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

Optimal. Leaf size=114 \[ -\frac {\left (a^2-8 a b+4 b^2\right ) \tan (e+f x)}{4 f}+\frac {1}{8} x \left (3 a^2-24 a b+8 b^2\right )+\frac {a^2 \sin ^4(e+f x) \tan (e+f x)}{4 f}-\frac {a (a-8 b) \sin (e+f x) \cos (e+f x)}{8 f}+\frac {b^2 \tan ^3(e+f x)}{3 f} \]

[Out]

1/8*(3*a^2-24*a*b+8*b^2)*x-1/8*a*(a-8*b)*cos(f*x+e)*sin(f*x+e)/f-1/4*(a^2-8*a*b+4*b^2)*tan(f*x+e)/f+1/4*a^2*si
n(f*x+e)^4*tan(f*x+e)/f+1/3*b^2*tan(f*x+e)^3/f

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Rubi [A]  time = 0.12, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4132, 463, 455, 1153, 203} \[ -\frac {\left (a^2-8 a b+4 b^2\right ) \tan (e+f x)}{4 f}+\frac {1}{8} x \left (3 a^2-24 a b+8 b^2\right )+\frac {a^2 \sin ^4(e+f x) \tan (e+f x)}{4 f}-\frac {a (a-8 b) \sin (e+f x) \cos (e+f x)}{8 f}+\frac {b^2 \tan ^3(e+f x)}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a^2 - 24*a*b + 8*b^2)*x)/8 - (a*(a - 8*b)*Cos[e + f*x]*Sin[e + f*x])/(8*f) - ((a^2 - 8*a*b + 4*b^2)*Tan[e
+ f*x])/(4*f) + (a^2*Sin[e + f*x]^4*Tan[e + f*x])/(4*f) + (b^2*Tan[e + f*x]^3)/(3*f)

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])

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 463

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

Rule 1153

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]

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]

Rubi steps

\begin {align*} \int \left (a+b \sec ^2(e+f x)\right )^2 \sin ^4(e+f x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^4 \left (a+b+b x^2\right )^2}{\left (1+x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {a^2 \sin ^4(e+f x) \tan (e+f x)}{4 f}-\frac {\operatorname {Subst}\left (\int \frac {x^4 \left (5 a^2-4 (a+b)^2-4 b^2 x^2\right )}{\left (1+x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=-\frac {a (a-8 b) \cos (e+f x) \sin (e+f x)}{8 f}+\frac {a^2 \sin ^4(e+f x) \tan (e+f x)}{4 f}+\frac {\operatorname {Subst}\left (\int \frac {a (a-8 b)-2 a (a-8 b) x^2+8 b^2 x^4}{1+x^2} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=-\frac {a (a-8 b) \cos (e+f x) \sin (e+f x)}{8 f}+\frac {a^2 \sin ^4(e+f x) \tan (e+f x)}{4 f}+\frac {\operatorname {Subst}\left (\int \left (-2 \left (a^2-8 a b+4 b^2\right )+8 b^2 x^2+\frac {3 a^2-24 a b+8 b^2}{1+x^2}\right ) \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=-\frac {a (a-8 b) \cos (e+f x) \sin (e+f x)}{8 f}-\frac {\left (a^2-8 a b+4 b^2\right ) \tan (e+f x)}{4 f}+\frac {a^2 \sin ^4(e+f x) \tan (e+f x)}{4 f}+\frac {b^2 \tan ^3(e+f x)}{3 f}+\frac {\left (3 a^2-24 a b+8 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=\frac {1}{8} \left (3 a^2-24 a b+8 b^2\right ) x-\frac {a (a-8 b) \cos (e+f x) \sin (e+f x)}{8 f}-\frac {\left (a^2-8 a b+4 b^2\right ) \tan (e+f x)}{4 f}+\frac {a^2 \sin ^4(e+f x) \tan (e+f x)}{4 f}+\frac {b^2 \tan ^3(e+f x)}{3 f}\\ \end {align*}

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Mathematica [A]  time = 1.62, size = 153, normalized size = 1.34 \[ \frac {\sec ^3(e+f x) \left (a \cos ^2(e+f x)+b\right )^2 \left (3 \cos ^3(e+f x) \left (4 f x \left (3 a^2-24 a b+8 b^2\right )+a^2 \sin (4 (e+f x))-8 a (a-2 b) \sin (2 (e+f x))\right )+64 b (3 a-2 b) \sec (e) \sin (f x) \cos ^2(e+f x)+32 b^2 \tan (e) \cos (e+f x)+32 b^2 \sec (e) \sin (f x)\right )}{24 f (a \cos (2 (e+f x))+a+2 b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b + a*Cos[e + f*x]^2)^2*Sec[e + f*x]^3*(32*b^2*Sec[e]*Sin[f*x] + 64*(3*a - 2*b)*b*Cos[e + f*x]^2*Sec[e]*Sin[
f*x] + 3*Cos[e + f*x]^3*(4*(3*a^2 - 24*a*b + 8*b^2)*f*x - 8*a*(a - 2*b)*Sin[2*(e + f*x)] + a^2*Sin[4*(e + f*x)
]) + 32*b^2*Cos[e + f*x]*Tan[e]))/(24*f*(a + 2*b + a*Cos[2*(e + f*x)])^2)

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fricas [A]  time = 0.65, size = 107, normalized size = 0.94 \[ \frac {3 \, {\left (3 \, a^{2} - 24 \, a b + 8 \, b^{2}\right )} f x \cos \left (f x + e\right )^{3} + {\left (6 \, a^{2} \cos \left (f x + e\right )^{6} - 3 \, {\left (5 \, a^{2} - 8 \, a b\right )} \cos \left (f x + e\right )^{4} + 16 \, {\left (3 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 8 \, b^{2}\right )} \sin \left (f x + e\right )}{24 \, f \cos \left (f x + e\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(3*a^2 - 24*a*b + 8*b^2)*f*x*cos(f*x + e)^3 + (6*a^2*cos(f*x + e)^6 - 3*(5*a^2 - 8*a*b)*cos(f*x + e)^4
+ 16*(3*a*b - 2*b^2)*cos(f*x + e)^2 + 8*b^2)*sin(f*x + e))/(f*cos(f*x + e)^3)

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giac [A]  time = 0.40, size = 132, normalized size = 1.16 \[ \frac {8 \, b^{2} \tan \left (f x + e\right )^{3} + 48 \, a b \tan \left (f x + e\right ) - 24 \, b^{2} \tan \left (f x + e\right ) + 3 \, {\left (3 \, a^{2} - 24 \, a b + 8 \, b^{2}\right )} {\left (f x + e\right )} - \frac {3 \, {\left (5 \, a^{2} \tan \left (f x + e\right )^{3} - 8 \, a b \tan \left (f x + e\right )^{3} + 3 \, a^{2} \tan \left (f x + e\right ) - 8 \, a b \tan \left (f x + e\right )\right )}}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{2}}}{24 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(8*b^2*tan(f*x + e)^3 + 48*a*b*tan(f*x + e) - 24*b^2*tan(f*x + e) + 3*(3*a^2 - 24*a*b + 8*b^2)*(f*x + e)
- 3*(5*a^2*tan(f*x + e)^3 - 8*a*b*tan(f*x + e)^3 + 3*a^2*tan(f*x + e) - 8*a*b*tan(f*x + e))/(tan(f*x + e)^2 +
1)^2)/f

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maple [A]  time = 0.90, size = 123, normalized size = 1.08 \[ \frac {a^{2} \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )+2 a b \left (\frac {\sin ^{5}\left (f x +e \right )}{\cos \left (f x +e \right )}+\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )-\frac {3 f x}{2}-\frac {3 e}{2}\right )+b^{2} \left (\frac {\left (\tan ^{3}\left (f x +e \right )\right )}{3}-\tan \left (f x +e \right )+f x +e \right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.45, size = 120, normalized size = 1.05 \[ \frac {8 \, b^{2} \tan \left (f x + e\right )^{3} + 3 \, {\left (3 \, a^{2} - 24 \, a b + 8 \, b^{2}\right )} {\left (f x + e\right )} + 24 \, {\left (2 \, a b - b^{2}\right )} \tan \left (f x + e\right ) - \frac {3 \, {\left ({\left (5 \, a^{2} - 8 \, a b\right )} \tan \left (f x + e\right )^{3} + {\left (3 \, a^{2} - 8 \, a b\right )} \tan \left (f x + e\right )\right )}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}}{24 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(8*b^2*tan(f*x + e)^3 + 3*(3*a^2 - 24*a*b + 8*b^2)*(f*x + e) + 24*(2*a*b - b^2)*tan(f*x + e) - 3*((5*a^2
- 8*a*b)*tan(f*x + e)^3 + (3*a^2 - 8*a*b)*tan(f*x + e))/(tan(f*x + e)^4 + 2*tan(f*x + e)^2 + 1))/f

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mupad [B]  time = 4.33, size = 116, normalized size = 1.02 \[ x\,\left (\frac {3\,a^2}{8}-3\,a\,b+b^2\right )+\frac {\left (a\,b-\frac {5\,a^2}{8}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3+\left (a\,b-\frac {3\,a^2}{8}\right )\,\mathrm {tan}\left (e+f\,x\right )}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^4+2\,{\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}+\frac {b^2\,{\mathrm {tan}\left (e+f\,x\right )}^3}{3\,f}-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (3\,b^2-2\,b\,\left (a+b\right )\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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