3.21 \(\int (a+b \sec ^2(e+f x))^2 \sin ^6(e+f x) \, dx\)

Optimal. Leaf size=148 \[ -\frac {\left (a^2-12 a b+12 b^2\right ) \tan (e+f x)}{6 f}-\frac {\left (3 a^2-36 a b+8 b^2\right ) \sin (e+f x) \cos (e+f x)}{16 f}+\frac {5}{16} x \left (a^2-12 a b+8 b^2\right )+\frac {a^2 \sin ^6(e+f x) \tan (e+f x)}{6 f}+\frac {a (a-12 b) \sin (e+f x) \cos ^3(e+f x)}{24 f}+\frac {b^2 \tan ^3(e+f x)}{3 f} \]

[Out]

5/16*(a^2-12*a*b+8*b^2)*x-1/16*(3*a^2-36*a*b+8*b^2)*cos(f*x+e)*sin(f*x+e)/f+1/24*a*(a-12*b)*cos(f*x+e)^3*sin(f
*x+e)/f-1/6*(a^2-12*a*b+12*b^2)*tan(f*x+e)/f+1/6*a^2*sin(f*x+e)^6*tan(f*x+e)/f+1/3*b^2*tan(f*x+e)^3/f

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Rubi [A]  time = 0.18, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {4132, 463, 455, 1814, 1153, 203} \[ -\frac {\left (a^2-12 a b+12 b^2\right ) \tan (e+f x)}{6 f}-\frac {\left (3 a^2-36 a b+8 b^2\right ) \sin (e+f x) \cos (e+f x)}{16 f}+\frac {5}{16} x \left (a^2-12 a b+8 b^2\right )+\frac {a^2 \sin ^6(e+f x) \tan (e+f x)}{6 f}+\frac {a (a-12 b) \sin (e+f x) \cos ^3(e+f x)}{24 f}+\frac {b^2 \tan ^3(e+f x)}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[e + f*x]^2)^2*Sin[e + f*x]^6,x]

[Out]

(5*(a^2 - 12*a*b + 8*b^2)*x)/16 - ((3*a^2 - 36*a*b + 8*b^2)*Cos[e + f*x]*Sin[e + f*x])/(16*f) + (a*(a - 12*b)*
Cos[e + f*x]^3*Sin[e + f*x])/(24*f) - ((a^2 - 12*a*b + 12*b^2)*Tan[e + f*x])/(6*f) + (a^2*Sin[e + f*x]^6*Tan[e
+ f*x])/(6*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 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

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 ^6(e+f x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^6 \left (a+b+b x^2\right )^2}{\left (1+x^2\right )^4} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {a^2 \sin ^6(e+f x) \tan (e+f x)}{6 f}-\frac {\operatorname {Subst}\left (\int \frac {x^6 \left (7 a^2-6 (a+b)^2-6 b^2 x^2\right )}{\left (1+x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{6 f}\\ &=\frac {a (a-12 b) \cos ^3(e+f x) \sin (e+f x)}{24 f}+\frac {a^2 \sin ^6(e+f x) \tan (e+f x)}{6 f}+\frac {\operatorname {Subst}\left (\int \frac {-a (a-12 b)+4 a (a-12 b) x^2-4 a (a-12 b) x^4+24 b^2 x^6}{\left (1+x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{24 f}\\ &=-\frac {\left (3 a^2-36 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 f}+\frac {a (a-12 b) \cos ^3(e+f x) \sin (e+f x)}{24 f}+\frac {a^2 \sin ^6(e+f x) \tan (e+f x)}{6 f}-\frac {\operatorname {Subst}\left (\int \frac {-7 a^2+84 a b-24 b^2+8 \left (a^2-12 a b+6 b^2\right ) x^2-48 b^2 x^4}{1+x^2} \, dx,x,\tan (e+f x)\right )}{48 f}\\ &=-\frac {\left (3 a^2-36 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 f}+\frac {a (a-12 b) \cos ^3(e+f x) \sin (e+f x)}{24 f}+\frac {a^2 \sin ^6(e+f x) \tan (e+f x)}{6 f}-\frac {\operatorname {Subst}\left (\int \left (8 \left (a^2-12 a b+12 b^2\right )-48 b^2 x^2-\frac {15 \left (a^2-12 a b+8 b^2\right )}{1+x^2}\right ) \, dx,x,\tan (e+f x)\right )}{48 f}\\ &=-\frac {\left (3 a^2-36 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 f}+\frac {a (a-12 b) \cos ^3(e+f x) \sin (e+f x)}{24 f}-\frac {\left (a^2-12 a b+12 b^2\right ) \tan (e+f x)}{6 f}+\frac {a^2 \sin ^6(e+f x) \tan (e+f x)}{6 f}+\frac {b^2 \tan ^3(e+f x)}{3 f}+\frac {\left (5 \left (a^2-12 a b+8 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{16 f}\\ &=\frac {5}{16} \left (a^2-12 a b+8 b^2\right ) x-\frac {\left (3 a^2-36 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 f}+\frac {a (a-12 b) \cos ^3(e+f x) \sin (e+f x)}{24 f}-\frac {\left (a^2-12 a b+12 b^2\right ) \tan (e+f x)}{6 f}+\frac {a^2 \sin ^6(e+f x) \tan (e+f x)}{6 f}+\frac {b^2 \tan ^3(e+f x)}{3 f}\\ \end {align*}

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Mathematica [B]  time = 1.35, size = 499, normalized size = 3.37 \[ \frac {\sec (e) \sec ^3(e+f x) \left (a \cos ^2(e+f x)+b\right )^2 \left (360 f x \left (a^2-12 a b+8 b^2\right ) \cos (2 e+f x)+360 f x \left (a^2-12 a b+8 b^2\right ) \cos (f x)-81 a^2 \sin (2 e+f x)-109 a^2 \sin (2 e+3 f x)-109 a^2 \sin (4 e+3 f x)-21 a^2 \sin (4 e+5 f x)-21 a^2 \sin (6 e+5 f x)+6 a^2 \sin (6 e+7 f x)+6 a^2 \sin (8 e+7 f x)-a^2 \sin (8 e+9 f x)-a^2 \sin (10 e+9 f x)+120 a^2 f x \cos (2 e+3 f x)+120 a^2 f x \cos (4 e+3 f x)-81 a^2 \sin (f x)-1164 a b \sin (2 e+f x)+2076 a b \sin (2 e+3 f x)+540 a b \sin (4 e+3 f x)+156 a b \sin (4 e+5 f x)+156 a b \sin (6 e+5 f x)-12 a b \sin (6 e+7 f x)-12 a b \sin (8 e+7 f x)-1440 a b f x \cos (2 e+3 f x)-1440 a b f x \cos (4 e+3 f x)+3444 a b \sin (f x)+2208 b^2 \sin (2 e+f x)-1936 b^2 \sin (2 e+3 f x)-144 b^2 \sin (4 e+3 f x)-48 b^2 \sin (4 e+5 f x)-48 b^2 \sin (6 e+5 f x)+960 b^2 f x \cos (2 e+3 f x)+960 b^2 f x \cos (4 e+3 f x)-3168 b^2 \sin (f x)\right )}{768 f (a \cos (2 (e+f x))+a+2 b)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[e + f*x]^2)^2*Sin[e + f*x]^6,x]

[Out]

((b + a*Cos[e + f*x]^2)^2*Sec[e]*Sec[e + f*x]^3*(360*(a^2 - 12*a*b + 8*b^2)*f*x*Cos[f*x] + 360*(a^2 - 12*a*b +
8*b^2)*f*x*Cos[2*e + f*x] + 120*a^2*f*x*Cos[2*e + 3*f*x] - 1440*a*b*f*x*Cos[2*e + 3*f*x] + 960*b^2*f*x*Cos[2*
e + 3*f*x] + 120*a^2*f*x*Cos[4*e + 3*f*x] - 1440*a*b*f*x*Cos[4*e + 3*f*x] + 960*b^2*f*x*Cos[4*e + 3*f*x] - 81*
a^2*Sin[f*x] + 3444*a*b*Sin[f*x] - 3168*b^2*Sin[f*x] - 81*a^2*Sin[2*e + f*x] - 1164*a*b*Sin[2*e + f*x] + 2208*
b^2*Sin[2*e + f*x] - 109*a^2*Sin[2*e + 3*f*x] + 2076*a*b*Sin[2*e + 3*f*x] - 1936*b^2*Sin[2*e + 3*f*x] - 109*a^
2*Sin[4*e + 3*f*x] + 540*a*b*Sin[4*e + 3*f*x] - 144*b^2*Sin[4*e + 3*f*x] - 21*a^2*Sin[4*e + 5*f*x] + 156*a*b*S
in[4*e + 5*f*x] - 48*b^2*Sin[4*e + 5*f*x] - 21*a^2*Sin[6*e + 5*f*x] + 156*a*b*Sin[6*e + 5*f*x] - 48*b^2*Sin[6*
e + 5*f*x] + 6*a^2*Sin[6*e + 7*f*x] - 12*a*b*Sin[6*e + 7*f*x] + 6*a^2*Sin[8*e + 7*f*x] - 12*a*b*Sin[8*e + 7*f*
x] - a^2*Sin[8*e + 9*f*x] - a^2*Sin[10*e + 9*f*x]))/(768*f*(a + 2*b + a*Cos[2*(e + f*x)])^2)

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fricas [A]  time = 0.56, size = 131, normalized size = 0.89 \[ \frac {15 \, {\left (a^{2} - 12 \, a b + 8 \, b^{2}\right )} f x \cos \left (f x + e\right )^{3} - {\left (8 \, a^{2} \cos \left (f x + e\right )^{8} - 2 \, {\left (13 \, a^{2} - 12 \, a b\right )} \cos \left (f x + e\right )^{6} + 3 \, {\left (11 \, a^{2} - 36 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 16 \, {\left (6 \, a b - 7 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 16 \, b^{2}\right )} \sin \left (f x + e\right )}{48 \, f \cos \left (f x + e\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)^2)^2*sin(f*x+e)^6,x, algorithm="fricas")

[Out]

1/48*(15*(a^2 - 12*a*b + 8*b^2)*f*x*cos(f*x + e)^3 - (8*a^2*cos(f*x + e)^8 - 2*(13*a^2 - 12*a*b)*cos(f*x + e)^
6 + 3*(11*a^2 - 36*a*b + 8*b^2)*cos(f*x + e)^4 - 16*(6*a*b - 7*b^2)*cos(f*x + e)^2 - 16*b^2)*sin(f*x + e))/(f*
cos(f*x + e)^3)

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giac [A]  time = 1.27, size = 197, normalized size = 1.33 \[ \frac {16 \, b^{2} \tan \left (f x + e\right )^{3} + 96 \, a b \tan \left (f x + e\right ) - 96 \, b^{2} \tan \left (f x + e\right ) + 15 \, {\left (a^{2} - 12 \, a b + 8 \, b^{2}\right )} {\left (f x + e\right )} - \frac {33 \, a^{2} \tan \left (f x + e\right )^{5} - 108 \, a b \tan \left (f x + e\right )^{5} + 24 \, b^{2} \tan \left (f x + e\right )^{5} + 40 \, a^{2} \tan \left (f x + e\right )^{3} - 192 \, a b \tan \left (f x + e\right )^{3} + 48 \, b^{2} \tan \left (f x + e\right )^{3} + 15 \, a^{2} \tan \left (f x + e\right ) - 84 \, a b \tan \left (f x + e\right ) + 24 \, b^{2} \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{3}}}{48 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)^2)^2*sin(f*x+e)^6,x, algorithm="giac")

[Out]

1/48*(16*b^2*tan(f*x + e)^3 + 96*a*b*tan(f*x + e) - 96*b^2*tan(f*x + e) + 15*(a^2 - 12*a*b + 8*b^2)*(f*x + e)
- (33*a^2*tan(f*x + e)^5 - 108*a*b*tan(f*x + e)^5 + 24*b^2*tan(f*x + e)^5 + 40*a^2*tan(f*x + e)^3 - 192*a*b*ta
n(f*x + e)^3 + 48*b^2*tan(f*x + e)^3 + 15*a^2*tan(f*x + e) - 84*a*b*tan(f*x + e) + 24*b^2*tan(f*x + e))/(tan(f
*x + e)^2 + 1)^3)/f

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maple [A]  time = 0.92, size = 199, normalized size = 1.34 \[ \frac {a^{2} \left (-\frac {\left (\sin ^{5}\left (f x +e \right )+\frac {5 \left (\sin ^{3}\left (f x +e \right )\right )}{4}+\frac {15 \sin \left (f x +e \right )}{8}\right ) \cos \left (f x +e \right )}{6}+\frac {5 f x}{16}+\frac {5 e}{16}\right )+2 a b \left (\frac {\sin ^{7}\left (f x +e \right )}{\cos \left (f x +e \right )}+\left (\sin ^{5}\left (f x +e \right )+\frac {5 \left (\sin ^{3}\left (f x +e \right )\right )}{4}+\frac {15 \sin \left (f x +e \right )}{8}\right ) \cos \left (f x +e \right )-\frac {15 f x}{8}-\frac {15 e}{8}\right )+b^{2} \left (\frac {\sin ^{7}\left (f x +e \right )}{3 \cos \left (f x +e \right )^{3}}-\frac {4 \left (\sin ^{7}\left (f x +e \right )\right )}{3 \cos \left (f x +e \right )}-\frac {4 \left (\sin ^{5}\left (f x +e \right )+\frac {5 \left (\sin ^{3}\left (f x +e \right )\right )}{4}+\frac {15 \sin \left (f x +e \right )}{8}\right ) \cos \left (f x +e \right )}{3}+\frac {5 f x}{2}+\frac {5 e}{2}\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(f*x+e)^2)^2*sin(f*x+e)^6,x)

[Out]

1/f*(a^2*(-1/6*(sin(f*x+e)^5+5/4*sin(f*x+e)^3+15/8*sin(f*x+e))*cos(f*x+e)+5/16*f*x+5/16*e)+2*a*b*(sin(f*x+e)^7
/cos(f*x+e)+(sin(f*x+e)^5+5/4*sin(f*x+e)^3+15/8*sin(f*x+e))*cos(f*x+e)-15/8*f*x-15/8*e)+b^2*(1/3*sin(f*x+e)^7/
cos(f*x+e)^3-4/3*sin(f*x+e)^7/cos(f*x+e)-4/3*(sin(f*x+e)^5+5/4*sin(f*x+e)^3+15/8*sin(f*x+e))*cos(f*x+e)+5/2*f*
x+5/2*e))

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maxima [A]  time = 0.45, size = 164, normalized size = 1.11 \[ \frac {16 \, b^{2} \tan \left (f x + e\right )^{3} + 15 \, {\left (a^{2} - 12 \, a b + 8 \, b^{2}\right )} {\left (f x + e\right )} + 96 \, {\left (a b - b^{2}\right )} \tan \left (f x + e\right ) - \frac {3 \, {\left (11 \, a^{2} - 36 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )^{5} + 8 \, {\left (5 \, a^{2} - 24 \, a b + 6 \, b^{2}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (5 \, a^{2} - 28 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{6} + 3 \, \tan \left (f x + e\right )^{4} + 3 \, \tan \left (f x + e\right )^{2} + 1}}{48 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)^2)^2*sin(f*x+e)^6,x, algorithm="maxima")

[Out]

1/48*(16*b^2*tan(f*x + e)^3 + 15*(a^2 - 12*a*b + 8*b^2)*(f*x + e) + 96*(a*b - b^2)*tan(f*x + e) - (3*(11*a^2 -
36*a*b + 8*b^2)*tan(f*x + e)^5 + 8*(5*a^2 - 24*a*b + 6*b^2)*tan(f*x + e)^3 + 3*(5*a^2 - 28*a*b + 8*b^2)*tan(f
*x + e))/(tan(f*x + e)^6 + 3*tan(f*x + e)^4 + 3*tan(f*x + e)^2 + 1))/f

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mupad [B]  time = 4.79, size = 163, normalized size = 1.10 \[ x\,\left (\frac {5\,a^2}{16}-\frac {15\,a\,b}{4}+\frac {5\,b^2}{2}\right )-\frac {\left (\frac {11\,a^2}{16}-\frac {9\,a\,b}{4}+\frac {b^2}{2}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^5+\left (\frac {5\,a^2}{6}-4\,a\,b+b^2\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3+\left (\frac {5\,a^2}{16}-\frac {7\,a\,b}{4}+\frac {b^2}{2}\right )\,\mathrm {tan}\left (e+f\,x\right )}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^6+3\,{\mathrm {tan}\left (e+f\,x\right )}^4+3\,{\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}+\frac {b^2\,{\mathrm {tan}\left (e+f\,x\right )}^3}{3\,f}-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (4\,b^2-2\,b\,\left (a+b\right )\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^6*(a + b/cos(e + f*x)^2)^2,x)

[Out]

x*((5*a^2)/16 - (15*a*b)/4 + (5*b^2)/2) - (tan(e + f*x)*((5*a^2)/16 - (7*a*b)/4 + b^2/2) + tan(e + f*x)^3*((5*
a^2)/6 - 4*a*b + b^2) + tan(e + f*x)^5*((11*a^2)/16 - (9*a*b)/4 + b^2/2))/(f*(3*tan(e + f*x)^2 + 3*tan(e + f*x
)^4 + tan(e + f*x)^6 + 1)) + (b^2*tan(e + f*x)^3)/(3*f) - (tan(e + f*x)*(4*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)**6,x)

[Out]

Timed out

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