3.5.86 \(\int (a+b \sec (c+d x))^5 \, dx\) [486]

Optimal. Leaf size=158 \[ a^5 x+\frac {b \left (40 a^4+40 a^2 b^2+3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a b^2 \left (53 a^2+20 b^2\right ) \tan (c+d x)}{6 d}+\frac {b^3 \left (58 a^2+9 b^2\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {11 a b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac {b^2 (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d} \]

[Out]

a^5*x+1/8*b*(40*a^4+40*a^2*b^2+3*b^4)*arctanh(sin(d*x+c))/d+1/6*a*b^2*(53*a^2+20*b^2)*tan(d*x+c)/d+1/24*b^3*(5
8*a^2+9*b^2)*sec(d*x+c)*tan(d*x+c)/d+11/12*a*b^2*(a+b*sec(d*x+c))^2*tan(d*x+c)/d+1/4*b^2*(a+b*sec(d*x+c))^3*ta
n(d*x+c)/d

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Rubi [A]
time = 0.17, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3867, 4141, 4133, 3855, 3852, 8} \begin {gather*} a^5 x+\frac {a b^2 \left (53 a^2+20 b^2\right ) \tan (c+d x)}{6 d}+\frac {b^3 \left (58 a^2+9 b^2\right ) \tan (c+d x) \sec (c+d x)}{24 d}+\frac {b \left (40 a^4+40 a^2 b^2+3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {11 a b^2 \tan (c+d x) (a+b \sec (c+d x))^2}{12 d}+\frac {b^2 \tan (c+d x) (a+b \sec (c+d x))^3}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^5,x]

[Out]

a^5*x + (b*(40*a^4 + 40*a^2*b^2 + 3*b^4)*ArcTanh[Sin[c + d*x]])/(8*d) + (a*b^2*(53*a^2 + 20*b^2)*Tan[c + d*x])
/(6*d) + (b^3*(58*a^2 + 9*b^2)*Sec[c + d*x]*Tan[c + d*x])/(24*d) + (11*a*b^2*(a + b*Sec[c + d*x])^2*Tan[c + d*
x])/(12*d) + (b^2*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(4*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3855

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

Rule 3867

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[(-b^2)*Cot[c + d*x]*((a + b*Csc[c + d*x])^(
n - 2)/(d*(n - 1))), x] + Dist[1/(n - 1), Int[(a + b*Csc[c + d*x])^(n - 3)*Simp[a^3*(n - 1) + (b*(b^2*(n - 2)
+ 3*a^2*(n - 1)))*Csc[c + d*x] + (a*b^2*(3*n - 4))*Csc[c + d*x]^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ
[a^2 - b^2, 0] && GtQ[n, 2] && IntegerQ[2*n]

Rule 4133

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_)), x_Symbol] :> Simp[(-b)*C*Csc[e + f*x]*(Cot[e + f*x]/(2*f)), x] + Dist[1/2, Int[Simp[2*A*a + (2*B*a + b
*(2*A + C))*Csc[e + f*x] + 2*(a*C + B*b)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x]

Rule 4141

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_))^(m_.), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[1/(m + 1), I
nt[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*A*(m + 1) + ((A*b + a*B)*(m + 1) + b*C*m)*Csc[e + f*x] + (b*B*(m + 1) +
a*C*m)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && IGtQ[2*m, 0]

Rubi steps

\begin {align*} \int (a+b \sec (c+d x))^5 \, dx &=\frac {b^2 (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{4} \int (a+b \sec (c+d x))^2 \left (4 a^3+3 b \left (4 a^2+b^2\right ) \sec (c+d x)+11 a b^2 \sec ^2(c+d x)\right ) \, dx\\ &=\frac {11 a b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac {b^2 (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{12} \int (a+b \sec (c+d x)) \left (12 a^4+a b \left (48 a^2+31 b^2\right ) \sec (c+d x)+b^2 \left (58 a^2+9 b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {b^3 \left (58 a^2+9 b^2\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {11 a b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac {b^2 (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{24} \int \left (24 a^5+3 b \left (40 a^4+40 a^2 b^2+3 b^4\right ) \sec (c+d x)+4 a b^2 \left (53 a^2+20 b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=a^5 x+\frac {b^3 \left (58 a^2+9 b^2\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {11 a b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac {b^2 (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{6} \left (a b^2 \left (53 a^2+20 b^2\right )\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{8} \left (b \left (40 a^4+40 a^2 b^2+3 b^4\right )\right ) \int \sec (c+d x) \, dx\\ &=a^5 x+\frac {b \left (40 a^4+40 a^2 b^2+3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b^3 \left (58 a^2+9 b^2\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {11 a b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac {b^2 (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac {\left (a b^2 \left (53 a^2+20 b^2\right )\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 d}\\ &=a^5 x+\frac {b \left (40 a^4+40 a^2 b^2+3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a b^2 \left (53 a^2+20 b^2\right ) \tan (c+d x)}{6 d}+\frac {b^3 \left (58 a^2+9 b^2\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {11 a b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac {b^2 (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]
time = 0.62, size = 114, normalized size = 0.72 \begin {gather*} \frac {24 a^5 d x+3 b \left (40 a^4+40 a^2 b^2+3 b^4\right ) \tanh ^{-1}(\sin (c+d x))+3 b^2 \left (40 a \left (2 a^2+b^2\right )+b \left (40 a^2+3 b^2\right ) \sec (c+d x)+2 b^3 \sec ^3(c+d x)\right ) \tan (c+d x)+40 a b^4 \tan ^3(c+d x)}{24 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x])^5,x]

[Out]

(24*a^5*d*x + 3*b*(40*a^4 + 40*a^2*b^2 + 3*b^4)*ArcTanh[Sin[c + d*x]] + 3*b^2*(40*a*(2*a^2 + b^2) + b*(40*a^2
+ 3*b^2)*Sec[c + d*x] + 2*b^3*Sec[c + d*x]^3)*Tan[c + d*x] + 40*a*b^4*Tan[c + d*x]^3)/(24*d)

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Maple [A]
time = 0.10, size = 160, normalized size = 1.01

method	result	size
derivativedivides	\(\frac {a^{5} \left (d x +c \right )+5 b \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+10 b^{2} a^{3} \tan \left (d x +c \right )+10 b^{3} a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-5 b^{4} a \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+b^{5} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(160\)
default	\(\frac {a^{5} \left (d x +c \right )+5 b \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+10 b^{2} a^{3} \tan \left (d x +c \right )+10 b^{3} a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-5 b^{4} a \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+b^{5} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(160\)
norman	\(\frac {a^{5} x +a^{5} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 a^{5} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 a^{5} x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 a^{5} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {5 b^{2} \left (16 a^{3}-8 b \,a^{2}+8 b^{2} a -b^{3}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {5 b^{2} \left (16 a^{3}+8 b \,a^{2}+8 b^{2} a +b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {b^{2} \left (720 a^{3}-120 b \,a^{2}+200 b^{2} a +9 b^{3}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}-\frac {b^{2} \left (720 a^{3}+120 b \,a^{2}+200 b^{2} a -9 b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {b \left (40 a^{4}+40 b^{2} a^{2}+3 b^{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {b \left (40 a^{4}+40 b^{2} a^{2}+3 b^{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(329\)
risch	\(a^{5} x -\frac {i b^{2} \left (120 b \,a^{2} {\mathrm e}^{7 i \left (d x +c \right )}+9 b^{3} {\mathrm e}^{7 i \left (d x +c \right )}-240 a^{3} {\mathrm e}^{6 i \left (d x +c \right )}+120 a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}+33 b^{3} {\mathrm e}^{5 i \left (d x +c \right )}-720 a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-240 a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-120 b \,a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-33 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-720 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-320 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-120 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}-9 b^{3} {\mathrm e}^{i \left (d x +c \right )}-240 a^{3}-80 b^{2} a \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {5 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{4}}{d}-\frac {5 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2}}{d}-\frac {3 b^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}+\frac {5 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{4}}{d}+\frac {5 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2}}{d}+\frac {3 b^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}\)	\(363\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^5,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^5*(d*x+c)+5*b*a^4*ln(sec(d*x+c)+tan(d*x+c))+10*b^2*a^3*tan(d*x+c)+10*b^3*a^2*(1/2*sec(d*x+c)*tan(d*x+c)
+1/2*ln(sec(d*x+c)+tan(d*x+c)))-5*b^4*a*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+b^5*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*
x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c))))

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Maxima [A]
time = 0.27, size = 198, normalized size = 1.25 \begin {gather*} a^{5} x + \frac {5 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a b^{4}}{3 \, d} - \frac {b^{5} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{16 \, d} - \frac {5 \, a^{2} b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{2 \, d} + \frac {5 \, a^{4} b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right )}{d} + \frac {10 \, a^{3} b^{2} \tan \left (d x + c\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^5,x, algorithm="maxima")

[Out]

a^5*x + 5/3*(tan(d*x + c)^3 + 3*tan(d*x + c))*a*b^4/d - 1/16*b^5*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d
*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1))/d - 5/2*a^2*b^3*(2*sin(
d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1))/d + 5*a^4*b*log(sec(d*x + c) +
tan(d*x + c))/d + 10*a^3*b^2*tan(d*x + c)/d

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Fricas [A]
time = 2.18, size = 183, normalized size = 1.16 \begin {gather*} \frac {48 \, a^{5} d x \cos \left (d x + c\right )^{4} + 3 \, {\left (40 \, a^{4} b + 40 \, a^{2} b^{3} + 3 \, b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (40 \, a^{4} b + 40 \, a^{2} b^{3} + 3 \, b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (40 \, a b^{4} \cos \left (d x + c\right ) + 6 \, b^{5} + 80 \, {\left (3 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (40 \, a^{2} b^{3} + 3 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^5,x, algorithm="fricas")

[Out]

1/48*(48*a^5*d*x*cos(d*x + c)^4 + 3*(40*a^4*b + 40*a^2*b^3 + 3*b^5)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(
40*a^4*b + 40*a^2*b^3 + 3*b^5)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(40*a*b^4*cos(d*x + c) + 6*b^5 + 80*(
3*a^3*b^2 + a*b^4)*cos(d*x + c)^3 + 3*(40*a^2*b^3 + 3*b^5)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sec {\left (c + d x \right )}\right )^{5}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))**5,x)

[Out]

Integral((a + b*sec(c + d*x))**5, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 380 vs. \(2 (148) = 296\).
time = 0.47, size = 380, normalized size = 2.41 \begin {gather*} \frac {24 \, {\left (d x + c\right )} a^{5} + 3 \, {\left (40 \, a^{4} b + 40 \, a^{2} b^{3} + 3 \, b^{5}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (40 \, a^{4} b + 40 \, a^{2} b^{3} + 3 \, b^{5}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (240 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 120 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 120 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 720 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 120 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 200 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 720 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 200 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 240 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 120 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 120 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^5,x, algorithm="giac")

[Out]

1/24*(24*(d*x + c)*a^5 + 3*(40*a^4*b + 40*a^2*b^3 + 3*b^5)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(40*a^4*b +
40*a^2*b^3 + 3*b^5)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(240*a^3*b^2*tan(1/2*d*x + 1/2*c)^7 - 120*a^2*b^3*t
an(1/2*d*x + 1/2*c)^7 + 120*a*b^4*tan(1/2*d*x + 1/2*c)^7 - 15*b^5*tan(1/2*d*x + 1/2*c)^7 - 720*a^3*b^2*tan(1/2
*d*x + 1/2*c)^5 + 120*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 - 200*a*b^4*tan(1/2*d*x + 1/2*c)^5 - 9*b^5*tan(1/2*d*x +
1/2*c)^5 + 720*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + 120*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 + 200*a*b^4*tan(1/2*d*x + 1
/2*c)^3 - 9*b^5*tan(1/2*d*x + 1/2*c)^3 - 240*a^3*b^2*tan(1/2*d*x + 1/2*c) - 120*a^2*b^3*tan(1/2*d*x + 1/2*c) -
120*a*b^4*tan(1/2*d*x + 1/2*c) - 15*b^5*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

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Mupad [B]
time = 1.36, size = 274, normalized size = 1.73 \begin {gather*} \frac {2\,a^5\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,b^5\,\sin \left (c+d\,x\right )}{8\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {b^5\,\sin \left (c+d\,x\right )}{4\,d\,{\cos \left (c+d\,x\right )}^4}+\frac {10\,a\,b^4\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {5\,a\,b^4\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3}+\frac {10\,a^3\,b^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {5\,a^2\,b^3\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^2}-\frac {b^5\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,3{}\mathrm {i}}{4\,d}-\frac {a^2\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,10{}\mathrm {i}}{d}-\frac {a^4\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,10{}\mathrm {i}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(c + d*x))^5,x)

[Out]

(2*a^5*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d - (b^5*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*
3i)/(4*d) + (3*b^5*sin(c + d*x))/(8*d*cos(c + d*x)^2) + (b^5*sin(c + d*x))/(4*d*cos(c + d*x)^4) - (a^2*b^3*ata
n((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*10i)/d - (a^4*b*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))
*10i)/d + (10*a*b^4*sin(c + d*x))/(3*d*cos(c + d*x)) + (5*a*b^4*sin(c + d*x))/(3*d*cos(c + d*x)^3) + (10*a^3*b
^2*sin(c + d*x))/(d*cos(c + d*x)) + (5*a^2*b^3*sin(c + d*x))/(d*cos(c + d*x)^2)

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