3.1214 \(\int \cos ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)
Optimal. Leaf size=271 \[ \frac{8 a^4 (35 A+28 B+17 C) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{21 d}+\frac{2 (35 A+77 B+73 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{105 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{4 a^4 (175 A+287 B+253 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{105 d}+\frac{4 (175 A+238 B+197 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{105 d \sqrt{\cos (c+d x)}}-\frac{8 a^4 (7 B+8 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a (7 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{35 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{7 d \cos ^{\frac{7}{2}}(c+d x)} \]
[Out]
(-8*a^4*(7*B + 8*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (8*a^4*(35*A + 28*B + 17*C)*EllipticF[(c + d*x)/2, 2])/
(21*d) - (4*a^4*(175*A + 287*B + 253*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(105*d) + (2*a*(7*B + 8*C)*(a + a*Cos
[c + d*x])^3*Sin[c + d*x])/(35*d*Cos[c + d*x]^(5/2)) + (2*C*(a + a*Cos[c + d*x])^4*Sin[c + d*x])/(7*d*Cos[c +
d*x]^(7/2)) + (2*(35*A + 77*B + 73*C)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(105*d*Cos[c + d*x]^(3/2)) + (4
*(175*A + 238*B + 197*C)*(a^4 + a^4*Cos[c + d*x])*Sin[c + d*x])/(105*d*Sqrt[Cos[c + d*x]])
________________________________________________________________________________________
Rubi [A] time = 0.870524, antiderivative size = 271, normalized size of antiderivative = 1.,
number of steps used = 10, number of rules used = 8, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.186, Rules used
= {4112, 3043, 2975, 2968, 3023, 2748, 2641, 2639} \[ \frac{8 a^4 (35 A+28 B+17 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 (35 A+77 B+73 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{105 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{4 a^4 (175 A+287 B+253 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{105 d}+\frac{4 (175 A+238 B+197 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{105 d \sqrt{\cos (c+d x)}}-\frac{8 a^4 (7 B+8 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a (7 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{35 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{7 d \cos ^{\frac{7}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(-8*a^4*(7*B + 8*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (8*a^4*(35*A + 28*B + 17*C)*EllipticF[(c + d*x)/2, 2])/
(21*d) - (4*a^4*(175*A + 287*B + 253*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(105*d) + (2*a*(7*B + 8*C)*(a + a*Cos
[c + d*x])^3*Sin[c + d*x])/(35*d*Cos[c + d*x]^(5/2)) + (2*C*(a + a*Cos[c + d*x])^4*Sin[c + d*x])/(7*d*Cos[c +
d*x]^(7/2)) + (2*(35*A + 77*B + 73*C)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(105*d*Cos[c + d*x]^(3/2)) + (4
*(175*A + 238*B + 197*C)*(a^4 + a^4*Cos[c + d*x])*Sin[c + d*x])/(105*d*Sqrt[Cos[c + d*x]])
Rule 4112
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sec[(e_.)
+ (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[d^(m + 2), Int[(b + a*Cos[e + f*x])^m*(d*
Cos[e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}
, x] && !IntegerQ[n] && IntegerQ[m]
Rule 3043
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C - B*c*d + A*d^2)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*d*(n + 1)
*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + (c*C -
B*d)*(a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2,
0] && !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
Rule 2975
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d
, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n]
|| EqQ[c, 0])
Rule 2968
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Rule 3023
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Rule 2748
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Rule 2641
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]
Rule 2639
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]
Rubi steps
\begin{align*} \int \cos ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \frac{(a+a \cos (c+d x))^4 \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right )}{\cos ^{\frac{9}{2}}(c+d x)} \, dx\\ &=\frac{2 C (a+a \cos (c+d x))^4 \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 \int \frac{(a+a \cos (c+d x))^4 \left (\frac{1}{2} a (7 B+8 C)+\frac{1}{2} a (7 A-3 C) \cos (c+d x)\right )}{\cos ^{\frac{7}{2}}(c+d x)} \, dx}{7 a}\\ &=\frac{2 a (7 B+8 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{35 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 C (a+a \cos (c+d x))^4 \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{4 \int \frac{(a+a \cos (c+d x))^3 \left (\frac{1}{4} a^2 (35 A+77 B+73 C)+\frac{1}{4} a^2 (35 A-21 B-39 C) \cos (c+d x)\right )}{\cos ^{\frac{5}{2}}(c+d x)} \, dx}{35 a}\\ &=\frac{2 a (7 B+8 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{35 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 C (a+a \cos (c+d x))^4 \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 (35 A+77 B+73 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{105 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{8 \int \frac{(a+a \cos (c+d x))^2 \left (\frac{1}{4} a^3 (175 A+238 B+197 C)-\frac{21}{4} a^3 (7 B+8 C) \cos (c+d x)\right )}{\cos ^{\frac{3}{2}}(c+d x)} \, dx}{105 a}\\ &=\frac{2 a (7 B+8 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{35 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 C (a+a \cos (c+d x))^4 \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 (35 A+77 B+73 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{105 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4 (175 A+238 B+197 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{105 d \sqrt{\cos (c+d x)}}+\frac{16 \int \frac{(a+a \cos (c+d x)) \left (\frac{3}{8} a^4 (175 A+189 B+141 C)-\frac{3}{8} a^4 (175 A+287 B+253 C) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx}{105 a}\\ &=\frac{2 a (7 B+8 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{35 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 C (a+a \cos (c+d x))^4 \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 (35 A+77 B+73 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{105 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4 (175 A+238 B+197 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{105 d \sqrt{\cos (c+d x)}}+\frac{16 \int \frac{\frac{3}{8} a^5 (175 A+189 B+141 C)+\left (\frac{3}{8} a^5 (175 A+189 B+141 C)-\frac{3}{8} a^5 (175 A+287 B+253 C)\right ) \cos (c+d x)-\frac{3}{8} a^5 (175 A+287 B+253 C) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{105 a}\\ &=-\frac{4 a^4 (175 A+287 B+253 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{105 d}+\frac{2 a (7 B+8 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{35 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 C (a+a \cos (c+d x))^4 \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 (35 A+77 B+73 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{105 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4 (175 A+238 B+197 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{105 d \sqrt{\cos (c+d x)}}+\frac{32 \int \frac{\frac{15}{8} a^5 (35 A+28 B+17 C)-\frac{63}{8} a^5 (7 B+8 C) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{315 a}\\ &=-\frac{4 a^4 (175 A+287 B+253 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{105 d}+\frac{2 a (7 B+8 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{35 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 C (a+a \cos (c+d x))^4 \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 (35 A+77 B+73 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{105 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4 (175 A+238 B+197 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{105 d \sqrt{\cos (c+d x)}}-\frac{1}{5} \left (4 a^4 (7 B+8 C)\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{21} \left (4 a^4 (35 A+28 B+17 C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{8 a^4 (7 B+8 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{8 a^4 (35 A+28 B+17 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}-\frac{4 a^4 (175 A+287 B+253 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{105 d}+\frac{2 a (7 B+8 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{35 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 C (a+a \cos (c+d x))^4 \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 (35 A+77 B+73 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{105 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4 (175 A+238 B+197 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{105 d \sqrt{\cos (c+d x)}}\\ \end{align*}
Mathematica [C] time = 7.22779, size = 1454, normalized size = 5.37 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Cos[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(Cos[c + d*x]^(13/2)*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(-((-
20*A - 61*B - 64*C + 20*A*Cos[2*c] + 5*B*Cos[2*c])*Csc[c]*Sec[c])/(40*d) + (A*Cos[d*x]*Sin[c])/(12*d) + (A*Cos
[c]*Sin[d*x])/(12*d) + (C*Sec[c]*Sec[c + d*x]^4*Sin[d*x])/(28*d) + (Sec[c]*Sec[c + d*x]^3*(5*C*Sin[c] + 7*B*Si
n[d*x] + 28*C*Sin[d*x]))/(140*d) + (Sec[c]*Sec[c + d*x]^2*(21*B*Sin[c] + 84*C*Sin[c] + 35*A*Sin[d*x] + 140*B*S
in[d*x] + 235*C*Sin[d*x]))/(420*d) + (Sec[c]*Sec[c + d*x]*(35*A*Sin[c] + 140*B*Sin[c] + 235*C*Sin[c] + 420*A*S
in[d*x] + 693*B*Sin[d*x] + 672*C*Sin[d*x]))/(420*d)))/(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x]) - (5*A
*Cos[c + d*x]^6*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^8*
(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - Ar
cTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]
])/(3*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (4*B*Cos[c + d*x]^6*Csc[c]*Hyp
ergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A
+ B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt
[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(A + 2*C + 2*B*Cos
[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (17*C*Cos[c + d*x]^6*Csc[c]*HypergeometricPFQ[{1/4, 1/2}
, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[
c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin
[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(21*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c +
2*d*x])*Sqrt[1 + Cot[c]^2]) + (7*B*Cos[c + d*x]^6*Csc[c]*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Se
c[c + d*x] + C*Sec[c + d*x]^2)*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x +
ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*
Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1
+ Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Co
s[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(10*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (4*C*
Cos[c + d*x]^6*Csc[c]*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((Hy
pergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 -
Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 +
Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x
+ ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Ta
n[c]^2]]))/(5*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x]))
________________________________________________________________________________________
Maple [B] time = 10.01, size = 1535, normalized size = 5.7 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^(3/2)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
[Out]
8/105*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^4/(16*sin(1/2*d*x+1/2*c)^8-32*sin(1/2*d*x+1/
2*c)^6+24*sin(1/2*d*x+1/2*c)^4-8*sin(1/2*d*x+1/2*c)^2+1)/sin(1/2*d*x+1/2*c)^3*(280*A*cos(1/2*d*x+1/2*c)*sin(1/
2*d*x+1/2*c)^10-2240*A*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)-2772*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-
2380*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+3010*A*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-140*B*(sin(1/2*d
*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-147*B*(sin(1/2*d*x+1
/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-85*C*(sin(1/2*d*x+1/2*c)
^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-168*C*(sin(1/2*d*x+1/2*c)^2)^
(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-175*A*(sin(1/2*d*x+1/2*c)^2)^(1/2
)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+427*B*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*
x+1/2*c)+1050*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(
1/2))*sin(1/2*d*x+1/2*c)^2+503*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*C+1120*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*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^6-1764*B*(2*sin(1/2*d
*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4+1
400*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)^6+1176*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)^6+680*C*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)^6+1344*C*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)^6+840*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*E
llipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+1008*C*(2*sin(1/2*d*x+1
/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+882*B
*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d
*x+1/2*c)^2+510*C*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*sin(1/2*d*x+1/2*c)^2-2688*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-1020*C*(2*sin(1/2*d*x+1/2*c)^2-1)^
(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-2100*A*(2*sin(1/
2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^
4-1680*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*s
in(1/2*d*x+1/2*c)^4-2016*C*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x
+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-1470*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+245*sin(1/2*d*x+1/2*c)^2*
cos(1/2*d*x+1/2*c)*A-2570*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+4438*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*
c)^6+4502*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*c
os(1/2*d*x+1/2*c)^2-1)^(1/2)/d
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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(cos(d*x+c)^(3/2)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")
[Out]
Timed out
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C a^{4} \cos \left (d x + c\right ) \sec \left (d x + c\right )^{6} +{\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right ) \sec \left (d x + c\right )^{5} +{\left (A + 4 \, B + 6 \, C\right )} a^{4} \cos \left (d x + c\right ) \sec \left (d x + c\right )^{4} + 2 \,{\left (2 \, A + 3 \, B + 2 \, C\right )} a^{4} \cos \left (d x + c\right ) \sec \left (d x + c\right )^{3} +{\left (6 \, A + 4 \, B + C\right )} a^{4} \cos \left (d x + c\right ) \sec \left (d x + c\right )^{2} +{\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right ) \sec \left (d x + c\right ) + A a^{4} \cos \left (d x + c\right )\right )} \sqrt{\cos \left (d x + c\right )}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(3/2)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")
[Out]
integral((C*a^4*cos(d*x + c)*sec(d*x + c)^6 + (B + 4*C)*a^4*cos(d*x + c)*sec(d*x + c)^5 + (A + 4*B + 6*C)*a^4*
cos(d*x + c)*sec(d*x + c)^4 + 2*(2*A + 3*B + 2*C)*a^4*cos(d*x + c)*sec(d*x + c)^3 + (6*A + 4*B + C)*a^4*cos(d*
x + c)*sec(d*x + c)^2 + (4*A + B)*a^4*cos(d*x + c)*sec(d*x + c) + A*a^4*cos(d*x + c))*sqrt(cos(d*x + c)), x)
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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(cos(d*x+c)**(3/2)*(a+a*sec(d*x+c))**4*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\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 )}^{4} \cos \left (d x + c\right )^{\frac{3}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(3/2)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^4*cos(d*x + c)^(3/2), x)