3.1484 \(\int \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx\)

Optimal. Leaf size=103 \[ \frac {3 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a \tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3 a \tan (c+d x) \sec (c+d x)}{8 d}+\frac {b \tan ^4(c+d x)}{4 d}-\frac {b \tan ^2(c+d x)}{2 d}-\frac {b \log (\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2834, 2611, 3770, 3473, 3475} \[ \frac {3 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a \tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3 a \tan (c+d x) \sec (c+d x)}{8 d}+\frac {b \tan ^4(c+d x)}{4 d}-\frac {b \tan ^2(c+d x)}{2 d}-\frac {b \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]*(a + b*Sin[c + d*x])*Tan[c + d*x]^4,x]

[Out]

(3*a*ArcTanh[Sin[c + d*x]])/(8*d) - (b*Log[Cos[c + d*x]])/d - (3*a*Sec[c + d*x]*Tan[c + d*x])/(8*d) - (b*Tan[c
+ d*x]^2)/(2*d) + (a*Sec[c + d*x]*Tan[c + d*x]^3)/(4*d) + (b*Tan[c + d*x]^4)/(4*d)

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2834

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]),
x_Symbol] :> Dist[a, Int[Cos[e + f*x]^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[Cos[e + f*x]^p*(d*Sin[e +
f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2] && IntegerQ[n] && ((LtQ[p, 0]
&& NeQ[a^2 - b^2, 0]) || LtQ[0, n, p - 1] || LtQ[p + 1, -n, 2*p + 1])

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3475

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

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 \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx &=a \int \sec (c+d x) \tan ^4(c+d x) \, dx+b \int \tan ^5(c+d x) \, dx\\ &=\frac {a \sec (c+d x) \tan ^3(c+d x)}{4 d}+\frac {b \tan ^4(c+d x)}{4 d}-\frac {1}{4} (3 a) \int \sec (c+d x) \tan ^2(c+d x) \, dx-b \int \tan ^3(c+d x) \, dx\\ &=-\frac {3 a \sec (c+d x) \tan (c+d x)}{8 d}-\frac {b \tan ^2(c+d x)}{2 d}+\frac {a \sec (c+d x) \tan ^3(c+d x)}{4 d}+\frac {b \tan ^4(c+d x)}{4 d}+\frac {1}{8} (3 a) \int \sec (c+d x) \, dx+b \int \tan (c+d x) \, dx\\ &=\frac {3 a \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac {b \log (\cos (c+d x))}{d}-\frac {3 a \sec (c+d x) \tan (c+d x)}{8 d}-\frac {b \tan ^2(c+d x)}{2 d}+\frac {a \sec (c+d x) \tan ^3(c+d x)}{4 d}+\frac {b \tan ^4(c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.31, size = 106, normalized size = 1.03 \[ \frac {a \tan ^3(c+d x) \sec (c+d x)}{d}-\frac {a \left (6 \tan (c+d x) \sec ^3(c+d x)-3 \left (\tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x)\right )\right )}{8 d}-\frac {b \left (-\tan ^4(c+d x)+2 \tan ^2(c+d x)+4 \log (\cos (c+d x))\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]*(a + b*Sin[c + d*x])*Tan[c + d*x]^4,x]

[Out]

(a*Sec[c + d*x]*Tan[c + d*x]^3)/d - (b*(4*Log[Cos[c + d*x]] + 2*Tan[c + d*x]^2 - Tan[c + d*x]^4))/(4*d) - (a*(
6*Sec[c + d*x]^3*Tan[c + d*x] - 3*(ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*Tan[c + d*x])))/(8*d)

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fricas [A]  time = 0.45, size = 104, normalized size = 1.01 \[ \frac {{\left (3 \, a - 8 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, a + 8 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 16 \, b \cos \left (d x + c\right )^{2} - 2 \, {\left (5 \, a \cos \left (d x + c\right )^{2} - 2 \, a\right )} \sin \left (d x + c\right ) + 4 \, b}{16 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^4*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/16*((3*a - 8*b)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (3*a + 8*b)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) - 1
6*b*cos(d*x + c)^2 - 2*(5*a*cos(d*x + c)^2 - 2*a)*sin(d*x + c) + 4*b)/(d*cos(d*x + c)^4)

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giac [A]  time = 0.29, size = 100, normalized size = 0.97 \[ \frac {{\left (3 \, a - 8 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (3 \, a + 8 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (6 \, b \sin \left (d x + c\right )^{4} + 5 \, a \sin \left (d x + c\right )^{3} - 4 \, b \sin \left (d x + c\right )^{2} - 3 \, a \sin \left (d x + c\right )\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^4*(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/16*((3*a - 8*b)*log(abs(sin(d*x + c) + 1)) - (3*a + 8*b)*log(abs(sin(d*x + c) - 1)) + 2*(6*b*sin(d*x + c)^4
+ 5*a*sin(d*x + c)^3 - 4*b*sin(d*x + c)^2 - 3*a*sin(d*x + c))/(sin(d*x + c)^2 - 1)^2)/d

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maple [A]  time = 0.24, size = 133, normalized size = 1.29 \[ \frac {a \left (\sin ^{5}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {a \left (\sin ^{5}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}-\frac {a \left (\sin ^{3}\left (d x +c \right )\right )}{8 d}-\frac {3 a \sin \left (d x +c \right )}{8 d}+\frac {3 a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {b \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}-\frac {b \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {b \ln \left (\cos \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*sin(d*x+c)^4*(a+b*sin(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.41, size = 100, normalized size = 0.97 \[ \frac {{\left (3 \, a - 8 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, a + 8 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + \frac {2 \, {\left (5 \, a \sin \left (d x + c\right )^{3} + 8 \, b \sin \left (d x + c\right )^{2} - 3 \, a \sin \left (d x + c\right ) - 6 \, b\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^4*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/16*((3*a - 8*b)*log(sin(d*x + c) + 1) - (3*a + 8*b)*log(sin(d*x + c) - 1) + 2*(5*a*sin(d*x + c)^3 + 8*b*sin(
d*x + c)^2 - 3*a*sin(d*x + c) - 6*b)/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d

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mupad [B]  time = 12.03, size = 221, normalized size = 2.15 \[ \frac {b\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-8\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {3\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (\frac {3\,a}{8}+b\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {3\,a}{8}-b\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)^4*(a + b*sin(c + d*x)))/cos(c + d*x)^5,x)

[Out]

(b*log(tan(c/2 + (d*x)/2)^2 + 1))/d - ((3*a*tan(c/2 + (d*x)/2))/4 - (11*a*tan(c/2 + (d*x)/2)^3)/4 - (11*a*tan(
c/2 + (d*x)/2)^5)/4 + (3*a*tan(c/2 + (d*x)/2)^7)/4 + 2*b*tan(c/2 + (d*x)/2)^2 - 8*b*tan(c/2 + (d*x)/2)^4 + 2*b
*tan(c/2 + (d*x)/2)^6)/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2
+ (d*x)/2)^8 + 1)) - (log(tan(c/2 + (d*x)/2) - 1)*((3*a)/8 + b))/d + (log(tan(c/2 + (d*x)/2) + 1)*((3*a)/8 - b
))/d

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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(sec(d*x+c)**5*sin(d*x+c)**4*(a+b*sin(d*x+c)),x)

[Out]

Timed out

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