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

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

[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])

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Rubi [A]  time = 0.222787, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, 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 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]

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 12

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

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 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 \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*}

Mathematica [C]  time = 3.31646, size = 325, normalized size = 1.74 \[ \frac{\sin (2 (e+f x)) \sec ^2(e+f x) \left (6 \sqrt{2} a \left (a^2+3 a b-4 b^2\right ) \sqrt{\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}}}{\sqrt{2}}\right ),1\right )-(a-b) \left (\left (6 a^2-2 a b-4 b^2\right ) \cos (2 (e+f x))+7 a^2-(a-b)^2 \cos (4 (e+f x))+48 a b+5 b^2\right )-6 \sqrt{2} a^2 (a+4 b) \sqrt{\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} \Pi \left (-\frac{b}{a-b};\left .\sin ^{-1}\left (\frac{\sqrt{\frac{(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt{2}}\right )\right |1\right )\right )}{32 \sqrt{2} f (a-b)^4 \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}} \]

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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Maple [C]  time = 1.602, size = 6025, normalized size = 32.2 \begin{align*} \text{output too large to display} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin ^{4}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)