3.197 \(\int \frac{\sin ^5(c+d x)}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=152 \[ \frac{\left (2 a^2-b^2\right ) \cos ^3(c+d x)}{3 a^3 d}-\frac{b \left (2 a^2-b^2\right ) \cos ^2(c+d x)}{2 a^4 d}-\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{a^5 d}+\frac{b \left (a^2-b^2\right )^2 \log (a \cos (c+d x)+b)}{a^6 d}+\frac{b \cos ^4(c+d x)}{4 a^2 d}-\frac{\cos ^5(c+d x)}{5 a d} \]

[Out]

-(((a^2 - b^2)^2*Cos[c + d*x])/(a^5*d)) - (b*(2*a^2 - b^2)*Cos[c + d*x]^2)/(2*a^4*d) + ((2*a^2 - b^2)*Cos[c +
d*x]^3)/(3*a^3*d) + (b*Cos[c + d*x]^4)/(4*a^2*d) - Cos[c + d*x]^5/(5*a*d) + (b*(a^2 - b^2)^2*Log[b + a*Cos[c +
d*x]])/(a^6*d)

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Rubi [A]  time = 0.194132, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3872, 2837, 12, 772} \[ \frac{\left (2 a^2-b^2\right ) \cos ^3(c+d x)}{3 a^3 d}-\frac{b \left (2 a^2-b^2\right ) \cos ^2(c+d x)}{2 a^4 d}-\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{a^5 d}+\frac{b \left (a^2-b^2\right )^2 \log (a \cos (c+d x)+b)}{a^6 d}+\frac{b \cos ^4(c+d x)}{4 a^2 d}-\frac{\cos ^5(c+d x)}{5 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a^2 - b^2)^2*Cos[c + d*x])/(a^5*d)) - (b*(2*a^2 - b^2)*Cos[c + d*x]^2)/(2*a^4*d) + ((2*a^2 - b^2)*Cos[c +
d*x]^3)/(3*a^3*d) + (b*Cos[c + d*x]^4)/(4*a^2*d) - Cos[c + d*x]^5/(5*a*d) + (b*(a^2 - b^2)^2*Log[b + a*Cos[c +
d*x]])/(a^6*d)

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2837

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 772

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IGtQ[p, 0]

Rubi steps

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

Mathematica [A]  time = 0.362462, size = 172, normalized size = 1.13 \[ \frac{-40 a^3 b^2 \cos (3 (c+d x))-60 a \left (-14 a^2 b^2+5 a^4+8 b^4\right ) \cos (c+d x)-60 \left (3 a^4 b-2 a^2 b^3\right ) \cos (2 (c+d x))-960 a^2 b^3 \log (a \cos (c+d x)+b)+15 a^4 b \cos (4 (c+d x))+480 a^4 b \log (a \cos (c+d x)+b)+50 a^5 \cos (3 (c+d x))-6 a^5 \cos (5 (c+d x))+480 b^5 \log (a \cos (c+d x)+b)}{480 a^6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-60*a*(5*a^4 - 14*a^2*b^2 + 8*b^4)*Cos[c + d*x] - 60*(3*a^4*b - 2*a^2*b^3)*Cos[2*(c + d*x)] + 50*a^5*Cos[3*(c
+ d*x)] - 40*a^3*b^2*Cos[3*(c + d*x)] + 15*a^4*b*Cos[4*(c + d*x)] - 6*a^5*Cos[5*(c + d*x)] + 480*a^4*b*Log[b
+ a*Cos[c + d*x]] - 960*a^2*b^3*Log[b + a*Cos[c + d*x]] + 480*b^5*Log[b + a*Cos[c + d*x]])/(480*a^6*d)

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Maple [A]  time = 0.043, size = 216, normalized size = 1.4 \begin{align*} -{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5\,ad}}+{\frac{b \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{4\,{a}^{2}d}}+{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3\,ad}}-{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}{b}^{2}}{3\,d{a}^{3}}}-{\frac{b \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{{a}^{2}d}}+{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{2}{b}^{3}}{2\,d{a}^{4}}}-{\frac{\cos \left ( dx+c \right ) }{ad}}+2\,{\frac{{b}^{2}\cos \left ( dx+c \right ) }{d{a}^{3}}}-{\frac{{b}^{4}\cos \left ( dx+c \right ) }{d{a}^{5}}}+{\frac{b\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{{a}^{2}d}}-2\,{\frac{{b}^{3}\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d{a}^{4}}}+{\frac{{b}^{5}\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d{a}^{6}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*cos(d*x+c)^5/a/d+1/4*b*cos(d*x+c)^4/a^2/d+2/3*cos(d*x+c)^3/a/d-1/3/d/a^3*cos(d*x+c)^3*b^2-b*cos(d*x+c)^2/
a^2/d+1/2/d/a^4*cos(d*x+c)^2*b^3-cos(d*x+c)/a/d+2/d/a^3*b^2*cos(d*x+c)-1/d/a^5*b^4*cos(d*x+c)+b*ln(b+a*cos(d*x
+c))/a^2/d-2/d*b^3/a^4*ln(b+a*cos(d*x+c))+1/d*b^5/a^6*ln(b+a*cos(d*x+c))

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Maxima [A]  time = 1.06019, size = 190, normalized size = 1.25 \begin{align*} -\frac{\frac{12 \, a^{4} \cos \left (d x + c\right )^{5} - 15 \, a^{3} b \cos \left (d x + c\right )^{4} - 20 \,{\left (2 \, a^{4} - a^{2} b^{2}\right )} \cos \left (d x + c\right )^{3} + 30 \,{\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{2} + 60 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )}{a^{5}} - \frac{60 \,{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{6}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*((12*a^4*cos(d*x + c)^5 - 15*a^3*b*cos(d*x + c)^4 - 20*(2*a^4 - a^2*b^2)*cos(d*x + c)^3 + 30*(2*a^3*b -
a*b^3)*cos(d*x + c)^2 + 60*(a^4 - 2*a^2*b^2 + b^4)*cos(d*x + c))/a^5 - 60*(a^4*b - 2*a^2*b^3 + b^5)*log(a*cos(
d*x + c) + b)/a^6)/d

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Fricas [A]  time = 1.9067, size = 327, normalized size = 2.15 \begin{align*} -\frac{12 \, a^{5} \cos \left (d x + c\right )^{5} - 15 \, a^{4} b \cos \left (d x + c\right )^{4} - 20 \,{\left (2 \, a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )^{3} + 30 \,{\left (2 \, a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} + 60 \,{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right ) - 60 \,{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{60 \, a^{6} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(12*a^5*cos(d*x + c)^5 - 15*a^4*b*cos(d*x + c)^4 - 20*(2*a^5 - a^3*b^2)*cos(d*x + c)^3 + 30*(2*a^4*b - a
^2*b^3)*cos(d*x + c)^2 + 60*(a^5 - 2*a^3*b^2 + a*b^4)*cos(d*x + c) - 60*(a^4*b - 2*a^2*b^3 + b^5)*log(a*cos(d*
x + c) + b))/(a^6*d)

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

[Out]

Timed out

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Giac [B]  time = 1.31959, size = 1170, normalized size = 7.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(60*(a^5*b - a^4*b^2 - 2*a^3*b^3 + 2*a^2*b^4 + a*b^5 - b^6)*log(abs(a + b + a*(cos(d*x + c) - 1)/(cos(d*x
+ c) + 1) - b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1)))/(a^7 - a^6*b) - 60*(a^4*b - 2*a^2*b^3 + b^5)*log(abs(-(
cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/a^6 + (64*a^5 - 137*a^4*b - 200*a^3*b^2 + 274*a^2*b^3 + 120*a*b^4 -
137*b^5 - 320*a^5*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 805*a^4*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 8
80*a^3*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1490*a^2*b^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 480*a*
b^4*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 685*b^5*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 640*a^5*(cos(d*x +
c) - 1)^2/(cos(d*x + c) + 1)^2 - 1970*a^4*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 1280*a^3*b^2*(cos(d*x
+ c) - 1)^2/(cos(d*x + c) + 1)^2 + 3100*a^2*b^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 720*a*b^4*(cos(d*
x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 1370*b^5*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 1970*a^4*b*(cos(d*x
+ c) - 1)^3/(cos(d*x + c) + 1)^3 + 720*a^3*b^2*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 3100*a^2*b^3*(cos(d
*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 480*a*b^4*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 1370*b^5*(cos(d*x
+ c) - 1)^3/(cos(d*x + c) + 1)^3 - 805*a^4*b*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 - 120*a^3*b^2*(cos(d*x
+ c) - 1)^4/(cos(d*x + c) + 1)^4 + 1490*a^2*b^3*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 + 120*a*b^4*(cos(d*x
+ c) - 1)^4/(cos(d*x + c) + 1)^4 - 685*b^5*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 + 137*a^4*b*(cos(d*x + c
) - 1)^5/(cos(d*x + c) + 1)^5 - 274*a^2*b^3*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 + 137*b^5*(cos(d*x + c)
- 1)^5/(cos(d*x + c) + 1)^5)/(a^6*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)^5))/d