3.5.79 \(\int (a+b \sec (c+d x))^4 \, dx\) [479]

Optimal. Leaf size=107 \[ a^4 x+\frac {2 a b \left (2 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b^2 \left (17 a^2+2 b^2\right ) \tan (c+d x)}{3 d}+\frac {4 a b^3 \sec (c+d x) \tan (c+d x)}{3 d}+\frac {b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d} \]

[Out]

a^4*x+2*a*b*(2*a^2+b^2)*arctanh(sin(d*x+c))/d+1/3*b^2*(17*a^2+2*b^2)*tan(d*x+c)/d+4/3*a*b^3*sec(d*x+c)*tan(d*x
+c)/d+1/3*b^2*(a+b*sec(d*x+c))^2*tan(d*x+c)/d

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Rubi [A]
time = 0.09, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3867, 4133, 3855, 3852, 8} \begin {gather*} a^4 x+\frac {b^2 \left (17 a^2+2 b^2\right ) \tan (c+d x)}{3 d}+\frac {2 a b \left (2 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {4 a b^3 \tan (c+d x) \sec (c+d x)}{3 d}+\frac {b^2 \tan (c+d x) (a+b \sec (c+d x))^2}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^4,x]

[Out]

a^4*x + (2*a*b*(2*a^2 + b^2)*ArcTanh[Sin[c + d*x]])/d + (b^2*(17*a^2 + 2*b^2)*Tan[c + d*x])/(3*d) + (4*a*b^3*S
ec[c + d*x]*Tan[c + d*x])/(3*d) + (b^2*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(3*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]

Rubi steps

\begin {align*} \int (a+b \sec (c+d x))^4 \, dx &=\frac {b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {1}{3} \int (a+b \sec (c+d x)) \left (3 a^3+b \left (9 a^2+2 b^2\right ) \sec (c+d x)+8 a b^2 \sec ^2(c+d x)\right ) \, dx\\ &=\frac {4 a b^3 \sec (c+d x) \tan (c+d x)}{3 d}+\frac {b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {1}{6} \int \left (6 a^4+12 a b \left (2 a^2+b^2\right ) \sec (c+d x)+2 b^2 \left (17 a^2+2 b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=a^4 x+\frac {4 a b^3 \sec (c+d x) \tan (c+d x)}{3 d}+\frac {b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\left (2 a b \left (2 a^2+b^2\right )\right ) \int \sec (c+d x) \, dx+\frac {1}{3} \left (b^2 \left (17 a^2+2 b^2\right )\right ) \int \sec ^2(c+d x) \, dx\\ &=a^4 x+\frac {2 a b \left (2 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {4 a b^3 \sec (c+d x) \tan (c+d x)}{3 d}+\frac {b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}-\frac {\left (b^2 \left (17 a^2+2 b^2\right )\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=a^4 x+\frac {2 a b \left (2 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b^2 \left (17 a^2+2 b^2\right ) \tan (c+d x)}{3 d}+\frac {4 a b^3 \sec (c+d x) \tan (c+d x)}{3 d}+\frac {b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]
time = 0.32, size = 77, normalized size = 0.72 \begin {gather*} \frac {3 a^4 d x+6 a b \left (2 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))+3 b^2 \left (6 a^2+b^2+2 a b \sec (c+d x)\right ) \tan (c+d x)+b^4 \tan ^3(c+d x)}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x])^4,x]

[Out]

(3*a^4*d*x + 6*a*b*(2*a^2 + b^2)*ArcTanh[Sin[c + d*x]] + 3*b^2*(6*a^2 + b^2 + 2*a*b*Sec[c + d*x])*Tan[c + d*x]
+ b^4*Tan[c + d*x]^3)/(3*d)

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Maple [A]
time = 0.08, size = 109, normalized size = 1.02

method	result	size
derivativedivides	\(\frac {a^{4} \left (d x +c \right )+4 b \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 b^{2} a^{2} \tan \left (d x +c \right )+4 b^{3} a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-b^{4} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(109\)
default	\(\frac {a^{4} \left (d x +c \right )+4 b \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 b^{2} a^{2} \tan \left (d x +c \right )+4 b^{3} a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-b^{4} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(109\)
risch	\(a^{4} x -\frac {4 i b^{2} \left (3 b a \,{\mathrm e}^{5 i \left (d x +c \right )}-9 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-18 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 b a \,{\mathrm e}^{i \left (d x +c \right )}-9 a^{2}-b^{2}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {4 b \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {2 b^{3} a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {4 b \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {2 b^{3} a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}\)	\(196\)
norman	\(\frac {a^{4} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a^{4} x +3 a^{4} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 a^{4} x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {4 b^{2} \left (18 a^{2}+b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 b^{2} \left (6 a^{2}-2 b a +b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 b^{2} \left (6 a^{2}+2 b a +b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {2 b a \left (2 a^{2}+b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {2 b a \left (2 a^{2}+b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\)	\(221\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^4*(d*x+c)+4*b*a^3*ln(sec(d*x+c)+tan(d*x+c))+6*b^2*a^2*tan(d*x+c)+4*b^3*a*(1/2*sec(d*x+c)*tan(d*x+c)+1/2
*ln(sec(d*x+c)+tan(d*x+c)))-b^4*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c))

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Maxima [A]
time = 0.26, size = 121, normalized size = 1.13 \begin {gather*} a^{4} x + \frac {{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} b^{4}}{3 \, d} - \frac {a 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 )}}{d} + \frac {4 \, a^{3} b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right )}{d} + \frac {6 \, a^{2} 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))^4,x, algorithm="maxima")

[Out]

a^4*x + 1/3*(tan(d*x + c)^3 + 3*tan(d*x + c))*b^4/d - a*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 + 4*a^3*b*log(sec(d*x + c) + tan(d*x + c))/d + 6*a^2*b^2*tan(d*x + c)/d

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Fricas [A]
time = 2.47, size = 138, normalized size = 1.29 \begin {gather*} \frac {3 \, a^{4} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (6 \, a b^{3} \cos \left (d x + c\right ) + b^{4} + 2 \, {\left (9 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

1/3*(3*a^4*d*x*cos(d*x + c)^3 + 3*(2*a^3*b + a*b^3)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(2*a^3*b + a*b^3)
*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + (6*a*b^3*cos(d*x + c) + b^4 + 2*(9*a^2*b^2 + b^4)*cos(d*x + c)^2)*sin
(d*x + c))/(d*cos(d*x + c)^3)

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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 )^{4}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))**4,x)

[Out]

Integral((a + b*sec(c + d*x))**4, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (101) = 202\).
time = 0.46, size = 221, normalized size = 2.07 \begin {gather*} \frac {3 \, {\left (d x + c\right )} a^{4} + 6 \, {\left (2 \, a^{3} b + a b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 6 \, {\left (2 \, a^{3} b + a b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (18 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 36 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, b^{4} \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 )}^{3}}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^4,x, algorithm="giac")

[Out]

1/3*(3*(d*x + c)*a^4 + 6*(2*a^3*b + a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 6*(2*a^3*b + a*b^3)*log(abs(ta
n(1/2*d*x + 1/2*c) - 1)) - 2*(18*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 6*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 3*b^4*tan(1
/2*d*x + 1/2*c)^5 - 36*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 2*b^4*tan(1/2*d*x + 1/2*c)^3 + 18*a^2*b^2*tan(1/2*d*x
+ 1/2*c) + 6*a*b^3*tan(1/2*d*x + 1/2*c) + 3*b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d

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Mupad [B]
time = 1.03, size = 185, normalized size = 1.73 \begin {gather*} \frac {2\,a^4\,\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 {2\,b^4\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {b^4\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3}+\frac {4\,a\,b^3\,\mathrm {atanh}\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 {8\,a^3\,b\,\mathrm {atanh}\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 {2\,a\,b^3\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^2}+\frac {6\,a^2\,b^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(c + d*x))^4,x)

[Out]

(2*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*b^4*sin(c + d*x))/(3*d*cos(c + d*x)) + (b^4*sin(c +
d*x))/(3*d*cos(c + d*x)^3) + (4*a*b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (8*a^3*b*atanh(sin(c/
2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*a*b^3*sin(c + d*x))/(d*cos(c + d*x)^2) + (6*a^2*b^2*sin(c + d*x))/(d*
cos(c + d*x))

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