3.134 \(\int \frac {\sin ^4(e+f x)}{(a+b \tan ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=187 \[ -\frac {b (13 a+2 b) \tan (e+f x)}{8 f (a-b)^3 \sqrt {a+b \tan ^2(e+f x)}}+\frac {3 a (a+4 b) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{8 f (a-b)^{7/2}}+\frac {\sin (e+f x) \cos ^3(e+f x)}{4 f (a-b) \sqrt {a+b \tan ^2(e+f x)}}-\frac {5 a \sin (e+f x) \cos (e+f x)}{8 f (a-b)^2 \sqrt {a+b \tan ^2(e+f x)}} \]

[Out]

3/8*a*(a+4*b)*arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/(a-b)^(7/2)/f-5/8*a*cos(f*x+e)*sin(f*x+e
)/(a-b)^2/f/(a+b*tan(f*x+e)^2)^(1/2)+1/4*cos(f*x+e)^3*sin(f*x+e)/(a-b)/f/(a+b*tan(f*x+e)^2)^(1/2)-1/8*b*(13*a+
2*b)*tan(f*x+e)/(a-b)^3/f/(a+b*tan(f*x+e)^2)^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3663, 470, 527, 12, 377, 203} \[ \frac {3 a (a+4 b) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{8 f (a-b)^{7/2}}-\frac {b (13 a+2 b) \tan (e+f x)}{8 f (a-b)^3 \sqrt {a+b \tan ^2(e+f x)}}+\frac {\sin (e+f x) \cos ^3(e+f x)}{4 f (a-b) \sqrt {a+b \tan ^2(e+f x)}}-\frac {5 a \sin (e+f x) \cos (e+f x)}{8 f (a-b)^2 \sqrt {a+b \tan ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^4/(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

(3*a*(a + 4*b)*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(8*(a - b)^(7/2)*f) - (5*a*Cos[e
+ f*x]*Sin[e + f*x])/(8*(a - b)^2*f*Sqrt[a + b*Tan[e + f*x]^2]) + (Cos[e + f*x]^3*Sin[e + f*x])/(4*(a - b)*f*
Sqrt[a + b*Tan[e + f*x]^2]) - (b*(13*a + 2*b)*Tan[e + f*x])/(8*(a - b)^3*f*Sqrt[a + b*Tan[e + f*x]^2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 377

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

Rule 470

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

Rule 527

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

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 \frac {\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^4}{\left (1+x^2\right )^3 \left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \sqrt {a+b \tan ^2(e+f x)}}-\frac {\operatorname {Subst}\left (\int \frac {a-4 a x^2}{\left (1+x^2\right )^2 \left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{4 (a-b) f}\\ &=-\frac {5 a \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \sqrt {a+b \tan ^2(e+f x)}}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \sqrt {a+b \tan ^2(e+f x)}}+\frac {\operatorname {Subst}\left (\int \frac {a (3 a+2 b)-10 a b x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{8 (a-b)^2 f}\\ &=-\frac {5 a \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \sqrt {a+b \tan ^2(e+f x)}}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \sqrt {a+b \tan ^2(e+f x)}}-\frac {b (13 a+2 b) \tan (e+f x)}{8 (a-b)^3 f \sqrt {a+b \tan ^2(e+f x)}}+\frac {\operatorname {Subst}\left (\int \frac {3 a^2 (a+4 b)}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{8 a (a-b)^3 f}\\ &=-\frac {5 a \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \sqrt {a+b \tan ^2(e+f x)}}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \sqrt {a+b \tan ^2(e+f x)}}-\frac {b (13 a+2 b) \tan (e+f x)}{8 (a-b)^3 f \sqrt {a+b \tan ^2(e+f x)}}+\frac {(3 a (a+4 b)) \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{8 (a-b)^3 f}\\ &=-\frac {5 a \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \sqrt {a+b \tan ^2(e+f x)}}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \sqrt {a+b \tan ^2(e+f x)}}-\frac {b (13 a+2 b) \tan (e+f x)}{8 (a-b)^3 f \sqrt {a+b \tan ^2(e+f x)}}+\frac {(3 a (a+4 b)) \operatorname {Subst}\left (\int \frac {1}{1-(-a+b) x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{8 (a-b)^3 f}\\ &=\frac {3 a (a+4 b) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{8 (a-b)^{7/2} f}-\frac {5 a \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \sqrt {a+b \tan ^2(e+f x)}}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \sqrt {a+b \tan ^2(e+f x)}}-\frac {b (13 a+2 b) \tan (e+f x)}{8 (a-b)^3 f \sqrt {a+b \tan ^2(e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 3.59, size = 325, normalized size = 1.74 \[ \frac {\sin (2 (e+f x)) \sec ^2(e+f x) \left (-(a-b) \left (\left (6 a^2-2 a b-4 b^2\right ) \cos (2 (e+f x))+7 a^2-(a-b)^2 \cos (4 (e+f x))+48 a b+5 b^2\right )+6 \sqrt {2} a \left (a^2+3 a b-4 b^2\right ) \sqrt {\frac {\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right )-6 \sqrt {2} a^2 (a+4 b) \sqrt {\frac {\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} \Pi \left (-\frac {b}{a-b};\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right )\right )}{32 \sqrt {2} f (a-b)^4 \sqrt {\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^4/(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

((-((a - b)*(7*a^2 + 48*a*b + 5*b^2 + (6*a^2 - 2*a*b - 4*b^2)*Cos[2*(e + f*x)] - (a - b)^2*Cos[4*(e + f*x)]))
+ 6*Sqrt[2]*a*(a^2 + 3*a*b - 4*b^2)*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*EllipticF[ArcS
in[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1] - 6*Sqrt[2]*a^2*(a + 4*b)*Sqrt[((a
+ b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*EllipticPi[-(b/(a - b)), ArcSin[Sqrt[((a + b + (a - b)*Cos
[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1])*Sec[e + f*x]^2*Sin[2*(e + f*x)])/(32*Sqrt[2]*(a - b)^4*f*Sqrt[
(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])

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fricas [B]  time = 128.62, size = 1046, normalized size = 5.59 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[1/64*(3*(a^2*b + 4*a*b^2 + (a^3 + 3*a^2*b - 4*a*b^2)*cos(f*x + e)^2)*sqrt(-a + b)*log(128*(a^4 - 4*a^3*b + 6*
a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^8 - 256*(a^4 - 5*a^3*b + 9*a^2*b^2 - 7*a*b^3 + 2*b^4)*cos(f*x + e)^6 + 3
2*(5*a^4 - 34*a^3*b + 77*a^2*b^2 - 72*a*b^3 + 24*b^4)*cos(f*x + e)^4 + a^4 - 32*a^3*b + 160*a^2*b^2 - 256*a*b^
3 + 128*b^4 - 32*(a^4 - 11*a^3*b + 34*a^2*b^2 - 40*a*b^3 + 16*b^4)*cos(f*x + e)^2 - 8*(16*(a^3 - 3*a^2*b + 3*a
*b^2 - b^3)*cos(f*x + e)^7 - 24*(a^3 - 4*a^2*b + 5*a*b^2 - 2*b^3)*cos(f*x + e)^5 + 2*(5*a^3 - 29*a^2*b + 48*a*
b^2 - 24*b^3)*cos(f*x + e)^3 - (a^3 - 10*a^2*b + 24*a*b^2 - 16*b^3)*cos(f*x + e))*sqrt(-a + b)*sqrt(((a - b)*c
os(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) + 8*(2*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^5 - 5*(a^
3 - 2*a^2*b + a*b^2)*cos(f*x + e)^3 - (13*a^2*b - 11*a*b^2 - 2*b^3)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2
+ b)/cos(f*x + e)^2)*sin(f*x + e))/((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*f*cos(f*x + e)^
2 + (a^4*b - 4*a^3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5)*f), 1/32*(3*(a^2*b + 4*a*b^2 + (a^3 + 3*a^2*b - 4*a*b^2)*c
os(f*x + e)^2)*sqrt(a - b)*arctan(-1/4*(8*(a^2 - 2*a*b + b^2)*cos(f*x + e)^5 - 8*(a^2 - 3*a*b + 2*b^2)*cos(f*x
+ e)^3 + (a^2 - 8*a*b + 8*b^2)*cos(f*x + e))*sqrt(a - b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((
2*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^4 - a^2*b + 3*a*b^2 - 2*b^3 - (a^3 - 6*a^2*b + 9*a*b^2 - 4*b^3)
*cos(f*x + e)^2)*sin(f*x + e))) + 4*(2*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^5 - 5*(a^3 - 2*a^2*b + a*b
^2)*cos(f*x + e)^3 - (13*a^2*b - 11*a*b^2 - 2*b^3)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e
)^2)*sin(f*x + e))/((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*f*cos(f*x + e)^2 + (a^4*b - 4*a^
3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5)*f)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (f x + e\right )^{4}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate(sin(f*x + e)^4/(b*tan(f*x + e)^2 + a)^(3/2), x)

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maple [C]  time = 16.16, size = 6027, normalized size = 32.23 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^4/(a+b*tan(f*x+e)^2)^(3/2),x)

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (f x + e\right )^{4}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(sin(f*x + e)^4/(b*tan(f*x + e)^2 + a)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\sin \left (e+f\,x\right )}^4}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^4/(a + b*tan(e + f*x)^2)^(3/2),x)

[Out]

int(sin(e + f*x)^4/(a + b*tan(e + f*x)^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{4}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**4/(a+b*tan(f*x+e)**2)**(3/2),x)

[Out]

Integral(sin(e + f*x)**4/(a + b*tan(e + f*x)**2)**(3/2), x)

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