\(\int \frac {\cos ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\) [1272]

Optimal result

Integrand size = 45, antiderivative size = 257 \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {2} (A-B+C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}-\frac {2 (43 A-91 B+35 C) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 (31 A-7 B+35 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}-\frac {2 (A-7 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \sec (c+d x)}}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}} \]

[Out]

(A-B+C)*arctanh(1/2*sin(d*x+c)*a^(1/2)*sec(d*x+c)^(1/2)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*2^(1/2)*cos(d*x+c)^(1/
2)*sec(d*x+c)^(1/2)/d/a^(1/2)-2/35*(A-7*B)*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/7*A*cos(d*x+
c)^(5/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-2/105*(43*A-91*B+35*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x
+c))^(1/2)+2/105*(31*A-7*B+35*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 1.05 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {4350, 4171, 4107, 4098, 3893, 212} \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {2} (A-B+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {2 (31 A-7 B+35 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{105 d \sqrt {a \sec (c+d x)+a}}-\frac {2 (43 A-91 B+35 C) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {2 (A-7 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{35 d \sqrt {a \sec (c+d x)+a}}+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}} \]

[In]

Int[(Cos[c + d*x]^(7/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[a + a*Sec[c + d*x]],x]

[Out]

(Sqrt[2]*(A - B + C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])]*Sqr
t[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(Sqrt[a]*d) - (2*(43*A - 91*B + 35*C)*Sin[c + d*x])/(105*d*Sqrt[Cos[c + d*
x]]*Sqrt[a + a*Sec[c + d*x]]) + (2*(31*A - 7*B + 35*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a + a*Sec[
c + d*x]]) - (2*(A - 7*B)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(35*d*Sqrt[a + a*Sec[c + d*x]]) + (2*A*Cos[c + d*x]
^(5/2)*Sin[c + d*x])/(7*d*Sqrt[a + a*Sec[c + d*x]])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3893

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2*b*(d/
(a*f)), Subst[Int[1/(2*b - d*x^2), x], x, b*(Cot[e + f*x]/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]]))], x
] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 4098

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Dist[(
a*A*m - b*B*n)/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A
, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] &&  !LeQ[m, -1]

Rule 4107

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Dist[1
/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*B*n - A*b*(m + n + 1)*Csc[e + f*x
], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]

Rule 4171

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[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*
Csc[e + f*x])^n/(f*n)), x] - Dist[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m -
b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 -
b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])

Rule 4350

Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx \\ & = \frac {2 A \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{2} a (A-7 B)+\frac {1}{2} a (6 A+7 C) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx}{7 a} \\ & = -\frac {2 (A-7 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \sec (c+d x)}}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{4} a^2 (31 A-7 B+35 C)-a^2 (A-7 B) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx}{35 a^2} \\ & = \frac {2 (31 A-7 B+35 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}-\frac {2 (A-7 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \sec (c+d x)}}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}+\frac {\left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{8} a^3 (43 A-91 B+35 C)+\frac {1}{4} a^3 (31 A-7 B+35 C) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx}{105 a^3} \\ & = -\frac {2 (43 A-91 B+35 C) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 (31 A-7 B+35 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}-\frac {2 (A-7 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \sec (c+d x)}}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}+\left ((A-B+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+a \sec (c+d x)}} \, dx \\ & = -\frac {2 (43 A-91 B+35 C) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 (31 A-7 B+35 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}-\frac {2 (A-7 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \sec (c+d x)}}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}-\frac {\left (2 (A-B+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d} \\ & = \frac {\sqrt {2} (A-B+C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}-\frac {2 (43 A-91 B+35 C) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 (31 A-7 B+35 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}-\frac {2 (A-7 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \sec (c+d x)}}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.05 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.69 \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\cos ^{\frac {5}{2}}(c+d x) \left (105 \sqrt {2} (A-B+C) \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec ^{\frac {7}{2}}(c+d x)+2 \sqrt {1-\sec (c+d x)} \left (-15 A+3 (A-7 B) \sec (c+d x)+(-31 A+7 (B-5 C)) \sec ^2(c+d x)+(43 A-91 B+35 C) \sec ^3(c+d x)\right )\right ) \sin (c+d x)}{105 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]

[In]

Integrate[(Cos[c + d*x]^(7/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[a + a*Sec[c + d*x]],x]

[Out]

-1/105*(Cos[c + d*x]^(5/2)*(105*Sqrt[2]*(A - B + C)*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]
]*Sec[c + d*x]^(7/2) + 2*Sqrt[1 - Sec[c + d*x]]*(-15*A + 3*(A - 7*B)*Sec[c + d*x] + (-31*A + 7*(B - 5*C))*Sec[
c + d*x]^2 + (43*A - 91*B + 35*C)*Sec[c + d*x]^3))*Sin[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c +
d*x])])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(516\) vs. \(2(218)=436\).

Time = 0.58 (sec) , antiderivative size = 517, normalized size of antiderivative = 2.01

method	result	size
default	\(\frac {\left (105 A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {2}\, \cos \left (d x +c \right )+30 A \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )-105 B \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {2}\, \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )+105 C \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {2}\, \cos \left (d x +c \right )+105 A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {2}-6 A \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )-105 B \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {2}\, \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+42 B \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+105 C \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {2}+62 A \cos \left (d x +c \right ) \sin \left (d x +c \right )-14 B \cos \left (d x +c \right ) \sin \left (d x +c \right )+70 C \cos \left (d x +c \right ) \sin \left (d x +c \right )-86 A \sin \left (d x +c \right )+182 B \sin \left (d x +c \right )-70 C \sin \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{105 d \left (1+\cos \left (d x +c \right )\right ) a}\)	\(517\)

[In]

int(cos(d*x+c)^(7/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/105/d*(105*A*arctan(1/2*sin(d*x+c)*2^(1/2)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))*(-1/(1+cos(d*x+c)))^(1/
2)*2^(1/2)*cos(d*x+c)+30*A*cos(d*x+c)^3*sin(d*x+c)-105*B*arctan(1/2*sin(d*x+c)*2^(1/2)/(1+cos(d*x+c))/(-1/(1+c
os(d*x+c)))^(1/2))*2^(1/2)*(-1/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+105*C*arctan(1/2*sin(d*x+c)*2^(1/2)/(1+cos(d*x
+c))/(-1/(1+cos(d*x+c)))^(1/2))*(-1/(1+cos(d*x+c)))^(1/2)*2^(1/2)*cos(d*x+c)+105*A*arctan(1/2*sin(d*x+c)*2^(1/
2)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))*(-1/(1+cos(d*x+c)))^(1/2)*2^(1/2)-6*A*cos(d*x+c)^2*sin(d*x+c)-105
*B*arctan(1/2*sin(d*x+c)*2^(1/2)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))*2^(1/2)*(-1/(1+cos(d*x+c)))^(1/2)+4
2*B*cos(d*x+c)^2*sin(d*x+c)+105*C*arctan(1/2*sin(d*x+c)*2^(1/2)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))*(-1/
(1+cos(d*x+c)))^(1/2)*2^(1/2)+62*A*cos(d*x+c)*sin(d*x+c)-14*B*cos(d*x+c)*sin(d*x+c)+70*C*cos(d*x+c)*sin(d*x+c)
-86*A*sin(d*x+c)+182*B*sin(d*x+c)-70*C*sin(d*x+c))*cos(d*x+c)^(1/2)*(a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x+c))/a

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 416, normalized size of antiderivative = 1.62 \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\left [\frac {4 \, {\left (15 \, A \cos \left (d x + c\right )^{3} - 3 \, {\left (A - 7 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (31 \, A - 7 \, B + 35 \, C\right )} \cos \left (d x + c\right ) - 43 \, A + 91 \, B - 35 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + \frac {105 \, \sqrt {2} {\left ({\left (A - B + C\right )} a \cos \left (d x + c\right ) + {\left (A - B + C\right )} a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - \frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{210 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}, -\frac {105 \, \sqrt {2} {\left ({\left (A - B + C\right )} a \cos \left (d x + c\right ) + {\left (A - B + C\right )} a\right )} \sqrt {-\frac {1}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \sqrt {\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) - 2 \, {\left (15 \, A \cos \left (d x + c\right )^{3} - 3 \, {\left (A - 7 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (31 \, A - 7 \, B + 35 \, C\right )} \cos \left (d x + c\right ) - 43 \, A + 91 \, B - 35 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{105 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}\right ] \]

[In]

integrate(cos(d*x+c)^(7/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/210*(4*(15*A*cos(d*x + c)^3 - 3*(A - 7*B)*cos(d*x + c)^2 + (31*A - 7*B + 35*C)*cos(d*x + c) - 43*A + 91*B -
35*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 105*sqrt(2)*((A - B + C)*a*co
s(d*x + c) + (A - B + C)*a)*log(-(cos(d*x + c)^2 - 2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(
d*x + c))*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1))/sqrt(a))/(a*d*cos(
d*x + c) + a*d), -1/105*(105*sqrt(2)*((A - B + C)*a*cos(d*x + c) + (A - B + C)*a)*sqrt(-1/a)*arctan(sqrt(2)*sq
rt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*sqrt(cos(d*x + c))/sin(d*x + c)) - 2*(15*A*cos(d*x + c)^3 - 3
*(A - 7*B)*cos(d*x + c)^2 + (31*A - 7*B + 35*C)*cos(d*x + c) - 43*A + 91*B - 35*C)*sqrt((a*cos(d*x + c) + a)/c
os(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x + c) + a*d)]

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**(7/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(1/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 970 vs. \(2 (218) = 436\).

Time = 0.55 (sec) , antiderivative size = 970, normalized size of antiderivative = 3.77 \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)^(7/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

-1/840*(sqrt(2)*(525*cos(6/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) - 175*c
os(4/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) + 21*cos(2/7*arctan2(sin(7/2*
d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) - 525*cos(7/2*d*x + 7/2*c)*sin(6/7*arctan2(sin(7/2*d
*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 175*cos(7/2*d*x + 7/2*c)*sin(4/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d
*x + 7/2*c))) - 21*cos(7/2*d*x + 7/2*c)*sin(2/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 420*log
(cos(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))^2 + sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/
2*d*x + 7/2*c)))^2 + 2*sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 1) + 420*log(cos(1/7*arc
tan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))^2 + sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*
c)))^2 - 2*sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 1) - 30*sin(7/2*d*x + 7/2*c) + 21*si
n(5/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 175*sin(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2
*d*x + 7/2*c))) + 525*sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))))*A/sqrt(a) + 28*(30*sqrt(2)
*cos(5/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(2*d*x + 2*c) - 3*(10*sqrt(2)*cos(2*d*x + 2*c) + sqrt
(2))*sin(5/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 15*sqrt(2)*log(cos(1/4*arctan2(sin(2*d*x + 2*c), c
os(2*d*x + 2*c)))^2 + sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/4*arctan2(sin(2*d*x + 2
*c), cos(2*d*x + 2*c))) + 1) - 15*sqrt(2)*log(cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/4
*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 - 2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1)
+ 5*sqrt(2)*sin(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 30*sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c)
, cos(2*d*x + 2*c))))*B/sqrt(a) - 140*(3*sqrt(2)*log(cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 +
sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))
) + 1) - 3*sqrt(2)*log(cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/4*arctan2(sin(2*d*x + 2*
c), cos(2*d*x + 2*c)))^2 - 2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + 2*sqrt(2)*sin(3/4*arc
tan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 6*sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*C/
sqrt(a))/d

Giac [F]

\[ \int \frac {\cos ^{\frac {7}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {7}{2}}}{\sqrt {a \sec \left (d x + c\right ) + a}} \,d x } \]

[In]

integrate(cos(d*x+c)^(7/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^(7/2)/sqrt(a*sec(d*x + c) + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{7/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \]

[In]

int((cos(c + d*x)^(7/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^(1/2),x)

[Out]

int((cos(c + d*x)^(7/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^(1/2), x)