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

Optimal. Leaf size=122 \[ -\frac {\left (a^2-10 a b+13 b^2\right ) \tan (e+f x)}{4 f}+\frac {1}{8} x \left (3 a^2-30 a b+35 b^2\right )+\frac {(a-b)^2 \sin ^4(e+f x) \tan (e+f x)}{4 f}-\frac {(a-9 b) (a-b) \sin (e+f x) \cos (e+f x)}{8 f}+\frac {b^2 \tan ^3(e+f x)}{3 f} \]

[Out]

1/8*(3*a^2-30*a*b+35*b^2)*x-1/8*(a-9*b)*(a-b)*cos(f*x+e)*sin(f*x+e)/f-1/4*(a^2-10*a*b+13*b^2)*tan(f*x+e)/f+1/4
*(a-b)^2*sin(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 = 0.13, antiderivative size = 122, 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 = {3663, 463, 455, 1153, 203} \[ -\frac {\left (a^2-10 a b+13 b^2\right ) \tan (e+f x)}{4 f}+\frac {1}{8} x \left (3 a^2-30 a b+35 b^2\right )+\frac {(a-b)^2 \sin ^4(e+f x) \tan (e+f x)}{4 f}-\frac {(a-9 b) (a-b) \sin (e+f x) \cos (e+f x)}{8 f}+\frac {b^2 \tan ^3(e+f x)}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a^2 - 30*a*b + 35*b^2)*x)/8 - ((a - 9*b)*(a - b)*Cos[e + f*x]*Sin[e + f*x])/(8*f) - ((a^2 - 10*a*b + 13*b^
2)*Tan[e + f*x])/(4*f) + ((a - b)^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 3663

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

Rubi steps

\begin {align*} \int \sin ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^4 \left (a+b x^2\right )^2}{\left (1+x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(a-b)^2 \sin ^4(e+f x) \tan (e+f x)}{4 f}-\frac {\operatorname {Subst}\left (\int \frac {x^4 \left (a^2-10 a b+5 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-9 b) (a-b) \cos (e+f x) \sin (e+f x)}{8 f}+\frac {(a-b)^2 \sin ^4(e+f x) \tan (e+f x)}{4 f}+\frac {\operatorname {Subst}\left (\int \frac {(a-9 b) (a-b)-2 (a-9 b) (a-b) x^2+8 b^2 x^4}{1+x^2} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=-\frac {(a-9 b) (a-b) \cos (e+f x) \sin (e+f x)}{8 f}+\frac {(a-b)^2 \sin ^4(e+f x) \tan (e+f x)}{4 f}+\frac {\operatorname {Subst}\left (\int \left (-2 \left (a^2-10 a b+13 b^2\right )+8 b^2 x^2+\frac {3 a^2-30 a b+35 b^2}{1+x^2}\right ) \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=-\frac {(a-9 b) (a-b) \cos (e+f x) \sin (e+f x)}{8 f}-\frac {\left (a^2-10 a b+13 b^2\right ) \tan (e+f x)}{4 f}+\frac {(a-b)^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^2-30 a b+35 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^2-30 a b+35 b^2\right ) x-\frac {(a-9 b) (a-b) \cos (e+f x) \sin (e+f x)}{8 f}-\frac {\left (a^2-10 a b+13 b^2\right ) \tan (e+f x)}{4 f}+\frac {(a-b)^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 = 1.46, size = 96, normalized size = 0.79 \[ \frac {12 \left (3 a^2-30 a b+35 b^2\right ) (e+f x)-24 \left (a^2-4 a b+3 b^2\right ) \sin (2 (e+f x))+3 (a-b)^2 \sin (4 (e+f x))+32 b \tan (e+f x) \left (6 a+b \sec ^2(e+f x)-10 b\right )}{96 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*(3*a^2 - 30*a*b + 35*b^2)*(e + f*x) - 24*(a^2 - 4*a*b + 3*b^2)*Sin[2*(e + f*x)] + 3*(a - b)^2*Sin[4*(e + f
*x)] + 32*b*(6*a - 10*b + b*Sec[e + f*x]^2)*Tan[e + f*x])/(96*f)

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fricas [A]  time = 0.44, size = 120, normalized size = 0.98 \[ \frac {3 \, {\left (3 \, a^{2} - 30 \, a b + 35 \, b^{2}\right )} f x \cos \left (f x + e\right )^{3} + {\left (6 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{6} - 3 \, {\left (5 \, a^{2} - 18 \, a b + 13 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 16 \, {\left (3 \, a b - 5 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 8 \, b^{2}\right )} \sin \left (f x + e\right )}{24 \, f \cos \left (f x + e\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(3*a^2 - 30*a*b + 35*b^2)*f*x*cos(f*x + e)^3 + (6*(a^2 - 2*a*b + b^2)*cos(f*x + e)^6 - 3*(5*a^2 - 18*a
*b + 13*b^2)*cos(f*x + e)^4 + 16*(3*a*b - 5*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 [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(sin(f*x+e)^4*(a+b*tan(f*x+e)^2)^2,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.78, size = 199, normalized size = 1.63 \[ \frac {a^{2} \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )+2 a b \left (\frac {\sin ^{7}\left (f x +e \right )}{\cos \left (f x +e \right )}+\left (\sin ^{5}\left (f x +e \right )+\frac {5 \left (\sin ^{3}\left (f x +e \right )\right )}{4}+\frac {15 \sin \left (f x +e \right )}{8}\right ) \cos \left (f x +e \right )-\frac {15 f x}{8}-\frac {15 e}{8}\right )+b^{2} \left (\frac {\sin ^{9}\left (f x +e \right )}{3 \cos \left (f x +e \right )^{3}}-\frac {2 \left (\sin ^{9}\left (f x +e \right )\right )}{\cos \left (f x +e \right )}-2 \left (\sin ^{7}\left (f x +e \right )+\frac {7 \left (\sin ^{5}\left (f x +e \right )\right )}{6}+\frac {35 \left (\sin ^{3}\left (f x +e \right )\right )}{24}+\frac {35 \sin \left (f x +e \right )}{16}\right ) \cos \left (f x +e \right )+\frac {35 f x}{8}+\frac {35 e}{8}\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^4*(a+b*tan(f*x+e)^2)^2,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)^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)^9/cos(f*x+e)^3-2*sin(f
*x+e)^9/cos(f*x+e)-2*(sin(f*x+e)^7+7/6*sin(f*x+e)^5+35/24*sin(f*x+e)^3+35/16*sin(f*x+e))*cos(f*x+e)+35/8*f*x+3
5/8*e))

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maxima [A]  time = 0.59, size = 130, normalized size = 1.07 \[ \frac {8 \, b^{2} \tan \left (f x + e\right )^{3} + 3 \, {\left (3 \, a^{2} - 30 \, a b + 35 \, b^{2}\right )} {\left (f x + e\right )} + 24 \, {\left (2 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right ) - \frac {3 \, {\left ({\left (5 \, a^{2} - 18 \, a b + 13 \, b^{2}\right )} \tan \left (f x + e\right )^{3} + {\left (3 \, a^{2} - 14 \, a b + 11 \, b^{2}\right )} \tan \left (f x + e\right )\right )}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}}{24 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(8*b^2*tan(f*x + e)^3 + 3*(3*a^2 - 30*a*b + 35*b^2)*(f*x + e) + 24*(2*a*b - 3*b^2)*tan(f*x + e) - 3*((5*a
^2 - 18*a*b + 13*b^2)*tan(f*x + e)^3 + (3*a^2 - 14*a*b + 11*b^2)*tan(f*x + e))/(tan(f*x + e)^4 + 2*tan(f*x + e
)^2 + 1))/f

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mupad [B]  time = 12.33, size = 128, normalized size = 1.05 \[ x\,\left (\frac {3\,a^2}{8}-\frac {15\,a\,b}{4}+\frac {35\,b^2}{8}\right )+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (2\,a\,b-3\,b^2\right )}{f}+\frac {b^2\,{\mathrm {tan}\left (e+f\,x\right )}^3}{3\,f}-\frac {\left (\frac {5\,a^2}{8}-\frac {9\,a\,b}{4}+\frac {13\,b^2}{8}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3+\left (\frac {3\,a^2}{8}-\frac {7\,a\,b}{4}+\frac {11\,b^2}{8}\right )\,\mathrm {tan}\left (e+f\,x\right )}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^4+2\,{\mathrm {tan}\left (e+f\,x\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*((3*a^2)/8 - (15*a*b)/4 + (35*b^2)/8) + (tan(e + f*x)*(2*a*b - 3*b^2))/f + (b^2*tan(e + f*x)^3)/(3*f) - (tan
(e + f*x)*((3*a^2)/8 - (7*a*b)/4 + (11*b^2)/8) + tan(e + f*x)^3*((5*a^2)/8 - (9*a*b)/4 + (13*b^2)/8))/(f*(2*ta
n(e + f*x)^2 + tan(e + f*x)^4 + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{2} \sin ^{4}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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