3.1271 \(\int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=353 \[ \frac {a^{5/2} (1304 A+1132 B+1015 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{512 d}+\frac {a^3 (1304 A+1132 B+1015 C) \sin (c+d x)}{512 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {a^3 (1304 A+1132 B+1015 C) \sin (c+d x)}{768 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {a^3 (680 A+628 B+545 C) \sin (c+d x)}{960 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {a^2 (120 A+156 B+115 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{480 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {a (12 B+5 C) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{60 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {C \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{6 d \cos ^{\frac {7}{2}}(c+d x)} \]

[Out]

1/60*a*(12*B+5*C)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/cos(d*x+c)^(7/2)+1/6*C*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)
/d/cos(d*x+c)^(7/2)+1/512*a^(5/2)*(1304*A+1132*B+1015*C)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))*co
s(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+1/960*a^3*(680*A+628*B+545*C)*sin(d*x+c)/d/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c))
^(1/2)+1/768*a^3*(1304*A+1132*B+1015*C)*sin(d*x+c)/d/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(1/2)+1/512*a^3*(1304*A
+1132*B+1015*C)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)+1/480*a^2*(120*A+156*B+115*C)*sin(d*x+c)*
(a+a*sec(d*x+c))^(1/2)/d/cos(d*x+c)^(7/2)

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Rubi [A]  time = 1.10, antiderivative size = 353, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {4265, 4088, 4018, 4016, 3803, 3801, 215} \[ \frac {a^3 (1304 A+1132 B+1015 C) \sin (c+d x)}{512 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {a^3 (1304 A+1132 B+1015 C) \sin (c+d x)}{768 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {a^3 (680 A+628 B+545 C) \sin (c+d x)}{960 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {a^2 (120 A+156 B+115 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{480 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {a^{5/2} (1304 A+1132 B+1015 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{512 d}+\frac {a (12 B+5 C) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{60 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {C \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{6 d \cos ^{\frac {7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Cos[c + d*x]^(5/2),x]

[Out]

(a^(5/2)*(1304*A + 1132*B + 1015*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]*Sqrt[Cos[c + d*x]
]*Sqrt[Sec[c + d*x]])/(512*d) + (a^3*(680*A + 628*B + 545*C)*Sin[c + d*x])/(960*d*Cos[c + d*x]^(7/2)*Sqrt[a +
a*Sec[c + d*x]]) + (a^3*(1304*A + 1132*B + 1015*C)*Sin[c + d*x])/(768*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Sec[c +
d*x]]) + (a^3*(1304*A + 1132*B + 1015*C)*Sin[c + d*x])/(512*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (
a^2*(120*A + 156*B + 115*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(480*d*Cos[c + d*x]^(7/2)) + (a*(12*B + 5*C
)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(60*d*Cos[c + d*x]^(7/2)) + (C*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d
*x])/(6*d*Cos[c + d*x]^(7/2))

Rule 215

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]

Rule 3801

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

Rule 3803

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] + Dist[(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] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a
^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]

Rule 4016

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*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]]
), x] + Dist[(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]

Rule 4018

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] + Dist[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] && Ne
Q[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1]

Rule 4088

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
*Csc[e + f*x])^n)/(f*(m + n + 1)), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Csc[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]

Rule 4265

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*} \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\\ &=\frac {C (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{6 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac {1}{2} a (12 A+5 C)+\frac {1}{2} a (12 B+5 C) \sec (c+d x)\right ) \, dx}{6 a}\\ &=\frac {a (12 B+5 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{60 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {C (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{6 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac {15}{4} a^2 (8 A+4 B+5 C)+\frac {1}{4} a^2 (120 A+156 B+115 C) \sec (c+d x)\right ) \, dx}{30 a}\\ &=\frac {a^2 (120 A+156 B+115 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{480 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {a (12 B+5 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{60 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {C (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{6 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {5}{8} a^3 (312 A+252 B+235 C)+\frac {3}{8} a^3 (680 A+628 B+545 C) \sec (c+d x)\right ) \, dx}{120 a}\\ &=\frac {a^3 (680 A+628 B+545 C) \sin (c+d x)}{960 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (120 A+156 B+115 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{480 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {a (12 B+5 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{60 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {C (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{6 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {1}{384} \left (a^2 (1304 A+1132 B+1015 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a^3 (680 A+628 B+545 C) \sin (c+d x)}{960 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (1304 A+1132 B+1015 C) \sin (c+d x)}{768 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (120 A+156 B+115 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{480 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {a (12 B+5 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{60 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {C (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{6 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {1}{512} \left (a^2 (1304 A+1132 B+1015 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a^3 (680 A+628 B+545 C) \sin (c+d x)}{960 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (1304 A+1132 B+1015 C) \sin (c+d x)}{768 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (1304 A+1132 B+1015 C) \sin (c+d x)}{512 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (120 A+156 B+115 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{480 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {a (12 B+5 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{60 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {C (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{6 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {\left (a^2 (1304 A+1132 B+1015 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx}{1024}\\ &=\frac {a^3 (680 A+628 B+545 C) \sin (c+d x)}{960 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (1304 A+1132 B+1015 C) \sin (c+d x)}{768 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (1304 A+1132 B+1015 C) \sin (c+d x)}{512 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (120 A+156 B+115 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{480 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {a (12 B+5 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{60 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {C (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{6 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {\left (a^2 (1304 A+1132 B+1015 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{a}}} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{512 d}\\ &=\frac {a^{5/2} (1304 A+1132 B+1015 C) \sinh ^{-1}\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)}}{512 d}+\frac {a^3 (680 A+628 B+545 C) \sin (c+d x)}{960 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (1304 A+1132 B+1015 C) \sin (c+d x)}{768 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (1304 A+1132 B+1015 C) \sin (c+d x)}{512 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (120 A+156 B+115 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{480 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {a (12 B+5 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{60 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {C (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{6 d \cos ^{\frac {7}{2}}(c+d x)}\\ \end {align*}

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Mathematica [B]  time = 6.49, size = 947, normalized size = 2.68 \[ \text {result too large to display} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Cos[c + d*x]^(5/2),x]

[Out]

(4*Sec[(c + d*x)/2]^5*(a*(1 + Sec[c + d*x]))^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sqrt[(1 - 2*Sin[(c
+ d*x)/2]^2)^(-1)]*Sqrt[1 - 2*Sin[(c + d*x)/2]^2]*((C*Sin[(c + d*x)/2])/(48*(1 - 2*Sin[(c + d*x)/2]^2)^6) + ((
B + 2*C)*Sin[(c + d*x)/2])/(40*(1 - 2*Sin[(c + d*x)/2]^2)^5) + ((A + 2*B + C)*Sin[(c + d*x)/2])/(32*(1 - 2*Sin
[(c + d*x)/2]^2)^4) + ((2*A + B)*Sin[(c + d*x)/2])/(24*(1 - 2*Sin[(c + d*x)/2]^2)^3) + (A*Sin[(c + d*x)/2])/(1
6*(1 - 2*Sin[(c + d*x)/2]^2)^2) + (3*A*(Sqrt[2]*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]] + (2*Sin[(c + d*x)/2])/(1 -
2*Sin[(c + d*x)/2]^2)))/64 + (5*(2*A + B)*((4*Sin[(c + d*x)/2])/(1 - 2*Sin[(c + d*x)/2]^2)^2 + 3*(Sqrt[2]*ArcT
anh[Sqrt[2]*Sin[(c + d*x)/2]] + (2*Sin[(c + d*x)/2])/(1 - 2*Sin[(c + d*x)/2]^2))))/384 + (7*(A + 2*B + C)*((16
*Sin[(c + d*x)/2])/(1 - 2*Sin[(c + d*x)/2]^2)^3 + 5*((4*Sin[(c + d*x)/2])/(1 - 2*Sin[(c + d*x)/2]^2)^2 + 3*(Sq
rt[2]*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]] + (2*Sin[(c + d*x)/2])/(1 - 2*Sin[(c + d*x)/2]^2)))))/3072 + (3*(B + 2
*C)*((96*Sin[(c + d*x)/2])/(1 - 2*Sin[(c + d*x)/2]^2)^4 + 7*((16*Sin[(c + d*x)/2])/(1 - 2*Sin[(c + d*x)/2]^2)^
3 + 5*((4*Sin[(c + d*x)/2])/(1 - 2*Sin[(c + d*x)/2]^2)^2 + 3*(Sqrt[2]*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]] + (2*S
in[(c + d*x)/2])/(1 - 2*Sin[(c + d*x)/2]^2))))))/10240 + (11*C*((256*Sin[(c + d*x)/2])/(1 - 2*Sin[(c + d*x)/2]
^2)^5 + 3*((96*Sin[(c + d*x)/2])/(1 - 2*Sin[(c + d*x)/2]^2)^4 + 7*((16*Sin[(c + d*x)/2])/(1 - 2*Sin[(c + d*x)/
2]^2)^3 + 5*((4*Sin[(c + d*x)/2])/(1 - 2*Sin[(c + d*x)/2]^2)^2 + 3*(Sqrt[2]*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]
+ (2*Sin[(c + d*x)/2])/(1 - 2*Sin[(c + d*x)/2]^2)))))))/122880))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c +
2*d*x])*Sec[c + d*x]^(9/2))

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fricas [A]  time = 0.96, size = 625, normalized size = 1.77 \[ \left [\frac {4 \, {\left (15 \, {\left (1304 \, A + 1132 \, B + 1015 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + 10 \, {\left (1304 \, A + 1132 \, B + 1015 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 8 \, {\left (920 \, A + 1132 \, B + 1015 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 48 \, {\left (40 \, A + 116 \, B + 145 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 128 \, {\left (12 \, B + 35 \, C\right )} a^{2} \cos \left (d x + c\right ) + 1280 \, C a^{2}\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 ) + 15 \, {\left ({\left (1304 \, A + 1132 \, B + 1015 \, C\right )} a^{2} \cos \left (d x + c\right )^{7} + {\left (1304 \, A + 1132 \, B + 1015 \, C\right )} a^{2} \cos \left (d x + c\right )^{6}\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 )}{30720 \, {\left (d \cos \left (d x + c\right )^{7} + d \cos \left (d x + c\right )^{6}\right )}}, \frac {2 \, {\left (15 \, {\left (1304 \, A + 1132 \, B + 1015 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + 10 \, {\left (1304 \, A + 1132 \, B + 1015 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 8 \, {\left (920 \, A + 1132 \, B + 1015 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 48 \, {\left (40 \, A + 116 \, B + 145 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 128 \, {\left (12 \, B + 35 \, C\right )} a^{2} \cos \left (d x + c\right ) + 1280 \, C a^{2}\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 ) + 15 \, {\left ({\left (1304 \, A + 1132 \, B + 1015 \, C\right )} a^{2} \cos \left (d x + c\right )^{7} + {\left (1304 \, A + 1132 \, B + 1015 \, C\right )} a^{2} \cos \left (d x + c\right )^{6}\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 )}{15360 \, {\left (d \cos \left (d x + c\right )^{7} + d \cos \left (d x + c\right )^{6}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

[1/30720*(4*(15*(1304*A + 1132*B + 1015*C)*a^2*cos(d*x + c)^5 + 10*(1304*A + 1132*B + 1015*C)*a^2*cos(d*x + c)
^4 + 8*(920*A + 1132*B + 1015*C)*a^2*cos(d*x + c)^3 + 48*(40*A + 116*B + 145*C)*a^2*cos(d*x + c)^2 + 128*(12*B
+ 35*C)*a^2*cos(d*x + c) + 1280*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c
) + 15*((1304*A + 1132*B + 1015*C)*a^2*cos(d*x + c)^7 + (1304*A + 1132*B + 1015*C)*a^2*cos(d*x + c)^6)*sqrt(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)^7 + d*cos(d*x
+ c)^6), 1/15360*(2*(15*(1304*A + 1132*B + 1015*C)*a^2*cos(d*x + c)^5 + 10*(1304*A + 1132*B + 1015*C)*a^2*cos(
d*x + c)^4 + 8*(920*A + 1132*B + 1015*C)*a^2*cos(d*x + c)^3 + 48*(40*A + 116*B + 145*C)*a^2*cos(d*x + c)^2 + 1
28*(12*B + 35*C)*a^2*cos(d*x + c) + 1280*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin
(d*x + c) + 15*((1304*A + 1132*B + 1015*C)*a^2*cos(d*x + c)^7 + (1304*A + 1132*B + 1015*C)*a^2*cos(d*x + c)^6)
*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)^7 + d*cos(d*x + c)^6)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\cos \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^(5/2)/cos(d*x + c)^(5/2), x)

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maple [B]  time = 2.03, size = 815, normalized size = 2.31 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2),x)

[Out]

-1/15360/d*a^2*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))*(19560*A*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/
2)*(cos(d*x+c)+1+sin(d*x+c))*2^(1/2))*2^(1/2)*cos(d*x+c)^6-19560*A*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d
*x+c)+1-sin(d*x+c))*2^(1/2))*2^(1/2)*cos(d*x+c)^6+16980*B*cos(d*x+c)^6*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(c
os(d*x+c)+1+sin(d*x+c))*2^(1/2))*2^(1/2)-16980*B*cos(d*x+c)^6*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)
+1-sin(d*x+c))*2^(1/2))*2^(1/2)+15225*C*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1+sin(d*x+c))*2^(1/2)
)*2^(1/2)*cos(d*x+c)^6-15225*C*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1-sin(d*x+c))*2^(1/2))*2^(1/2)
*cos(d*x+c)^6+39120*A*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*cos(d*x+c)^5+33960*B*cos(d*x+c)^5*(-2/(1+cos(d*x+c)
))^(1/2)*sin(d*x+c)+30450*C*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*cos(d*x+c)^5+26080*A*(-2/(1+cos(d*x+c)))^(1/2
)*cos(d*x+c)^4*sin(d*x+c)+22640*B*(-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^4*sin(d*x+c)+20300*C*(-2/(1+cos(d*x+c))
)^(1/2)*cos(d*x+c)^4*sin(d*x+c)+14720*A*cos(d*x+c)^3*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2)+18112*B*cos(d*x+c)^3
*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2)+16240*C*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*cos(d*x+c)^3+3840*A*sin(d*x
+c)*cos(d*x+c)^2*(-2/(1+cos(d*x+c)))^(1/2)+11136*B*sin(d*x+c)*cos(d*x+c)^2*(-2/(1+cos(d*x+c)))^(1/2)+13920*C*(
-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*cos(d*x+c)^2+3072*B*sin(d*x+c)*cos(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2)+8960*C
*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*cos(d*x+c)+2560*C*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c))/(-2/(1+cos(d*x+c
)))^(1/2)/cos(d*x+c)^(11/2)/sin(d*x+c)^2

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\cos \left (c+d\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + a/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/cos(c + d*x)^(5/2),x)

[Out]

int(((a + a/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/cos(c + d*x)^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/cos(d*x+c)**(5/2),x)

[Out]

Timed out

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