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

[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))

________________________________________________________________________________________

Rubi [A]  time = 1.10375, antiderivative size = 353, normalized size of antiderivative = 1., 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 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]

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 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 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 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 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 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]

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*}

Mathematica [B]  time = 6.58711, size = 947, normalized size = 2.68 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[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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Maple [B]  time = 0.414, size = 815, normalized size = 2.3 \begin{align*} \text{result too large to display} \end{align*}

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

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Fricas [A]  time = 1.66173, size = 1736, normalized size = 4.92 \begin{align*} \text{result too large to display} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)