3.9 \(\int (a+b \sec ^2(e+f x)) \sin ^4(e+f x) \, dx\)

Optimal. Leaf size=70 \[ -\frac{(5 a-4 b) \sin (e+f x) \cos (e+f x)}{8 f}+\frac{3}{8} x (a-4 b)+\frac{a \sin (e+f x) \cos ^3(e+f x)}{4 f}+\frac{b \tan (e+f x)}{f} \]

[Out]

(3*(a - 4*b)*x)/8 - ((5*a - 4*b)*Cos[e + f*x]*Sin[e + f*x])/(8*f) + (a*Cos[e + f*x]^3*Sin[e + f*x])/(4*f) + (b
*Tan[e + f*x])/f

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Rubi [A]  time = 0.0638165, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {4132, 455, 1157, 388, 203} \[ -\frac{(5 a-4 b) \sin (e+f x) \cos (e+f x)}{8 f}+\frac{3}{8} x (a-4 b)+\frac{a \sin (e+f x) \cos ^3(e+f x)}{4 f}+\frac{b \tan (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[e + f*x]^2)*Sin[e + f*x]^4,x]

[Out]

(3*(a - 4*b)*x)/8 - ((5*a - 4*b)*Cos[e + f*x]*Sin[e + f*x])/(8*f) + (a*Cos[e + f*x]^3*Sin[e + f*x])/(4*f) + (b
*Tan[e + f*x])/f

Rule 4132

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p)/(
1 + ff^2*x^2)^(m/2 + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/2]

Rule 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 203

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

Rubi steps

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

Mathematica [A]  time = 0.303466, size = 54, normalized size = 0.77 \[ \frac{12 (a-4 b) (e+f x)-8 (a-b) \sin (2 (e+f x))+a \sin (4 (e+f x))+32 b \tan (e+f x)}{32 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[e + f*x]^2)*Sin[e + f*x]^4,x]

[Out]

(12*(a - 4*b)*(e + f*x) - 8*(a - b)*Sin[2*(e + f*x)] + a*Sin[4*(e + f*x)] + 32*b*Tan[e + f*x])/(32*f)

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Maple [A]  time = 0.046, size = 92, normalized size = 1.3 \begin{align*}{\frac{1}{f} \left ( a \left ( -{\frac{\cos \left ( fx+e \right ) }{4} \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{3}+{\frac{3\,\sin \left ( fx+e \right ) }{2}} \right ) }+{\frac{3\,fx}{8}}+{\frac{3\,e}{8}} \right ) +b \left ({\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{5}}{\cos \left ( fx+e \right ) }}+ \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{3}+{\frac{3\,\sin \left ( fx+e \right ) }{2}} \right ) \cos \left ( fx+e \right ) -{\frac{3\,fx}{2}}-{\frac{3\,e}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(f*x+e)^2)*sin(f*x+e)^4,x)

[Out]

1/f*(a*(-1/4*(sin(f*x+e)^3+3/2*sin(f*x+e))*cos(f*x+e)+3/8*f*x+3/8*e)+b*(sin(f*x+e)^5/cos(f*x+e)+(sin(f*x+e)^3+
3/2*sin(f*x+e))*cos(f*x+e)-3/2*f*x-3/2*e))

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Maxima [A]  time = 1.50531, size = 111, normalized size = 1.59 \begin{align*} \frac{3 \,{\left (f x + e\right )}{\left (a - 4 \, b\right )} + 8 \, b \tan \left (f x + e\right ) - \frac{{\left (5 \, a - 4 \, b\right )} \tan \left (f x + e\right )^{3} +{\left (3 \, a - 4 \, b\right )} \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}}{8 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)^2)*sin(f*x+e)^4,x, algorithm="maxima")

[Out]

1/8*(3*(f*x + e)*(a - 4*b) + 8*b*tan(f*x + e) - ((5*a - 4*b)*tan(f*x + e)^3 + (3*a - 4*b)*tan(f*x + e))/(tan(f
*x + e)^4 + 2*tan(f*x + e)^2 + 1))/f

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Fricas [A]  time = 0.489777, size = 167, normalized size = 2.39 \begin{align*} \frac{3 \,{\left (a - 4 \, b\right )} f x \cos \left (f x + e\right ) +{\left (2 \, a \cos \left (f x + e\right )^{4} -{\left (5 \, a - 4 \, b\right )} \cos \left (f x + e\right )^{2} + 8 \, b\right )} \sin \left (f x + e\right )}{8 \, f \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)^2)*sin(f*x+e)^4,x, algorithm="fricas")

[Out]

1/8*(3*(a - 4*b)*f*x*cos(f*x + e) + (2*a*cos(f*x + e)^4 - (5*a - 4*b)*cos(f*x + e)^2 + 8*b)*sin(f*x + e))/(f*c
os(f*x + e))

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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((a+b*sec(f*x+e)**2)*sin(f*x+e)**4,x)

[Out]

Timed out

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Giac [A]  time = 1.15206, size = 120, normalized size = 1.71 \begin{align*} \frac{3 \,{\left (f x + e\right )}{\left (a - 4 \, b\right )} + 8 \, b \tan \left (f x + e\right ) - \frac{5 \, a \tan \left (f x + e\right )^{3} - 4 \, b \tan \left (f x + e\right )^{3} + 3 \, a \tan \left (f x + e\right ) - 4 \, b \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{2}}}{8 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)^2)*sin(f*x+e)^4,x, algorithm="giac")

[Out]

1/8*(3*(f*x + e)*(a - 4*b) + 8*b*tan(f*x + e) - (5*a*tan(f*x + e)^3 - 4*b*tan(f*x + e)^3 + 3*a*tan(f*x + e) -
4*b*tan(f*x + e))/(tan(f*x + e)^2 + 1)^2)/f