Integrand size = 45, antiderivative size = 253 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {a^{3/2} (112 A+88 B+75 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{64 d}+\frac {a^2 (48 A+56 B+39 C) \sin (c+d x)}{96 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (112 A+88 B+75 C) \sin (c+d x)}{64 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a (8 B+3 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{24 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {C (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{4 d \cos ^{\frac {5}{2}}(c+d x)} \]
1/4*C*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/cos(d*x+c)^(5/2)+1/64*a^(3/2)*(1 12*A+88*B+75*C)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))*cos(d*x +c)^(1/2)*sec(d*x+c)^(1/2)/d+1/96*a^2*(48*A+56*B+39*C)*sin(d*x+c)/d/cos(d* x+c)^(5/2)/(a+a*sec(d*x+c))^(1/2)+1/64*a^2*(112*A+88*B+75*C)*sin(d*x+c)/d/ cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)+1/24*a*(8*B+3*C)*sin(d*x+c)*(a+a*s ec(d*x+c))^(1/2)/d/cos(d*x+c)^(5/2)
Time = 8.19 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.87 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {a^2 \sqrt {\cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \left (3 (112 A+88 B+75 C) \arcsin \left (\sqrt {1-\sec (c+d x)}\right )+2 (48 A+88 B+75 C) \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x)+8 (8 B+15 C) \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x)+48 C \sqrt {1-\sec (c+d x)} \sec ^{\frac {7}{2}}(c+d x)+3 (112 A+88 B+75 C) \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}\right ) \sin (c+d x)}{192 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]
Integrate[((a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x] ^2))/Cos[c + d*x]^(3/2),x]
(a^2*Sqrt[Cos[c + d*x]]*Sec[c + d*x]^(3/2)*(3*(112*A + 88*B + 75*C)*ArcSin [Sqrt[1 - Sec[c + d*x]]] + 2*(48*A + 88*B + 75*C)*Sqrt[1 - Sec[c + d*x]]*S ec[c + d*x]^(3/2) + 8*(8*B + 15*C)*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(5/ 2) + 48*C*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(7/2) + 3*(112*A + 88*B + 75 *C)*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])])*Sin[c + d*x])/(192*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
Time = 1.52 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.02, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4753, 3042, 4576, 27, 3042, 4506, 27, 3042, 4504, 3042, 4290, 3042, 4288, 222}
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.
\(\displaystyle \int \frac {(a \sec (c+d x)+a)^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sec (c+d x)+a)^{3/2} \left (A+B \sec (c+d x)+C \sec (c+d x)^2\right )}{\cos (c+d x)^{3/2}}dx\) |
\(\Big \downarrow \) 4753 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^{3/2} \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right )dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+B \csc \left (c+d x+\frac {\pi }{2}\right )+A\right )dx\) |
\(\Big \downarrow \) 4576 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {1}{2} \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^{3/2} (a (8 A+3 C)+a (8 B+3 C) \sec (c+d x))dx}{4 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^{3/2} (a (8 A+3 C)+a (8 B+3 C) \sec (c+d x))dx}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (8 A+3 C)+a (8 B+3 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\right )\) |
\(\Big \downarrow \) 4506 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{3} \int \frac {1}{2} \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a} \left (3 (16 A+8 B+9 C) a^2+(48 A+56 B+39 C) \sec (c+d x) a^2\right )dx+\frac {a^2 (8 B+3 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a} \left (3 (16 A+8 B+9 C) a^2+(48 A+56 B+39 C) \sec (c+d x) a^2\right )dx+\frac {a^2 (8 B+3 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 (16 A+8 B+9 C) a^2+(48 A+56 B+39 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {a^2 (8 B+3 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\right )\) |
\(\Big \downarrow \) 4504 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} \left (\frac {3}{4} a^2 (112 A+88 B+75 C) \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a^3 (48 A+56 B+39 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (8 B+3 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} \left (\frac {3}{4} a^2 (112 A+88 B+75 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^3 (48 A+56 B+39 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (8 B+3 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\right )\) |
\(\Big \downarrow \) 4290 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} \left (\frac {3}{4} a^2 (112 A+88 B+75 C) \left (\frac {1}{2} \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (48 A+56 B+39 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (8 B+3 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} \left (\frac {3}{4} a^2 (112 A+88 B+75 C) \left (\frac {1}{2} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (48 A+56 B+39 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (8 B+3 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\right )\) |
\(\Big \downarrow \) 4288 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} \left (\frac {3}{4} a^2 (112 A+88 B+75 C) \left (\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a^3 (48 A+56 B+39 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (8 B+3 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\right )\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {a^2 (8 B+3 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}+\frac {1}{6} \left (\frac {a^3 (48 A+56 B+39 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}+\frac {3}{4} a^2 (112 A+88 B+75 C) \left (\frac {\sqrt {a} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )\right )}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\right )\) |
Int[((a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/C os[c + d*x]^(3/2),x]
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((C*Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(4*d) + ((a^2*(8*B + 3*C)*Sec[c + d*x]^(5/2)*S qrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(3*d) + ((a^3*(48*A + 56*B + 39*C)*S ec[c + d*x]^(5/2)*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]]) + (3*a^2*(1 12*A + 88*B + 75*C)*((Sqrt[a]*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Se c[c + d*x]]])/d + (a*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]])))/4)/6)/(8*a))
3.13.60.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(a/(b*f))*Sqrt[a*(d/b)] Subst[Int[1/Sqrt[1 + x^2/a], x], x, b*(Cot[e + f*x]/Sqrt[a + b*Csc[e + f*x]])], x] /; FreeQ[{a , b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[a*(d/b), 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*b*d*Cot[e + f*x]*((d*Csc[e + f*x])^(n - 1)/( f*(2*n - 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[2*a*d*((n - 1)/(b*(2*n - 1))) Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n - 1), x], x] /; Fre eQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[-2*b*B*C ot[e + f*x]*((d*Csc[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[(A*b*(2*n + 1) + 2*a*B*n)/(b*(2*n + 1)) Int[Sqrt[a + b*Csc[e + f* x]]*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ [A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] && !LtQ[n, 0]
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]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Cs c[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1)) Int[(a + b*Cs c[e + f*x])^m*(d*Csc[e + f*x])^n*Simp[A*b*(m + n + 1) + b*C*n + (a*C*m + b* B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m , n}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && !LtQ[n, -2^(-1)] && NeQ[m + n + 1, 0]
Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m Int[ActivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSecantIntegrandQ[u, x ]
Leaf count of result is larger than twice the leaf count of optimal. \(636\) vs. \(2(217)=434\).
Time = 1.03 (sec) , antiderivative size = 637, normalized size of antiderivative = 2.52
method | result | size |
default | \(\frac {a \left (336 A \cos \left (d x +c \right )^{4} \arctan \left (\frac {\cos \left (d x +c \right )-\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-336 A \cos \left (d x +c \right )^{4} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+672 A \cos \left (d x +c \right )^{3} \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+264 B \arctan \left (\frac {\cos \left (d x +c \right )-\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \cos \left (d x +c \right )^{4}-264 B \cos \left (d x +c \right )^{4} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+528 B \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3} \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+225 C \cos \left (d x +c \right )^{4} \arctan \left (\frac {\cos \left (d x +c \right )-\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-225 C \cos \left (d x +c \right )^{4} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+450 C \cos \left (d x +c \right )^{3} \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+192 A \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2} \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+352 B \cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+300 C \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2} \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+128 B \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+240 C \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+96 C \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{384 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )^{\frac {7}{2}}}\) | \(637\) |
int((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2 ),x,method=_RETURNVERBOSE)
1/384*a/d*(336*A*cos(d*x+c)^4*arctan(1/2*(cos(d*x+c)-sin(d*x+c)+1)/(1+cos( d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))-336*A*cos(d*x+c)^4*arctan(1/2*(cos(d*x+ c)+sin(d*x+c)+1)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))+672*A*cos(d*x+c )^3*sin(d*x+c)*(-1/(1+cos(d*x+c)))^(1/2)+264*B*arctan(1/2*(cos(d*x+c)-sin( d*x+c)+1)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^4-264*B*cos (d*x+c)^4*arctan(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(1+cos(d*x+c))/(-1/(1+cos(d *x+c)))^(1/2))+528*B*sin(d*x+c)*cos(d*x+c)^3*(-1/(1+cos(d*x+c)))^(1/2)+225 *C*cos(d*x+c)^4*arctan(1/2*(cos(d*x+c)-sin(d*x+c)+1)/(1+cos(d*x+c))/(-1/(1 +cos(d*x+c)))^(1/2))-225*C*cos(d*x+c)^4*arctan(1/2*(cos(d*x+c)+sin(d*x+c)+ 1)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))+450*C*cos(d*x+c)^3*sin(d*x+c) *(-1/(1+cos(d*x+c)))^(1/2)+192*A*sin(d*x+c)*cos(d*x+c)^2*(-1/(1+cos(d*x+c) ))^(1/2)+352*B*cos(d*x+c)^2*sin(d*x+c)*(-1/(1+cos(d*x+c)))^(1/2)+300*C*sin (d*x+c)*cos(d*x+c)^2*(-1/(1+cos(d*x+c)))^(1/2)+128*B*cos(d*x+c)*sin(d*x+c) *(-1/(1+cos(d*x+c)))^(1/2)+240*C*cos(d*x+c)*sin(d*x+c)*(-1/(1+cos(d*x+c))) ^(1/2)+96*C*sin(d*x+c)*(-1/(1+cos(d*x+c)))^(1/2))*(a*(1+sec(d*x+c)))^(1/2) /(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2)/cos(d*x+c)^(7/2)
Time = 0.53 (sec) , antiderivative size = 509, normalized size of antiderivative = 2.01 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\left [\frac {4 \, {\left (3 \, {\left (112 \, A + 88 \, B + 75 \, C\right )} a \cos \left (d x + c\right )^{3} + 2 \, {\left (48 \, A + 88 \, B + 75 \, C\right )} a \cos \left (d x + c\right )^{2} + 8 \, {\left (8 \, B + 15 \, C\right )} a \cos \left (d x + c\right ) + 48 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 3 \, {\left ({\left (112 \, A + 88 \, B + 75 \, C\right )} a \cos \left (d x + c\right )^{5} + {\left (112 \, A + 88 \, B + 75 \, C\right )} a \cos \left (d x + c\right )^{4}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 4 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (\cos \left (d x + c\right ) - 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{768 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}}, \frac {2 \, {\left (3 \, {\left (112 \, A + 88 \, B + 75 \, C\right )} a \cos \left (d x + c\right )^{3} + 2 \, {\left (48 \, A + 88 \, B + 75 \, C\right )} a \cos \left (d x + c\right )^{2} + 8 \, {\left (8 \, B + 15 \, C\right )} a \cos \left (d x + c\right ) + 48 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 3 \, {\left ({\left (112 \, A + 88 \, B + 75 \, C\right )} a \cos \left (d x + c\right )^{5} + {\left (112 \, A + 88 \, B + 75 \, C\right )} a \cos \left (d x + c\right )^{4}\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right )}{384 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}}\right ] \]
integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c )^(3/2),x, algorithm="fricas")
[1/768*(4*(3*(112*A + 88*B + 75*C)*a*cos(d*x + c)^3 + 2*(48*A + 88*B + 75* C)*a*cos(d*x + c)^2 + 8*(8*B + 15*C)*a*cos(d*x + c) + 48*C*a)*sqrt((a*cos( d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 3*((112*A + 88*B + 75*C)*a*cos(d*x + c)^5 + (112*A + 88*B + 75*C)*a*cos(d*x + c)^4)*sq rt(a)*log((a*cos(d*x + c)^3 - 4*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(cos(d*x + c) - 2)*sqrt(cos(d*x + c))*sin(d*x + c) - 7*a*cos(d*x + c )^2 + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)))/(d*cos(d*x + c)^5 + d*cos(d *x + c)^4), 1/384*(2*(3*(112*A + 88*B + 75*C)*a*cos(d*x + c)^3 + 2*(48*A + 88*B + 75*C)*a*cos(d*x + c)^2 + 8*(8*B + 15*C)*a*cos(d*x + c) + 48*C*a)*s qrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 3 *((112*A + 88*B + 75*C)*a*cos(d*x + c)^5 + (112*A + 88*B + 75*C)*a*cos(d*x + c)^4)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c) )*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a )))/(d*cos(d*x + c)^5 + d*cos(d*x + c)^4)]
Timed out. \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 8121 vs. \(2 (217) = 434\).
Time = 0.91 (sec) , antiderivative size = 8121, normalized size of antiderivative = 32.10 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \]
integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c )^(3/2),x, algorithm="maxima")
-1/768*(48*(56*sqrt(2)*a*cos(7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 24*sqrt(2)*a*cos(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) *sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 12*sqrt(2) *a*sin(3/2*d*x + 3/2*c) + 28*sqrt(2)*a*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c ), cos(3/2*d*x + 3/2*c))) - 4*(3*sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 7*sqrt(2 )*a*sin(7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 3*sqrt( 2)*a*sin(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 7*sqrt (2)*a*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*cos(8/ 3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 8*(3*sqrt(2)*a*si n(3/2*d*x + 3/2*c) - 7*sqrt(2)*a*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos (3/2*d*x + 3/2*c))))*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3 /2*c))) - 7*(a*cos(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)) )^2 + 4*a*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + a*sin(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 4*a*si n(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))*sin(4/3*arctan2 (sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 4*a*sin(4/3*arctan2(sin(3/ 2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*(2*a*cos(4/3*arctan2(sin(3/2* d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + a)*cos(8/3*arctan2(sin(3/2*d*x + 3/ 2*c), cos(3/2*d*x + 3/2*c))) + 4*a*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c)...
\[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c )^(3/2),x, algorithm="giac")
Timed out. \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\cos \left (c+d\,x\right )}^{3/2}} \,d x \]
int(((a + a/cos(c + d*x))^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/c os(c + d*x)^(3/2),x)