3.392 \(\int \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=99 \[ \frac{\left (3 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\sec ^2(c+d x) \left (\left (3 a^2-b^2\right ) \sin (c+d x)+2 a b\right )}{8 d}+\frac{\sec ^4(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.0878322, antiderivative size = 99, 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, 739, 639, 206} \[ \frac{\left (3 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\sec ^2(c+d x) \left (\left (3 a^2-b^2\right ) \sin (c+d x)+2 a b\right )}{8 d}+\frac{\sec ^4(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5*(a + b*Sin[c + d*x])^2,x]

[Out]

((3*a^2 - b^2)*ArcTanh[Sin[c + d*x]])/(8*d) + (Sec[c + d*x]^4*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x]))/(4*d)
+ (Sec[c + d*x]^2*(2*a*b + (3*a^2 - b^2)*Sin[c + d*x]))/(8*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 739

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[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^
2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 639

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

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

Mathematica [A]  time = 0.731291, size = 166, normalized size = 1.68 \[ \frac{4 \left (b^2-a^2\right ) \sec ^4(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^3+\left (b^2-3 a^2\right ) \left (\left (4 a b^3-6 a^3 b\right ) \tan ^2(c+d x)+\left (a^2-b^2\right )^2 (\log (1-\sin (c+d x))-\log (\sin (c+d x)+1))-2 \left (a^4-b^4\right ) \tan (c+d x) \sec (c+d x)+2 a^3 b \sec ^2(c+d x)\right )}{16 d \left (a^2-b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5*(a + b*Sin[c + d*x])^2,x]

[Out]

(4*(-a^2 + b^2)*Sec[c + d*x]^4*(b - a*Sin[c + d*x])*(a + b*Sin[c + d*x])^3 + (-3*a^2 + b^2)*((a^2 - b^2)^2*(Lo
g[1 - Sin[c + d*x]] - Log[1 + Sin[c + d*x]]) + 2*a^3*b*Sec[c + d*x]^2 - 2*(a^4 - b^4)*Sec[c + d*x]*Tan[c + d*x
] + (-6*a^3*b + 4*a*b^3)*Tan[c + d*x]^2))/(16*(a^2 - b^2)^2*d)

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Maple [A]  time = 0.058, size = 165, normalized size = 1.7 \begin{align*}{\frac{{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,{a}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{ab}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{b}^{2}\sin \left ( dx+c \right ) }{8\,d}}-{\frac{{b}^{2}\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*(a+b*sin(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 0.968193, size = 155, normalized size = 1.57 \begin{align*} \frac{{\left (3 \, a^{2} - b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (3 \, a^{2} - b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left ({\left (3 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{3} - 4 \, a b -{\left (5 \, a^{2} + 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))^2,x, algorithm="maxima")

[Out]

1/16*((3*a^2 - b^2)*log(sin(d*x + c) + 1) - (3*a^2 - b^2)*log(sin(d*x + c) - 1) - 2*((3*a^2 - b^2)*sin(d*x + c
)^3 - 4*a*b - (5*a^2 + 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 = 2.3231, size = 275, normalized size = 2.78 \begin{align*} \frac{{\left (3 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (3 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 8 \, a b + 2 \,{\left ({\left (3 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, a^{2} + 2 \, b^{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))^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.14714, size = 159, normalized size = 1.61 \begin{align*} \frac{{\left (3 \, a^{2} - b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) -{\left (3 \, a^{2} - b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (3 \, a^{2} \sin \left (d x + c\right )^{3} - b^{2} \sin \left (d x + c\right )^{3} - 5 \, a^{2} \sin \left (d x + c\right ) - b^{2} \sin \left (d x + c\right ) - 4 \, a b\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))^2,x, algorithm="giac")

[Out]

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