right )^{2}+3 C \cos \left (d x +c \right )^{2}+A \right ) \sqrt {b \cos \left (d x +c \right )}}{3 d \cos \left (d x +c \right )^{\frac {7}{2}}}\) | \(54\) |
parts | \(\frac {C \sin \left (d x +c \right ) \sqrt {b \cos \left (d x +c \right )}}{d \cos \left (d x +c \right )^{\frac {3}{2}}}+\frac {A \sin \left (d x +c \right ) \left (2 \cos \left (d x +c \right )^{2}+1\right ) \sqrt {b \cos \left (d x +c \right )}}{3 d \cos \left (d x +c \right )^{\frac {7}{2}}}\) | \(73\) |
risch | \(\frac {2 i \sqrt {b \cos \left (d x +c \right )}\, \left (3 C \,{\mathrm e}^{4 i \left (d x +c \right )}+6 A \,{\mathrm e}^{2 i \left (d x +c \right )}+6 C \,{\mathrm e}^{2 i \left (d x +c \right )}+2 A +3 C \right )}{3 \sqrt {\cos \left (d x +c \right )}\, d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}\) | \(81\) |
Input:
int((b*cos(d*x+c))^(1/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x,method=_RET URNVERBOSE)
Output:
1/3/d*sin(d*x+c)*(2*A*cos(d*x+c)^2+3*C*cos(d*x+c)^2+A)*(b*cos(d*x+c))^(1/2 )/cos(d*x+c)^(7/2)
Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.59 \[ \int \frac {\sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {{\left ({\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + A\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{\frac {7}{2}}} \] Input:
integrate((b*cos(d*x+c))^(1/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x, algo rithm="fricas")
Output:
1/3*((2*A + 3*C)*cos(d*x + c)^2 + A)*sqrt(b*cos(d*x + c))*sin(d*x + c)/(d* cos(d*x + c)^(7/2))
Timed out. \[ \int \frac {\sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((b*cos(d*x+c))**(1/2)*(A+C*cos(d*x+c)**2)/cos(d*x+c)**(9/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 350 vs. \(2 (67) = 134\).
Time = 0.27 (sec) , antiderivative size = 350, normalized size of antiderivative = 4.43 \[ \int \frac {\sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 \, {\left (\frac {2 \, {\left ({\left (3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (6 \, d x + 6 \, c\right ) + 3 \, {\left (3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (4 \, d x + 4 \, c\right ) - 3 \, \cos \left (6 \, d x + 6 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) - 9 \, \cos \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right )\right )} A \sqrt {b}}{2 \, {\left (3 \, \cos \left (4 \, d x + 4 \, c\right ) + 3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (6 \, d x + 6 \, c\right ) + \cos \left (6 \, d x + 6 \, c\right )^{2} + 6 \, {\left (3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (4 \, d x + 4 \, c\right ) + 9 \, \cos \left (4 \, d x + 4 \, c\right )^{2} + 9 \, \cos \left (2 \, d x + 2 \, c\right )^{2} + 6 \, {\left (\sin \left (4 \, d x + 4 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right )\right )} \sin \left (6 \, d x + 6 \, c\right ) + \sin \left (6 \, d x + 6 \, c\right )^{2} + 9 \, \sin \left (4 \, d x + 4 \, c\right )^{2} + 18 \, \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 9 \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 6 \, \cos \left (2 \, d x + 2 \, c\right ) + 1} + \frac {3 \, C \sqrt {b} \sin \left (2 \, d x + 2 \, c\right )}{\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1}\right )}}{3 \, d} \] Input:
integrate((b*cos(d*x+c))^(1/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x, algo rithm="maxima")
Output:
2/3*(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)*sin(2*d*x + 2*c) - 9*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*A*sqrt(b)/(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) + 3*C*sqrt(b)*sin(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))/d
Time = 0.20 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.65 \[ \int \frac {\sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {b}}{3 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} d} \] Input:
integrate((b*cos(d*x+c))^(1/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x, algo rithm="giac")
Output:
-2/3*(3*A*tan(1/2*d*x + 1/2*c)^5 + 3*C*tan(1/2*d*x + 1/2*c)^5 - 2*A*tan(1/ 2*d*x + 1/2*c)^3 - 6*C*tan(1/2*d*x + 1/2*c)^3 + 3*A*tan(1/2*d*x + 1/2*c) + 3*C*tan(1/2*d*x + 1/2*c))*sqrt(b)/((tan(1/2*d*x + 1/2*c)^6 - 3*tan(1/2*d* x + 1/2*c)^4 + 3*tan(1/2*d*x + 1/2*c)^2 - 1)*d)
Time = 45.41 (sec) , antiderivative size = 217, normalized size of antiderivative = 2.75 \[ \int \frac {\sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (18\,A\,\sin \left (2\,c+2\,d\,x\right )+12\,A\,\sin \left (4\,c+4\,d\,x\right )+2\,A\,\sin \left (6\,c+6\,d\,x\right )+15\,C\,\sin \left (2\,c+2\,d\,x\right )+12\,C\,\sin \left (4\,c+4\,d\,x\right )+3\,C\,\sin \left (6\,c+6\,d\,x\right )+A\,20{}\mathrm {i}+C\,30{}\mathrm {i}+A\,\cos \left (2\,c+2\,d\,x\right )\,30{}\mathrm {i}+A\,\cos \left (4\,c+4\,d\,x\right )\,12{}\mathrm {i}+A\,\cos \left (6\,c+6\,d\,x\right )\,2{}\mathrm {i}+C\,\cos \left (2\,c+2\,d\,x\right )\,45{}\mathrm {i}+C\,\cos \left (4\,c+4\,d\,x\right )\,18{}\mathrm {i}+C\,\cos \left (6\,c+6\,d\,x\right )\,3{}\mathrm {i}\right )}{3\,d\,\sqrt {\cos \left (c+d\,x\right )}\,\left (15\,\cos \left (2\,c+2\,d\,x\right )+6\,\cos \left (4\,c+4\,d\,x\right )+\cos \left (6\,c+6\,d\,x\right )+10\right )} \] Input:
int(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(1/2))/cos(c + d*x)^(9/2),x)
Output:
((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*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))
Time = 0.16 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {\sqrt {b}\, \sin \left (d x +c \right ) \left (2 \sin \left (d x +c \right )^{2} a +3 \sin \left (d x +c \right )^{2} c -3 a -3 c \right )}{3 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int((b*cos(d*x+c))^(1/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x)
Output:
(sqrt(b)*sin(c + d*x)*(2*sin(c + d*x)**2*a + 3*sin(c + d*x)**2*c - 3*a - 3 *c))/(3*cos(c + d*x)*d*(sin(c + d*x)**2 - 1))