3.1483 \(\int (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx\)

Optimal. Leaf size=116 \[ -\frac{(8 a+15 b) \log (1-\sin (c+d x))}{16 d}-\frac{(8 a-15 b) \log (\sin (c+d x)+1)}{16 d}+\frac{\tan ^4(c+d x) (a+b \sin (c+d x))}{4 d}-\frac{\tan ^2(c+d x) (4 a+5 b \sin (c+d x))}{8 d}-\frac{15 b \sin (c+d x)}{8 d} \]

[Out]

-((8*a + 15*b)*Log[1 - Sin[c + d*x]])/(16*d) - ((8*a - 15*b)*Log[1 + Sin[c + d*x]])/(16*d) - (15*b*Sin[c + d*x
])/(8*d) - ((4*a + 5*b*Sin[c + d*x])*Tan[c + d*x]^2)/(8*d) + ((a + b*Sin[c + d*x])*Tan[c + d*x]^4)/(4*d)

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Rubi [A]  time = 0.108635, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {2721, 819, 774, 633, 31} \[ -\frac{(8 a+15 b) \log (1-\sin (c+d x))}{16 d}-\frac{(8 a-15 b) \log (\sin (c+d x)+1)}{16 d}+\frac{\tan ^4(c+d x) (a+b \sin (c+d x))}{4 d}-\frac{\tan ^2(c+d x) (4 a+5 b \sin (c+d x))}{8 d}-\frac{15 b \sin (c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((8*a + 15*b)*Log[1 - Sin[c + d*x]])/(16*d) - ((8*a - 15*b)*Log[1 + Sin[c + d*x]])/(16*d) - (15*b*Sin[c + d*x
])/(8*d) - ((4*a + 5*b*Sin[c + d*x])*Tan[c + d*x]^2)/(8*d) + ((a + b*Sin[c + d*x])*Tan[c + d*x]^4)/(4*d)

Rule 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rule 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 774

Int[(((d_.) + (e_.)*(x_))*((f_) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*g*x)/c, x] + Dist[1
/c, Int[(c*d*f - a*e*g + c*(e*f + d*g)*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x]

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^5 (a+x)}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{(a+b \sin (c+d x)) \tan ^4(c+d x)}{4 d}-\frac{\operatorname{Subst}\left (\int \frac{x^3 \left (4 a b^2+5 b^2 x\right )}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 b^2 d}\\ &=-\frac{(4 a+5 b \sin (c+d x)) \tan ^2(c+d x)}{8 d}+\frac{(a+b \sin (c+d x)) \tan ^4(c+d x)}{4 d}+\frac{\operatorname{Subst}\left (\int \frac{x \left (8 a b^4+15 b^4 x\right )}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^4 d}\\ &=-\frac{15 b \sin (c+d x)}{8 d}-\frac{(4 a+5 b \sin (c+d x)) \tan ^2(c+d x)}{8 d}+\frac{(a+b \sin (c+d x)) \tan ^4(c+d x)}{4 d}-\frac{\operatorname{Subst}\left (\int \frac{-15 b^6-8 a b^4 x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^4 d}\\ &=-\frac{15 b \sin (c+d x)}{8 d}-\frac{(4 a+5 b \sin (c+d x)) \tan ^2(c+d x)}{8 d}+\frac{(a+b \sin (c+d x)) \tan ^4(c+d x)}{4 d}+\frac{(8 a-15 b) \operatorname{Subst}\left (\int \frac{1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac{(8 a+15 b) \operatorname{Subst}\left (\int \frac{1}{b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}\\ &=-\frac{(8 a+15 b) \log (1-\sin (c+d x))}{16 d}-\frac{(8 a-15 b) \log (1+\sin (c+d x))}{16 d}-\frac{15 b \sin (c+d x)}{8 d}-\frac{(4 a+5 b \sin (c+d x)) \tan ^2(c+d x)}{8 d}+\frac{(a+b \sin (c+d x)) \tan ^4(c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.304658, size = 123, normalized size = 1.06 \[ -\frac{a \left (-\tan ^4(c+d x)+2 \tan ^2(c+d x)+4 \log (\cos (c+d x))\right )}{4 d}-\frac{b \sin (c+d x) \tan ^4(c+d x)}{d}-\frac{5 b \left (6 \tan (c+d x) \sec ^3(c+d x)-8 \tan ^3(c+d x) \sec (c+d x)-3 \left (\tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x)\right )\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.047, size = 147, normalized size = 1.3 \begin{align*}{\frac{a \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{a \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{a\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{b \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{3\,b \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,b \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{8\,d}}-{\frac{5\,b \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d}}-{\frac{15\,b\sin \left ( dx+c \right ) }{8\,d}}+{\frac{15\,b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*tan(d*x+c)^4/d-1/2*a*tan(d*x+c)^2/d-a*ln(cos(d*x+c))/d+1/4/d*b*sin(d*x+c)^7/cos(d*x+c)^4-3/8/d*b*sin(d*x
+c)^7/cos(d*x+c)^2-3/8*b*sin(d*x+c)^5/d-5/8*b*sin(d*x+c)^3/d-15/8*b*sin(d*x+c)/d+15/8/d*b*ln(sec(d*x+c)+tan(d*
x+c))

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Maxima [A]  time = 0.967047, size = 146, normalized size = 1.26 \begin{align*} -\frac{{\left (8 \, a - 15 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (8 \, a + 15 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 16 \, b \sin \left (d x + c\right ) - \frac{2 \,{\left (9 \, b \sin \left (d x + c\right )^{3} + 8 \, a \sin \left (d x + c\right )^{2} - 7 \, b \sin \left (d x + c\right ) - 6 \, a\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*((8*a - 15*b)*log(sin(d*x + c) + 1) + (8*a + 15*b)*log(sin(d*x + c) - 1) + 16*b*sin(d*x + c) - 2*(9*b*si
n(d*x + c)^3 + 8*a*sin(d*x + c)^2 - 7*b*sin(d*x + c) - 6*a)/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d

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Fricas [A]  time = 2.09253, size = 302, normalized size = 2.6 \begin{align*} -\frac{{\left (8 \, a - 15 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (8 \, a + 15 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 16 \, a \cos \left (d x + c\right )^{2} + 2 \,{\left (8 \, b \cos \left (d x + c\right )^{4} + 9 \, b \cos \left (d x + c\right )^{2} - 2 \, b\right )} \sin \left (d x + c\right ) - 4 \, a}{16 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*((8*a - 15*b)*cos(d*x + c)^4*log(sin(d*x + c) + 1) + (8*a + 15*b)*cos(d*x + c)^4*log(-sin(d*x + c) + 1)
+ 16*a*cos(d*x + c)^2 + 2*(8*b*cos(d*x + c)^4 + 9*b*cos(d*x + c)^2 - 2*b)*sin(d*x + c) - 4*a)/(d*cos(d*x + c)^
4)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.22834, size = 146, normalized size = 1.26 \begin{align*} -\frac{{\left (8 \, a - 15 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) +{\left (8 \, a + 15 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + 16 \, b \sin \left (d x + c\right ) - \frac{2 \,{\left (6 \, a \sin \left (d x + c\right )^{4} + 9 \, b \sin \left (d x + c\right )^{3} - 4 \, a \sin \left (d x + c\right )^{2} - 7 \, b \sin \left (d x + c\right )\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*((8*a - 15*b)*log(abs(sin(d*x + c) + 1)) + (8*a + 15*b)*log(abs(sin(d*x + c) - 1)) + 16*b*sin(d*x + c) -
2*(6*a*sin(d*x + c)^4 + 9*b*sin(d*x + c)^3 - 4*a*sin(d*x + c)^2 - 7*b*sin(d*x + c))/(sin(d*x + c)^2 - 1)^2)/d