3.188 \(\int (a+a \cos (x))^4 (A+B \sec (x)) \, dx\)

Optimal. Leaf size=104 \[ \frac {1}{8} a^4 x (35 A+48 B)+\frac {5}{8} a^4 (7 A+8 B) \sin (x)+\frac {1}{24} (35 A+32 B) \sin (x) \left (a^4 \cos (x)+a^4\right )+a^4 B \tanh ^{-1}(\sin (x))+\frac {1}{12} (7 A+4 B) \sin (x) \left (a^2 \cos (x)+a^2\right )^2+\frac {1}{4} a A \sin (x) (a \cos (x)+a)^3 \]

[Out]

1/8*a^4*(35*A+48*B)*x+a^4*B*arctanh(sin(x))+5/8*a^4*(7*A+8*B)*sin(x)+1/4*a*A*(a+a*cos(x))^3*sin(x)+1/12*(7*A+4
*B)*(a^2+a^2*cos(x))^2*sin(x)+1/24*(35*A+32*B)*(a^4+a^4*cos(x))*sin(x)

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Rubi [A]  time = 0.40, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2828, 2976, 2968, 3023, 2735, 3770} \[ \frac {1}{8} a^4 x (35 A+48 B)+\frac {5}{8} a^4 (7 A+8 B) \sin (x)+\frac {1}{12} (7 A+4 B) \sin (x) \left (a^2 \cos (x)+a^2\right )^2+\frac {1}{24} (35 A+32 B) \sin (x) \left (a^4 \cos (x)+a^4\right )+a^4 B \tanh ^{-1}(\sin (x))+\frac {1}{4} a A \sin (x) (a \cos (x)+a)^3 \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[x])^4*(A + B*Sec[x]),x]

[Out]

(a^4*(35*A + 48*B)*x)/8 + a^4*B*ArcTanh[Sin[x]] + (5*a^4*(7*A + 8*B)*Sin[x])/8 + (a*A*(a + a*Cos[x])^3*Sin[x])
/4 + ((7*A + 4*B)*(a^2 + a^2*Cos[x])^2*Sin[x])/12 + ((35*A + 32*B)*(a^4 + a^4*Cos[x])*Sin[x])/24

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2828

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> In
t[((a + b*Sin[e + f*x])^m*(d + c*Sin[e + f*x])^n)/Sin[e + f*x]^n, x] /; FreeQ[{a, b, c, d, e, f, m}, x] && Int
egerQ[n]

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 2976

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*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])
^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*S
in[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int (a+a \cos (x))^4 (A+B \sec (x)) \, dx &=\int (a+a \cos (x))^4 (B+A \cos (x)) \sec (x) \, dx\\ &=\frac {1}{4} a A (a+a \cos (x))^3 \sin (x)+\frac {1}{4} \int (a+a \cos (x))^3 (4 a B+a (7 A+4 B) \cos (x)) \sec (x) \, dx\\ &=\frac {1}{4} a A (a+a \cos (x))^3 \sin (x)+\frac {1}{12} (7 A+4 B) \left (a^2+a^2 \cos (x)\right )^2 \sin (x)+\frac {1}{12} \int (a+a \cos (x))^2 \left (12 a^2 B+a^2 (35 A+32 B) \cos (x)\right ) \sec (x) \, dx\\ &=\frac {1}{4} a A (a+a \cos (x))^3 \sin (x)+\frac {1}{12} (7 A+4 B) \left (a^2+a^2 \cos (x)\right )^2 \sin (x)+\frac {1}{24} (35 A+32 B) \left (a^4+a^4 \cos (x)\right ) \sin (x)+\frac {1}{24} \int (a+a \cos (x)) \left (24 a^3 B+15 a^3 (7 A+8 B) \cos (x)\right ) \sec (x) \, dx\\ &=\frac {1}{4} a A (a+a \cos (x))^3 \sin (x)+\frac {1}{12} (7 A+4 B) \left (a^2+a^2 \cos (x)\right )^2 \sin (x)+\frac {1}{24} (35 A+32 B) \left (a^4+a^4 \cos (x)\right ) \sin (x)+\frac {1}{24} \int \left (24 a^4 B+\left (24 a^4 B+15 a^4 (7 A+8 B)\right ) \cos (x)+15 a^4 (7 A+8 B) \cos ^2(x)\right ) \sec (x) \, dx\\ &=\frac {5}{8} a^4 (7 A+8 B) \sin (x)+\frac {1}{4} a A (a+a \cos (x))^3 \sin (x)+\frac {1}{12} (7 A+4 B) \left (a^2+a^2 \cos (x)\right )^2 \sin (x)+\frac {1}{24} (35 A+32 B) \left (a^4+a^4 \cos (x)\right ) \sin (x)+\frac {1}{24} \int \left (24 a^4 B+3 a^4 (35 A+48 B) \cos (x)\right ) \sec (x) \, dx\\ &=\frac {1}{8} a^4 (35 A+48 B) x+\frac {5}{8} a^4 (7 A+8 B) \sin (x)+\frac {1}{4} a A (a+a \cos (x))^3 \sin (x)+\frac {1}{12} (7 A+4 B) \left (a^2+a^2 \cos (x)\right )^2 \sin (x)+\frac {1}{24} (35 A+32 B) \left (a^4+a^4 \cos (x)\right ) \sin (x)+\left (a^4 B\right ) \int \sec (x) \, dx\\ &=\frac {1}{8} a^4 (35 A+48 B) x+a^4 B \tanh ^{-1}(\sin (x))+\frac {5}{8} a^4 (7 A+8 B) \sin (x)+\frac {1}{4} a A (a+a \cos (x))^3 \sin (x)+\frac {1}{12} (7 A+4 B) \left (a^2+a^2 \cos (x)\right )^2 \sin (x)+\frac {1}{24} (35 A+32 B) \left (a^4+a^4 \cos (x)\right ) \sin (x)\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 97, normalized size = 0.93 \[ \frac {1}{96} a^4 \left (24 (28 A+27 B) \sin (x)+24 (7 A+4 B) \sin (2 x)+420 A x+32 A \sin (3 x)+3 A \sin (4 x)+576 B x+8 B \sin (3 x)-96 B \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+96 B \log \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[x])^4*(A + B*Sec[x]),x]

[Out]

(a^4*(420*A*x + 576*B*x - 96*B*Log[Cos[x/2] - Sin[x/2]] + 96*B*Log[Cos[x/2] + Sin[x/2]] + 24*(28*A + 27*B)*Sin
[x] + 24*(7*A + 4*B)*Sin[2*x] + 32*A*Sin[3*x] + 8*B*Sin[3*x] + 3*A*Sin[4*x]))/96

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fricas [A]  time = 0.70, size = 89, normalized size = 0.86 \[ \frac {1}{8} \, {\left (35 \, A + 48 \, B\right )} a^{4} x + \frac {1}{2} \, B a^{4} \log \left (\sin \relax (x) + 1\right ) - \frac {1}{2} \, B a^{4} \log \left (-\sin \relax (x) + 1\right ) + \frac {1}{24} \, {\left (6 \, A a^{4} \cos \relax (x)^{3} + 8 \, {\left (4 \, A + B\right )} a^{4} \cos \relax (x)^{2} + 3 \, {\left (27 \, A + 16 \, B\right )} a^{4} \cos \relax (x) + 160 \, {\left (A + B\right )} a^{4}\right )} \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(x))^4*(A+B*sec(x)),x, algorithm="fricas")

[Out]

1/8*(35*A + 48*B)*a^4*x + 1/2*B*a^4*log(sin(x) + 1) - 1/2*B*a^4*log(-sin(x) + 1) + 1/24*(6*A*a^4*cos(x)^3 + 8*
(4*A + B)*a^4*cos(x)^2 + 3*(27*A + 16*B)*a^4*cos(x) + 160*(A + B)*a^4)*sin(x)

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giac [A]  time = 0.16, size = 149, normalized size = 1.43 \[ B a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right ) - B a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right ) + \frac {1}{8} \, {\left (35 \, A a^{4} + 48 \, B a^{4}\right )} x + \frac {105 \, A a^{4} \tan \left (\frac {1}{2} \, x\right )^{7} + 120 \, B a^{4} \tan \left (\frac {1}{2} \, x\right )^{7} + 385 \, A a^{4} \tan \left (\frac {1}{2} \, x\right )^{5} + 424 \, B a^{4} \tan \left (\frac {1}{2} \, x\right )^{5} + 511 \, A a^{4} \tan \left (\frac {1}{2} \, x\right )^{3} + 520 \, B a^{4} \tan \left (\frac {1}{2} \, x\right )^{3} + 279 \, A a^{4} \tan \left (\frac {1}{2} \, x\right ) + 216 \, B a^{4} \tan \left (\frac {1}{2} \, x\right )}{12 \, {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(x))^4*(A+B*sec(x)),x, algorithm="giac")

[Out]

B*a^4*log(abs(tan(1/2*x) + 1)) - B*a^4*log(abs(tan(1/2*x) - 1)) + 1/8*(35*A*a^4 + 48*B*a^4)*x + 1/12*(105*A*a^
4*tan(1/2*x)^7 + 120*B*a^4*tan(1/2*x)^7 + 385*A*a^4*tan(1/2*x)^5 + 424*B*a^4*tan(1/2*x)^5 + 511*A*a^4*tan(1/2*
x)^3 + 520*B*a^4*tan(1/2*x)^3 + 279*A*a^4*tan(1/2*x) + 216*B*a^4*tan(1/2*x))/(tan(1/2*x)^2 + 1)^4

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maple [A]  time = 0.12, size = 103, normalized size = 0.99 \[ \frac {A \,a^{4} \sin \relax (x ) \left (\cos ^{3}\relax (x )\right )}{4}+\frac {27 A \,a^{4} \sin \relax (x ) \cos \relax (x )}{8}+\frac {35 A \,a^{4} x}{8}+\frac {B \,a^{4} \left (2+\cos ^{2}\relax (x )\right ) \sin \relax (x )}{3}+\frac {4 A \,a^{4} \left (2+\cos ^{2}\relax (x )\right ) \sin \relax (x )}{3}+2 B \,a^{4} \sin \relax (x ) \cos \relax (x )+6 B \,a^{4} x +6 B \,a^{4} \sin \relax (x )+4 A \,a^{4} \sin \relax (x )+B \,a^{4} \ln \left (\sec \relax (x )+\tan \relax (x )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(x))^4*(A+B*sec(x)),x)

[Out]

1/4*A*a^4*sin(x)*cos(x)^3+27/8*A*a^4*sin(x)*cos(x)+35/8*A*a^4*x+1/3*B*a^4*(2+cos(x)^2)*sin(x)+4/3*A*a^4*(2+cos
(x)^2)*sin(x)+2*B*a^4*sin(x)*cos(x)+6*B*a^4*x+6*B*a^4*sin(x)+4*A*a^4*sin(x)+B*a^4*ln(sec(x)+tan(x))

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maxima [A]  time = 0.31, size = 118, normalized size = 1.13 \[ -\frac {4}{3} \, {\left (\sin \relax (x)^{3} - 3 \, \sin \relax (x)\right )} A a^{4} - \frac {1}{3} \, {\left (\sin \relax (x)^{3} - 3 \, \sin \relax (x)\right )} B a^{4} + \frac {1}{32} \, A a^{4} {\left (12 \, x + \sin \left (4 \, x\right ) + 8 \, \sin \left (2 \, x\right )\right )} + \frac {3}{2} \, A a^{4} {\left (2 \, x + \sin \left (2 \, x\right )\right )} + B a^{4} {\left (2 \, x + \sin \left (2 \, x\right )\right )} + A a^{4} x + 4 \, B a^{4} x + B a^{4} \log \left (\sec \relax (x) + \tan \relax (x)\right ) + 4 \, A a^{4} \sin \relax (x) + 6 \, B a^{4} \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(x))^4*(A+B*sec(x)),x, algorithm="maxima")

[Out]

-4/3*(sin(x)^3 - 3*sin(x))*A*a^4 - 1/3*(sin(x)^3 - 3*sin(x))*B*a^4 + 1/32*A*a^4*(12*x + sin(4*x) + 8*sin(2*x))
+ 3/2*A*a^4*(2*x + sin(2*x)) + B*a^4*(2*x + sin(2*x)) + A*a^4*x + 4*B*a^4*x + B*a^4*log(sec(x) + tan(x)) + 4*
A*a^4*sin(x) + 6*B*a^4*sin(x)

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mupad [B]  time = 2.52, size = 460, normalized size = 4.42 \[ \frac {\left (\frac {35\,A\,a^4}{4}+10\,B\,a^4\right )\,{\mathrm {tan}\left (\frac {x}{2}\right )}^7+\left (\frac {385\,A\,a^4}{12}+\frac {106\,B\,a^4}{3}\right )\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5+\left (\frac {511\,A\,a^4}{12}+\frac {130\,B\,a^4}{3}\right )\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+\left (\frac {93\,A\,a^4}{4}+18\,B\,a^4\right )\,\mathrm {tan}\left (\frac {x}{2}\right )}{{\mathrm {tan}\left (\frac {x}{2}\right )}^8+4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1}+\frac {a^4\,\mathrm {atan}\left (\frac {42875\,A^3\,a^{12}\,\mathrm {tan}\left (\frac {x}{2}\right )}{8\,\left (\frac {42875\,A^3\,a^{12}}{8}+22050\,A^2\,B\,a^{12}+30520\,A\,B^2\,a^{12}+14208\,B^3\,a^{12}\right )}+\frac {14208\,B^3\,a^{12}\,\mathrm {tan}\left (\frac {x}{2}\right )}{\frac {42875\,A^3\,a^{12}}{8}+22050\,A^2\,B\,a^{12}+30520\,A\,B^2\,a^{12}+14208\,B^3\,a^{12}}+\frac {30520\,A\,B^2\,a^{12}\,\mathrm {tan}\left (\frac {x}{2}\right )}{\frac {42875\,A^3\,a^{12}}{8}+22050\,A^2\,B\,a^{12}+30520\,A\,B^2\,a^{12}+14208\,B^3\,a^{12}}+\frac {22050\,A^2\,B\,a^{12}\,\mathrm {tan}\left (\frac {x}{2}\right )}{\frac {42875\,A^3\,a^{12}}{8}+22050\,A^2\,B\,a^{12}+30520\,A\,B^2\,a^{12}+14208\,B^3\,a^{12}}\right )\,\left (35\,A+48\,B\right )}{4}+2\,B\,a^4\,\mathrm {atanh}\left (\frac {2368\,B^3\,a^{12}\,\mathrm {tan}\left (\frac {x}{2}\right )}{1225\,A^2\,B\,a^{12}+3360\,A\,B^2\,a^{12}+2368\,B^3\,a^{12}}+\frac {3360\,A\,B^2\,a^{12}\,\mathrm {tan}\left (\frac {x}{2}\right )}{1225\,A^2\,B\,a^{12}+3360\,A\,B^2\,a^{12}+2368\,B^3\,a^{12}}+\frac {1225\,A^2\,B\,a^{12}\,\mathrm {tan}\left (\frac {x}{2}\right )}{1225\,A^2\,B\,a^{12}+3360\,A\,B^2\,a^{12}+2368\,B^3\,a^{12}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*cos(x))^4*(A + B/cos(x)),x)

[Out]

(tan(x/2)^7*((35*A*a^4)/4 + 10*B*a^4) + tan(x/2)^5*((385*A*a^4)/12 + (106*B*a^4)/3) + tan(x/2)^3*((511*A*a^4)/
12 + (130*B*a^4)/3) + tan(x/2)*((93*A*a^4)/4 + 18*B*a^4))/(4*tan(x/2)^2 + 6*tan(x/2)^4 + 4*tan(x/2)^6 + tan(x/
2)^8 + 1) + (a^4*atan((42875*A^3*a^12*tan(x/2))/(8*((42875*A^3*a^12)/8 + 14208*B^3*a^12 + 30520*A*B^2*a^12 + 2
2050*A^2*B*a^12)) + (14208*B^3*a^12*tan(x/2))/((42875*A^3*a^12)/8 + 14208*B^3*a^12 + 30520*A*B^2*a^12 + 22050*
A^2*B*a^12) + (30520*A*B^2*a^12*tan(x/2))/((42875*A^3*a^12)/8 + 14208*B^3*a^12 + 30520*A*B^2*a^12 + 22050*A^2*
B*a^12) + (22050*A^2*B*a^12*tan(x/2))/((42875*A^3*a^12)/8 + 14208*B^3*a^12 + 30520*A*B^2*a^12 + 22050*A^2*B*a^
12))*(35*A + 48*B))/4 + 2*B*a^4*atanh((2368*B^3*a^12*tan(x/2))/(2368*B^3*a^12 + 3360*A*B^2*a^12 + 1225*A^2*B*a
^12) + (3360*A*B^2*a^12*tan(x/2))/(2368*B^3*a^12 + 3360*A*B^2*a^12 + 1225*A^2*B*a^12) + (1225*A^2*B*a^12*tan(x
/2))/(2368*B^3*a^12 + 3360*A*B^2*a^12 + 1225*A^2*B*a^12))

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sympy [A]  time = 15.26, size = 116, normalized size = 1.12 \[ \frac {35 A a^{4} x}{8} - \frac {4 A a^{4} \sin ^{3}{\relax (x )}}{3} + 8 A a^{4} \sin {\relax (x )} + \frac {7 A a^{4} \sin {\left (2 x \right )}}{4} + \frac {A a^{4} \sin {\left (4 x \right )}}{32} + 6 B a^{4} x + B a^{4} \log {\left (\tan {\relax (x )} + \sec {\relax (x )} \right )} - \frac {B a^{4} \sin ^{3}{\relax (x )}}{3} + 2 B a^{4} \sin {\relax (x )} \cos {\relax (x )} + 7 B a^{4} \sin {\relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(x))**4*(A+B*sec(x)),x)

[Out]

35*A*a**4*x/8 - 4*A*a**4*sin(x)**3/3 + 8*A*a**4*sin(x) + 7*A*a**4*sin(2*x)/4 + A*a**4*sin(4*x)/32 + 6*B*a**4*x
+ B*a**4*log(tan(x) + sec(x)) - B*a**4*sin(x)**3/3 + 2*B*a**4*sin(x)*cos(x) + 7*B*a**4*sin(x)

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