Integrand size = 33, antiderivative size = 211 \[ \int \frac {(a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=-\frac {4 a^3 (5 A+9 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^3 (5 A+3 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {4 a^3 (20 A+21 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a B \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {2 (5 A+9 B) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d} \] Output:
-4/5*a^3*(5*A+9*B)*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+4/3*a^3*(5*A+3*B)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2* d*x+1/2*c,2^(1/2))*sec(d*x+c)^(1/2)/d+4/15*a^3*(20*A+21*B)*sec(d*x+c)^(1/2 )*sin(d*x+c)/d+2/5*a*B*sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^2*sin(d*x+c)/d+2/ 15*(5*A+9*B)*sec(d*x+c)^(1/2)*(a^3+a^3*sec(d*x+c))*sin(d*x+c)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 3.13 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.16 \[ \int \frac {(a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\frac {a^3 e^{-i d x} \sec ^{\frac {5}{2}}(c+d x) (\cos (d x)+i \sin (d x)) \left (-90 i A \cos (c+d x)-162 i B \cos (c+d x)-30 i A \cos (3 (c+d x))-54 i B \cos (3 (c+d x))+40 (5 A+3 B) \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+2 i (5 A+9 B) e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+45 A \sin (c+d x)+66 B \sin (c+d x)+10 A \sin (2 (c+d x))+30 B \sin (2 (c+d x))+45 A \sin (3 (c+d x))+54 B \sin (3 (c+d x))\right )}{30 d} \] Input:
Integrate[((a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x]))/Sqrt[Sec[c + d*x]] ,x]
Output:
(a^3*Sec[c + d*x]^(5/2)*(Cos[d*x] + I*Sin[d*x])*((-90*I)*A*Cos[c + d*x] - (162*I)*B*Cos[c + d*x] - (30*I)*A*Cos[3*(c + d*x)] - (54*I)*B*Cos[3*(c + d *x)] + 40*(5*A + 3*B)*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + ((2*I )*(5*A + 9*B)*(1 + E^((2*I)*(c + d*x)))^(5/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^(I*(c + d*x)) + 45*A*Sin[c + d*x] + 66*B*Sin [c + d*x] + 10*A*Sin[2*(c + d*x)] + 30*B*Sin[2*(c + d*x)] + 45*A*Sin[3*(c + d*x)] + 54*B*Sin[3*(c + d*x)]))/(30*d*E^(I*d*x))
Time = 1.25 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.02, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 4506, 27, 3042, 4506, 3042, 4485, 27, 3042, 4274, 3042, 4258, 3042, 3119, 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.
\(\displaystyle \int \frac {(a \sec (c+d x)+a)^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 4506 |
\(\displaystyle \frac {2}{5} \int \frac {(\sec (c+d x) a+a)^2 (a (5 A-B)+a (5 A+9 B) \sec (c+d x))}{2 \sqrt {\sec (c+d x)}}dx+\frac {2 a B \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {(\sec (c+d x) a+a)^2 (a (5 A-B)+a (5 A+9 B) \sec (c+d x))}{\sqrt {\sec (c+d x)}}dx+\frac {2 a B \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (a (5 A-B)+a (5 A+9 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a B \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
\(\Big \downarrow \) 4506 |
\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {(\sec (c+d x) a+a) \left ((5 A-6 B) a^2+(20 A+21 B) \sec (c+d x) a^2\right )}{\sqrt {\sec (c+d x)}}dx+\frac {2 (5 A+9 B) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}\right )+\frac {2 a B \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left ((5 A-6 B) a^2+(20 A+21 B) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (5 A+9 B) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}\right )+\frac {2 a B \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
\(\Big \downarrow \) 4485 |
\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (2 \int -\frac {3 a^3 (5 A+9 B)-5 a^3 (5 A+3 B) \sec (c+d x)}{2 \sqrt {\sec (c+d x)}}dx+\frac {2 a^3 (20 A+21 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {2 (5 A+9 B) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}\right )+\frac {2 a B \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (\frac {2 a^3 (20 A+21 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {3 a^3 (5 A+9 B)-5 a^3 (5 A+3 B) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx\right )+\frac {2 (5 A+9 B) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}\right )+\frac {2 a B \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (\frac {2 a^3 (20 A+21 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {3 a^3 (5 A+9 B)-5 a^3 (5 A+3 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 (5 A+9 B) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}\right )+\frac {2 a B \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
\(\Big \downarrow \) 4274 |
\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (-3 a^3 (5 A+9 B) \int \frac {1}{\sqrt {\sec (c+d x)}}dx+5 a^3 (5 A+3 B) \int \sqrt {\sec (c+d x)}dx+\frac {2 a^3 (20 A+21 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {2 (5 A+9 B) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}\right )+\frac {2 a B \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (-3 a^3 (5 A+9 B) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+5 a^3 (5 A+3 B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a^3 (20 A+21 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {2 (5 A+9 B) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}\right )+\frac {2 a B \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (5 a^3 (5 A+3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx-3 a^3 (5 A+9 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx+\frac {2 a^3 (20 A+21 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {2 (5 A+9 B) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}\right )+\frac {2 a B \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (5 a^3 (5 A+3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 a^3 (5 A+9 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a^3 (20 A+21 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {2 (5 A+9 B) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}\right )+\frac {2 a B \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (5 a^3 (5 A+3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^3 (20 A+21 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {6 a^3 (5 A+9 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 (5 A+9 B) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}\right )+\frac {2 a B \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {1}{5} \left (\frac {2 (5 A+9 B) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}+\frac {2}{3} \left (\frac {2 a^3 (20 A+21 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {10 a^3 (5 A+3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {6 a^3 (5 A+9 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {2 a B \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
Input:
Int[((a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x]))/Sqrt[Sec[c + d*x]],x]
Output:
(2*a*B*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(5*d) + ((2 *(5*A + 9*B)*Sqrt[Sec[c + d*x]]*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(3* d) + (2*((-6*a^3*(5*A + 9*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]* Sqrt[Sec[c + d*x]])/d + (10*a^3*(5*A + 3*B)*Sqrt[Cos[c + d*x]]*EllipticF[( c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*a^3*(20*A + 21*B)*Sqrt[Sec[c + d *x]]*Sin[c + d*x])/d))/3)/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[(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_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[ e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Simp[1/(n + 1) Int[(d*Csc [e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x ], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && !LeQ[ n, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B* Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x] )^n*Simp[a*A*d*(m + n) + B*(b*d*n) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))* Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(915\) vs. \(2(190)=380\).
Time = 5.79 (sec) , antiderivative size = 916, normalized size of antiderivative = 4.34
method | result | size |
default | \(\text {Expression too large to display}\) | \(916\) |
parts | \(\text {Expression too large to display}\) | \(1061\) |
Input:
int((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x,method=_RETURNV ERBOSE)
Output:
-4/15*a^3*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(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)^3*(180*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-100*A*Ellipt icF(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)^4-60*A*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)*si n(1/2*d*x+1/2*c)^4+216*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-60*B*Elli pticF(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)^4-108*B*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)^4-190*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+100*A* EllipticF(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+60*A*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)^2-246*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+60* B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si n(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2+108*B*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+50*A*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-2 5*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*...
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.15 \[ \int \frac {(a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (5 \, A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (5 \, A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} {\left (5 \, A + 9 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} {\left (5 \, A + 9 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (9 \, {\left (5 \, A + 6 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 5 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right ) + 3 \, B a^{3}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{15 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x, algorith m="fricas")
Output:
-2/15*(5*I*sqrt(2)*(5*A + 3*B)*a^3*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(2)*(5*A + 3*B)*a^3*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*I*sqrt( 2)*(5*A + 9*B)*a^3*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInver se(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*I*sqrt(2)*(5*A + 9*B)*a^3*co s(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c ) - I*sin(d*x + c))) - (9*(5*A + 6*B)*a^3*cos(d*x + c)^2 + 5*(A + 3*B)*a^3 *cos(d*x + c) + 3*B*a^3)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^ 2)
Timed out. \[ \int \frac {(a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\text {Timed out} \] Input:
integrate((a+a*sec(d*x+c))**3*(A+B*sec(d*x+c))/sec(d*x+c)**(1/2),x)
Output:
Timed out
\[ \int \frac {(a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{3}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x, algorith m="maxima")
Output:
integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^3/sqrt(sec(d*x + c)), x)
\[ \int \frac {(a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{3}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x, algorith m="giac")
Output:
integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^3/sqrt(sec(d*x + c)), x)
Timed out. \[ \int \frac {(a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \] Input:
int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^3)/(1/cos(c + d*x))^(1/2),x )
Output:
int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^3)/(1/cos(c + d*x))^(1/2), x)
\[ \int \frac {(a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=a^{3} \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )}d x \right ) a +3 \left (\int \sqrt {\sec \left (d x +c \right )}d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right ) a +3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right ) b +3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) a +3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) b \right ) \] Input:
int((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x)
Output:
a**3*(int(sqrt(sec(c + d*x))/sec(c + d*x),x)*a + 3*int(sqrt(sec(c + d*x)), x)*a + int(sqrt(sec(c + d*x)),x)*b + int(sqrt(sec(c + d*x))*sec(c + d*x)** 3,x)*b + int(sqrt(sec(c + d*x))*sec(c + d*x)**2,x)*a + 3*int(sqrt(sec(c + d*x))*sec(c + d*x)**2,x)*b + 3*int(sqrt(sec(c + d*x))*sec(c + d*x),x)*a + 3*int(sqrt(sec(c + d*x))*sec(c + d*x),x)*b)