Integrand size = 23, antiderivative size = 189 \[ \int (a+b \cos (c+d x))^3 \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {6 a \left (a^2+5 b^2\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 b \left (a^2+b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {6 a \left (a^2+5 b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {8 a^2 b \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a^2 \sec ^{\frac {3}{2}}(c+d x) (b+a \sec (c+d x)) \sin (c+d x)}{5 d} \] Output:
-6/5*a*(a^2+5*b^2)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))* sec(d*x+c)^(1/2)/d+2*b*(a^2+b^2)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+ 1/2*c,2^(1/2))*sec(d*x+c)^(1/2)/d+6/5*a*(a^2+5*b^2)*sec(d*x+c)^(1/2)*sin(d *x+c)/d+8/5*a^2*b*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*a^2*sec(d*x+c)^(3/2)*( b+a*sec(d*x+c))*sin(d*x+c)/d
Time = 1.48 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.71 \[ \int (a+b \cos (c+d x))^3 \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-3 a \left (a^2+5 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 b \left (a^2+b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\frac {a \left (5 \left (a^2+3 b^2\right )+10 a b \cos (c+d x)+3 \left (a^2+5 b^2\right ) \cos (2 (c+d x))\right ) \sin (c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x)}\right )}{5 d} \] Input:
Integrate[(a + b*Cos[c + d*x])^3*Sec[c + d*x]^(7/2),x]
Output:
(2*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-3*a*(a^2 + 5*b^2)*EllipticE[(c + d*x)/2, 2] + 5*b*(a^2 + b^2)*EllipticF[(c + d*x)/2, 2] + (a*(5*(a^2 + 3* b^2) + 10*a*b*Cos[c + d*x] + 3*(a^2 + 5*b^2)*Cos[2*(c + d*x)])*Sin[c + d*x ])/(2*Cos[c + d*x]^(5/2))))/(5*d)
Time = 1.21 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.96, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.783, Rules used = {3042, 3717, 3042, 4329, 27, 3042, 4535, 3042, 4255, 3042, 4258, 3042, 3119, 4534, 3042, 4258, 3042, 3120}
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.
| \(\displaystyle \int \sec ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3dx\) |
| \(\Big \downarrow \) 3717 |
| \(\displaystyle \int \sqrt {\sec (c+d x)} (a \sec (c+d x)+b)^3dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+b\right )^3dx\) |
| \(\Big \downarrow \) 4329 |
| \(\displaystyle \frac {2}{5} \int \frac {1}{2} \sqrt {\sec (c+d x)} \left (12 a^2 b \sec ^2(c+d x)+3 a \left (a^2+5 b^2\right ) \sec (c+d x)+b \left (a^2+5 b^2\right )\right )dx+\frac {2 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \sqrt {\sec (c+d x)} \left (12 a^2 b \sec ^2(c+d x)+3 a \left (a^2+5 b^2\right ) \sec (c+d x)+b \left (a^2+5 b^2\right )\right )dx+\frac {2 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (12 a^2 b \csc \left (c+d x+\frac {\pi }{2}\right )^2+3 a \left (a^2+5 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+b \left (a^2+5 b^2\right )\right )dx+\frac {2 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {1}{5} \left (3 a \left (a^2+5 b^2\right ) \int \sec ^{\frac {3}{2}}(c+d x)dx+\int \sqrt {\sec (c+d x)} \left (12 a^2 b \sec ^2(c+d x)+b \left (a^2+5 b^2\right )\right )dx\right )+\frac {2 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (3 a \left (a^2+5 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx+\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (12 a^2 b \csc \left (c+d x+\frac {\pi }{2}\right )^2+b \left (a^2+5 b^2\right )\right )dx\right )+\frac {2 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{5} \left (\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (12 a^2 b \csc \left (c+d x+\frac {\pi }{2}\right )^2+b \left (a^2+5 b^2\right )\right )dx+3 a \left (a^2+5 b^2\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )\right )+\frac {2 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (12 a^2 b \csc \left (c+d x+\frac {\pi }{2}\right )^2+b \left (a^2+5 b^2\right )\right )dx+3 a \left (a^2+5 b^2\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )\right )+\frac {2 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {1}{5} \left (\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (12 a^2 b \csc \left (c+d x+\frac {\pi }{2}\right )^2+b \left (a^2+5 b^2\right )\right )dx+3 a \left (a^2+5 b^2\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )\right )+\frac {2 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (12 a^2 b \csc \left (c+d x+\frac {\pi }{2}\right )^2+b \left (a^2+5 b^2\right )\right )dx+3 a \left (a^2+5 b^2\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )\right )+\frac {2 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{5} \left (\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (12 a^2 b \csc \left (c+d x+\frac {\pi }{2}\right )^2+b \left (a^2+5 b^2\right )\right )dx+3 a \left (a^2+5 b^2\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {2 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \frac {1}{5} \left (5 b \left (a^2+b^2\right ) \int \sqrt {\sec (c+d x)}dx+3 a \left (a^2+5 b^2\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {8 a^2 b \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (5 b \left (a^2+b^2\right ) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+3 a \left (a^2+5 b^2\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {8 a^2 b \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {1}{5} \left (5 b \left (a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 a \left (a^2+5 b^2\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {8 a^2 b \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (5 b \left (a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a \left (a^2+5 b^2\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {8 a^2 b \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{5} \left (\frac {10 b \left (a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+3 a \left (a^2+5 b^2\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {8 a^2 b \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
Input:
Int[(a + b*Cos[c + d*x])^3*Sec[c + d*x]^(7/2),x]
Output:
(2*a^2*Sec[c + d*x]^(3/2)*(b + a*Sec[c + d*x])*Sin[c + d*x])/(5*d) + ((10* b*(a^2 + b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d* x]])/d + (8*a^2*b*Sec[c + d*x]^(3/2)*Sin[c + d*x])/d + 3*a*(a^2 + 5*b^2)*( (-2*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + ( 2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d))/5
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]^(n_.))^(p_.), x_Symbol] :> Simp[d^(n*p) Int[(d*Csc[e + f*x])^(m - n*p )*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x] && !IntegerQ[m] && IntegersQ[n, p]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[(-b^2)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*((d*Csc[e + f*x])^n/(f*(m + n - 1))), x] + Simp[1/(d*(m + n - 1)) Int[ (a + b*Csc[e + f*x])^(m - 3)*(d*Csc[e + f*x])^n*Simp[a^3*d*(m + n - 1) + a* b^2*d*n + b*(b^2*d*(m + n - 2) + 3*a^2*d*(m + n - 1))*Csc[e + f*x] + a*b^2* d*(3*m + 2*n - 4)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, n}, x ] && NeQ[a^2 - b^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) && !(IGtQ[n, 2] && !IntegerQ[m])
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) )), x] + Simp[(C*m + A*(m + 1))/(m + 1) Int[(b*Csc[e + f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] && !LeQ[m, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* (B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b Int[(b*Cs c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) , x] /; FreeQ[{b, e, f, A, B, C, m}, x]
Leaf count of result is larger than twice the leaf count of optimal. \(710\) vs. \(2(170)=340\).
Time = 983.10 (sec) , antiderivative size = 711, normalized size of antiderivative = 3.76
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(711\) |
| parts | \(\text {Expression too large to display}\) | \(898\) |
Input:
int((a+cos(d*x+c)*b)^3*sec(d*x+c)^(7/2),x,method=_RETURNVERBOSE)
Output:
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*b^3*(sin(1/2 *d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2 *c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+2/ 5*a^3/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c) ^2-1)/sin(1/2*d*x+1/2*c)^2*(24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12* EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin( 1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^4*cos (1/2*d*x+1/2*c)+12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2* c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2+8*sin(1/ 2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/ 2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*(-2*sin(1/2 *d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+6*a^2*b*(-1/6*cos(1/2*d*x+1/2*c) *(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^ 2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2 )/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d *x+1/2*c),2^(1/2)))+6*b^2*a/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1 )*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2* c)^2*cos(1/2*d*x+1/2*c)-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c) ^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))))/sin(1/2*d*x+1/2*c)/(2* cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.29 \[ \int (a+b \cos (c+d x))^3 \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {5 \, \sqrt {2} {\left (i \, a^{2} b + i \, b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, a^{2} b - i \, b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (i \, a^{3} + 5 i \, a b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (-i \, a^{3} - 5 i \, a b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (5 \, a^{2} b \cos \left (d x + c\right ) + a^{3} + 3 \, {\left (a^{3} + 5 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{5 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate((a+b*cos(d*x+c))^3*sec(d*x+c)^(7/2),x, algorithm="fricas")
Output:
-1/5*(5*sqrt(2)*(I*a^2*b + I*b^3)*cos(d*x + c)^2*weierstrassPInverse(-4, 0 , cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-I*a^2*b - I*b^3)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*sqrt(2) *(I*a^3 + 5*I*a*b^2)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInv erse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(2)*(-I*a^3 - 5*I*a*b^ 2)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d* x + c) - I*sin(d*x + c))) - 2*(5*a^2*b*cos(d*x + c) + a^3 + 3*(a^3 + 5*a*b ^2)*cos(d*x + c)^2)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2)
Timed out. \[ \int (a+b \cos (c+d x))^3 \sec ^{\frac {7}{2}}(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))**3*sec(d*x+c)**(7/2),x)
Output:
Timed out
\[ \int (a+b \cos (c+d x))^3 \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^3*sec(d*x+c)^(7/2),x, algorithm="maxima")
Output:
integrate((b*cos(d*x + c) + a)^3*sec(d*x + c)^(7/2), x)
\[ \int (a+b \cos (c+d x))^3 \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^3*sec(d*x+c)^(7/2),x, algorithm="giac")
Output:
integrate((b*cos(d*x + c) + a)^3*sec(d*x + c)^(7/2), x)
Timed out. \[ \int (a+b \cos (c+d x))^3 \sec ^{\frac {7}{2}}(c+d x) \, dx=\int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3 \,d x \] Input:
int((1/cos(c + d*x))^(7/2)*(a + b*cos(c + d*x))^3,x)
Output:
int((1/cos(c + d*x))^(7/2)*(a + b*cos(c + d*x))^3, x)
\[ \int (a+b \cos (c+d x))^3 \sec ^{\frac {7}{2}}(c+d x) \, dx=3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{3}d x \right ) a^{2} b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{3}d x \right ) b^{3}+3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3}d x \right ) a \,b^{2}+\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right ) a^{3} \] Input:
int((a+b*cos(d*x+c))^3*sec(d*x+c)^(7/2),x)
Output:
3*int(sqrt(sec(c + d*x))*cos(c + d*x)*sec(c + d*x)**3,x)*a**2*b + int(sqrt (sec(c + d*x))*cos(c + d*x)**3*sec(c + d*x)**3,x)*b**3 + 3*int(sqrt(sec(c + d*x))*cos(c + d*x)**2*sec(c + d*x)**3,x)*a*b**2 + int(sqrt(sec(c + d*x)) *sec(c + d*x)**3,x)*a**3