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 \[ \frac {8 a^4 (35 A+28 B+17 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {4 a^4 (175 A+287 B+253 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{105 d}+\frac {4 (175 A+238 B+197 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{105 d \sqrt {\cos (c+d x)}}-\frac {8 a^4 (7 B+8 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 (35 A+77 B+73 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a (7 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{7 d \cos ^{\frac {7}{2}}(c+d x)} \]

[Out]

-8/5*a^4*(7*B+8*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+8/2
1*a^4*(35*A+28*B+17*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/d
+2/35*a*(7*B+8*C)*(a+a*cos(d*x+c))^3*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2/7*C*(a+a*cos(d*x+c))^4*sin(d*x+c)/d/cos(d
*x+c)^(7/2)+2/105*(35*A+77*B+73*C)*(a^2+a^2*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(3/2)+4/105*(175*A+238*B+197
*C)*(a^4+a^4*cos(d*x+c))*sin(d*x+c)/d/cos(d*x+c)^(1/2)-4/105*a^4*(175*A+287*B+253*C)*sin(d*x+c)*cos(d*x+c)^(1/
2)/d

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Rubi [A]  time = 0.87, antiderivative size = 271, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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 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 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 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]

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*}

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Mathematica [C]  time = 7.16, 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)*(-1/4
0*((-20*A - 61*B - 64*C + 20*A*Cos[2*c] + 5*B*Cos[2*c])*Csc[c]*Sec[c])/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]))

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C a^{4} \cos \left (d x + c\right ) \sec \left (d x + c\right )^{6} + {\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right ) \sec \left (d x + c\right )^{5} + {\left (A + 4 \, B + 6 \, C\right )} a^{4} \cos \left (d x + c\right ) \sec \left (d x + c\right )^{4} + 2 \, {\left (2 \, A + 3 \, B + 2 \, C\right )} a^{4} \cos \left (d x + c\right ) \sec \left (d x + c\right )^{3} + {\left (6 \, A + 4 \, B + C\right )} a^{4} \cos \left (d x + c\right ) \sec \left (d x + c\right )^{2} + {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right ) \sec \left (d x + c\right ) + A a^{4} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{4} \cos \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]

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)

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maple [B]  time = 19.55, size = 1535, normalized size = 5.66 \[ \text {result too large to display} \]

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*(-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)^4-2240*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8
+245*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+1400*A*EllipticF(cos(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)^2)^(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))*
(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^6+1120*B*EllipticF(cos(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)^2)^(1/2)*sin(1/2*d*x+1/2*c)^6+1344*C*El
lipticE(cos(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)^2)^(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))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(
1/2)*sin(1/2*d*x+1/2*c)^6-85*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))+4502*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-2570*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*
c)^4+503*C*cos(1/2*d*x+1/2*c)*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+3010*A*cos(1
/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-1470*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-168*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))-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))-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))-147*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si
n(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-2016*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4+1008*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+280*A*co
s(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^10+427*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-2100*A*EllipticF(cos(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)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-1764*B*
EllipticE(cos(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)^2)^(1/2)*sin(1/2*d*
x+1/2*c)^4-1680*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)^4+840*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)^2+1050*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)^2+510*C*EllipticF(cos(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)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+4438*B*cos(1/2*d*x+1/2*c)*sin
(1/2*d*x+1/2*c)^6+882*B*EllipticE(cos(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)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2-1020*C*EllipticF(cos(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)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*
cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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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(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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mupad [B]  time = 9.53, size = 560, normalized size = 2.07 \[ \frac {2\,\left (12\,A\,a^4\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+19\,A\,a^4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+A\,a^4\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )\right )}{3\,d}+\frac {2\,\left (B\,a^4\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+4\,B\,a^4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{d}+\frac {2\,\left (\frac {34\,B\,a^4\,\sin \left (c+d\,x\right )}{\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {B\,a^4\,\sin \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5\,d}+\frac {2\,C\,a^4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {8\,A\,a^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,A\,a^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {8\,B\,a^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {8\,B\,a^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {7}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{15\,d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {8\,C\,a^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {4\,C\,a^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {8\,C\,a^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,C\,a^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{4},\frac {1}{2};\ -\frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^(3/2)*(a + a/cos(c + d*x))^4*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)

[Out]

(2*(12*A*a^4*ellipticE(c/2 + (d*x)/2, 2) + 19*A*a^4*ellipticF(c/2 + (d*x)/2, 2) + A*a^4*cos(c + d*x)^(1/2)*sin
(c + d*x)))/(3*d) + (2*(B*a^4*ellipticE(c/2 + (d*x)/2, 2) + 4*B*a^4*ellipticF(c/2 + (d*x)/2, 2)))/d + (2*((34*
B*a^4*sin(c + d*x))/(cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2)) + (B*a^4*sin(c + d*x))/(cos(c + d*x)^(5/2)*(si
n(c + d*x)^2)^(1/2)))*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(5*d) + (2*C*a^4*ellipticF(c/2 + (d*x)/2, 2
))/d + (8*A*a^4*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^
2)^(1/2)) + (2*A*a^4*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(3*d*cos(c + d*x)^(3/2)*(sin(c
+ d*x)^2)^(1/2)) + (8*B*a^4*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(3*d*cos(c + d*x)^(3/2)*
(sin(c + d*x)^2)^(1/2)) - (8*B*a^4*sin(c + d*x)*hypergeom([-1/4, 1/2], 7/4, cos(c + d*x)^2))/(15*d*cos(c + d*x
)^(1/2)*(sin(c + d*x)^2)^(1/2)) + (8*C*a^4*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c
+ d*x)^(1/2)*(sin(c + d*x)^2)^(1/2)) + (4*C*a^4*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(d*c
os(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2)) + (8*C*a^4*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2)
)/(5*d*cos(c + d*x)^(5/2)*(sin(c + d*x)^2)^(1/2)) + (2*C*a^4*sin(c + d*x)*hypergeom([-7/4, 1/2], -3/4, cos(c +
d*x)^2))/(7*d*cos(c + d*x)^(7/2)*(sin(c + d*x)^2)^(1/2))

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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(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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