3.11 \(\int \sin ^5(c+d x) (a+b \tan (c+d x)) \, dx\)

Optimal. Leaf size=101 \[ -\frac {a \cos ^5(c+d x)}{5 d}+\frac {2 a \cos ^3(c+d x)}{3 d}-\frac {a \cos (c+d x)}{d}-\frac {b \sin ^5(c+d x)}{5 d}-\frac {b \sin ^3(c+d x)}{3 d}-\frac {b \sin (c+d x)}{d}+\frac {b \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

b*arctanh(sin(d*x+c))/d-a*cos(d*x+c)/d+2/3*a*cos(d*x+c)^3/d-1/5*a*cos(d*x+c)^5/d-b*sin(d*x+c)/d-1/3*b*sin(d*x+
c)^3/d-1/5*b*sin(d*x+c)^5/d

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Rubi [A]  time = 0.08, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3517, 2633, 2592, 302, 206} \[ -\frac {a \cos ^5(c+d x)}{5 d}+\frac {2 a \cos ^3(c+d x)}{3 d}-\frac {a \cos (c+d x)}{d}-\frac {b \sin ^5(c+d x)}{5 d}-\frac {b \sin ^3(c+d x)}{3 d}-\frac {b \sin (c+d x)}{d}+\frac {b \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^5*(a + b*Tan[c + d*x]),x]

[Out]

(b*ArcTanh[Sin[c + d*x]])/d - (a*Cos[c + d*x])/d + (2*a*Cos[c + d*x]^3)/(3*d) - (a*Cos[c + d*x]^5)/(5*d) - (b*
Sin[c + d*x])/d - (b*Sin[c + d*x]^3)/(3*d) - (b*Sin[c + d*x]^5)/(5*d)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 2633

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

Rule 3517

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[Expand[Sin[e
+ f*x]^m*(a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \sin ^5(c+d x) (a+b \tan (c+d x)) \, dx &=\int \left (a \sin ^5(c+d x)+b \sin ^5(c+d x) \tan (c+d x)\right ) \, dx\\ &=a \int \sin ^5(c+d x) \, dx+b \int \sin ^5(c+d x) \tan (c+d x) \, dx\\ &=-\frac {a \operatorname {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac {b \operatorname {Subst}\left (\int \frac {x^6}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac {a \cos (c+d x)}{d}+\frac {2 a \cos ^3(c+d x)}{3 d}-\frac {a \cos ^5(c+d x)}{5 d}+\frac {b \operatorname {Subst}\left (\int \left (-1-x^2-x^4+\frac {1}{1-x^2}\right ) \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac {a \cos (c+d x)}{d}+\frac {2 a \cos ^3(c+d x)}{3 d}-\frac {a \cos ^5(c+d x)}{5 d}-\frac {b \sin (c+d x)}{d}-\frac {b \sin ^3(c+d x)}{3 d}-\frac {b \sin ^5(c+d x)}{5 d}+\frac {b \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {a \cos (c+d x)}{d}+\frac {2 a \cos ^3(c+d x)}{3 d}-\frac {a \cos ^5(c+d x)}{5 d}-\frac {b \sin (c+d x)}{d}-\frac {b \sin ^3(c+d x)}{3 d}-\frac {b \sin ^5(c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 103, normalized size = 1.02 \[ -\frac {5 a \cos (c+d x)}{8 d}+\frac {5 a \cos (3 (c+d x))}{48 d}-\frac {a \cos (5 (c+d x))}{80 d}-\frac {b \sin ^5(c+d x)}{5 d}-\frac {b \sin ^3(c+d x)}{3 d}-\frac {b \sin (c+d x)}{d}+\frac {b \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^5*(a + b*Tan[c + d*x]),x]

[Out]

(b*ArcTanh[Sin[c + d*x]])/d - (5*a*Cos[c + d*x])/(8*d) + (5*a*Cos[3*(c + d*x)])/(48*d) - (a*Cos[5*(c + d*x)])/
(80*d) - (b*Sin[c + d*x])/d - (b*Sin[c + d*x]^3)/(3*d) - (b*Sin[c + d*x]^5)/(5*d)

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fricas [A]  time = 0.51, size = 97, normalized size = 0.96 \[ -\frac {6 \, a \cos \left (d x + c\right )^{5} - 20 \, a \cos \left (d x + c\right )^{3} + 30 \, a \cos \left (d x + c\right ) - 15 \, b \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, b \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (3 \, b \cos \left (d x + c\right )^{4} - 11 \, b \cos \left (d x + c\right )^{2} + 23 \, b\right )} \sin \left (d x + c\right )}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(6*a*cos(d*x + c)^5 - 20*a*cos(d*x + c)^3 + 30*a*cos(d*x + c) - 15*b*log(sin(d*x + c) + 1) + 15*b*log(-s
in(d*x + c) + 1) + 2*(3*b*cos(d*x + c)^4 - 11*b*cos(d*x + c)^2 + 23*b)*sin(d*x + c))/d

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giac [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(sin(d*x+c)^5*(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.32, size = 113, normalized size = 1.12 \[ -\frac {8 a \cos \left (d x +c \right )}{15 d}-\frac {a \cos \left (d x +c \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{5 d}-\frac {4 a \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{15 d}-\frac {b \left (\sin ^{5}\left (d x +c \right )\right )}{5 d}-\frac {b \left (\sin ^{3}\left (d x +c \right )\right )}{3 d}-\frac {b \sin \left (d x +c \right )}{d}+\frac {b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/15*a*cos(d*x+c)/d-1/5/d*a*cos(d*x+c)*sin(d*x+c)^4-4/15/d*a*cos(d*x+c)*sin(d*x+c)^2-1/5*b*sin(d*x+c)^5/d-1/3
*b*sin(d*x+c)^3/d-b*sin(d*x+c)/d+1/d*b*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.37, size = 91, normalized size = 0.90 \[ -\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 10 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )\right )} a + {\left (6 \, \sin \left (d x + c\right )^{5} + 10 \, \sin \left (d x + c\right )^{3} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 30 \, \sin \left (d x + c\right )\right )} b}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(2*(3*cos(d*x + c)^5 - 10*cos(d*x + c)^3 + 15*cos(d*x + c))*a + (6*sin(d*x + c)^5 + 10*sin(d*x + c)^3 -
15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1) + 30*sin(d*x + c))*b)/d

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mupad [B]  time = 3.80, size = 121, normalized size = 1.20 \[ \frac {2\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,a\,{\cos \left (c+d\,x\right )}^3}{3\,d}-\frac {a\,{\cos \left (c+d\,x\right )}^5}{5\,d}-\frac {a\,\cos \left (c+d\,x\right )}{d}-\frac {23\,b\,\sin \left (c+d\,x\right )}{15\,d}+\frac {11\,b\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{15\,d}-\frac {b\,{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{5\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*a*cos(c + d*x)^3)/(3*d) - (a*cos(c + d*x)^5)/(5*d) -
(a*cos(c + d*x))/d - (23*b*sin(c + d*x))/(15*d) + (11*b*cos(c + d*x)^2*sin(c + d*x))/(15*d) - (b*cos(c + d*x)
^4*sin(c + d*x))/(5*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan {\left (c + d x \right )}\right ) \sin ^{5}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*tan(c + d*x))*sin(c + d*x)**5, x)

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