3.405 \(\int \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=94 \[ \frac{3 a \left (a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{3 a \sec ^2(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))}{8 d}+\frac{\tan (c+d x) \sec ^3(c+d x) (a+b \sin (c+d x))^3}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.0800275, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2668, 729, 723, 206} \[ \frac{3 a \left (a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{3 a \sec ^2(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))}{8 d}+\frac{\tan (c+d x) \sec ^3(c+d x) (a+b \sin (c+d x))^3}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5*(a + b*Sin[c + d*x])^3,x]

[Out]

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

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(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] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 729

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^m*(2*c*x)*(a + c*x^2)^(
p + 1))/(4*a*c*(p + 1)), x] - Dist[(m*(2*c*d))/(4*a*c*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x],
x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 3, 0] && LtQ[p, -1]

Rule 723

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(a*e - c*d*x)*(a
+ c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] + Dist[((2*p + 3)*(c*d^2 + a*e^2))/(2*a*c*(p + 1)), Int[(d + e*x)^(m -
2)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 2, 0] && Lt
Q[p, -1]

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])

Rubi steps

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

Mathematica [B]  time = 4.11311, size = 318, normalized size = 3.38 \[ \frac{16 a^4 b \left (3 a^2-2 b^2\right ) \tan ^2(c+d x)+8 b^3 \left (-5 a^2 b^2+4 a^4+b^4\right ) \tan ^4(c+d x)-6 a \left (a^2-b^2\right )^3 (\log (1-\sin (c+d x))-\log (\sin (c+d x)+1))+a b \sec ^4(c+d x) \left (\left (-11 a^2 b^3+18 a^4 b+5 b^5\right ) \sin (3 (c+d x))+8 a^3 b^2-8 a^5\right )+16 a^2 b \sec ^2(c+d x) \left (\left (-5 a^2 b^2+2 a^4+3 b^4\right ) \tan ^2(c+d x)-a^4\right )+a \left (-22 a^4 b^2+29 a^2 b^4+8 a^6-3 b^6\right ) \tan (c+d x) \sec ^3(c+d x)+4 a \tan (c+d x) \sec (c+d x) \left (4 b^2 \left (-5 a^2 b^2+3 a^4+2 b^4\right ) \tan ^2(c+d x)+3 \left (a^6-5 a^4 b^2\right )\right )}{32 d \left (a^2-b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5*(a + b*Sin[c + d*x])^3,x]

[Out]

(-6*a*(a^2 - b^2)^3*(Log[1 - Sin[c + d*x]] - Log[1 + Sin[c + d*x]]) + a*b*Sec[c + d*x]^4*(-8*a^5 + 8*a^3*b^2 +
(18*a^4*b - 11*a^2*b^3 + 5*b^5)*Sin[3*(c + d*x)]) + a*(8*a^6 - 22*a^4*b^2 + 29*a^2*b^4 - 3*b^6)*Sec[c + d*x]^
3*Tan[c + d*x] + 16*a^4*b*(3*a^2 - 2*b^2)*Tan[c + d*x]^2 + 8*b^3*(4*a^4 - 5*a^2*b^2 + b^4)*Tan[c + d*x]^4 + 4*
a*Sec[c + d*x]*Tan[c + d*x]*(3*(a^6 - 5*a^4*b^2) + 4*b^2*(3*a^4 - 5*a^2*b^2 + 2*b^4)*Tan[c + d*x]^2) + 16*a^2*
b*Sec[c + d*x]^2*(-a^4 + (2*a^4 - 5*a^2*b^2 + 3*b^4)*Tan[c + d*x]^2))/(32*(a^2 - b^2)^2*d)

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Maple [B]  time = 0.072, size = 195, normalized size = 2.1 \begin{align*}{\frac{{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{3\,{a}^{2}b}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{3\,a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{3\,a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{3\,a{b}^{2}\sin \left ( dx+c \right ) }{8\,d}}-{\frac{3\,a{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*(a+b*sin(d*x+c))^3,x)

[Out]

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

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Maxima [A]  time = 0.950487, size = 184, normalized size = 1.96 \begin{align*} \frac{3 \,{\left (a^{3} - a b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (a^{3} - a b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + \frac{2 \,{\left (4 \, b^{3} \sin \left (d x + c\right )^{2} - 3 \,{\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )^{3} + 6 \, a^{2} b - 2 \, b^{3} +{\left (5 \, a^{3} + 3 \, a b^{2}\right )} \sin \left (d x + c\right )\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*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.40031, size = 332, normalized size = 3.53 \begin{align*} \frac{3 \,{\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 8 \, b^{3} \cos \left (d x + c\right )^{2} + 12 \, a^{2} b + 4 \, b^{3} + 2 \,{\left (2 \, a^{3} + 6 \, a b^{2} + 3 \,{\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{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*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/16*(3*(a^3 - a*b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(a^3 - a*b^2)*cos(d*x + c)^4*log(-sin(d*x + c)
+ 1) - 8*b^3*cos(d*x + c)^2 + 12*a^2*b + 4*b^3 + 2*(2*a^3 + 6*a*b^2 + 3*(a^3 - a*b^2)*cos(d*x + c)^2)*sin(d*x
+ c))/(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*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.1669, size = 188, normalized size = 2. \begin{align*} \frac{3 \,{\left (a^{3} - a b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 3 \,{\left (a^{3} - a b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (3 \, a^{3} \sin \left (d x + c\right )^{3} - 3 \, a b^{2} \sin \left (d x + c\right )^{3} - 4 \, b^{3} \sin \left (d x + c\right )^{2} - 5 \, a^{3} \sin \left (d x + c\right ) - 3 \, a b^{2} \sin \left (d x + c\right ) - 6 \, a^{2} b + 2 \, b^{3}\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*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

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