3.2.46 \(\int \frac {\sin ^4(e+f x)}{(a+b \tan ^2(e+f x))^{5/2}} \, dx\) [146]

Optimal. Leaf size=246 \[ \frac {\left (3 a^2+24 a b+8 b^2\right ) \text {ArcTan}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{8 (a-b)^{9/2} f}-\frac {(5 a+2 b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {b (23 a+12 b) \tan (e+f x)}{24 (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {5 b (11 a+10 b) \tan (e+f x)}{24 (a-b)^4 f \sqrt {a+b \tan ^2(e+f x)}} \]

[Out]

1/8*(3*a^2+24*a*b+8*b^2)*arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/(a-b)^(9/2)/f-5/24*b*(11*a+10
*b)*tan(f*x+e)/(a-b)^4/f/(a+b*tan(f*x+e)^2)^(1/2)-1/8*(5*a+2*b)*cos(f*x+e)*sin(f*x+e)/(a-b)^2/f/(a+b*tan(f*x+e
)^2)^(3/2)+1/4*cos(f*x+e)^3*sin(f*x+e)/(a-b)/f/(a+b*tan(f*x+e)^2)^(3/2)-1/24*b*(23*a+12*b)*tan(f*x+e)/(a-b)^3/
f/(a+b*tan(f*x+e)^2)^(3/2)

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Rubi [A]
time = 0.23, antiderivative size = 246, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3744, 481, 541, 12, 385, 209} \begin {gather*} \frac {\left (3 a^2+24 a b+8 b^2\right ) \text {ArcTan}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{8 f (a-b)^{9/2}}-\frac {5 b (11 a+10 b) \tan (e+f x)}{24 f (a-b)^4 \sqrt {a+b \tan ^2(e+f x)}}-\frac {b (23 a+12 b) \tan (e+f x)}{24 f (a-b)^3 \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\sin (e+f x) \cos ^3(e+f x)}{4 f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {(5 a+2 b) \sin (e+f x) \cos (e+f x)}{8 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^4/(a + b*Tan[e + f*x]^2)^(5/2),x]

[Out]

((3*a^2 + 24*a*b + 8*b^2)*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(8*(a - b)^(9/2)*f) -
((5*a + 2*b)*Cos[e + f*x]*Sin[e + f*x])/(8*(a - b)^2*f*(a + b*Tan[e + f*x]^2)^(3/2)) + (Cos[e + f*x]^3*Sin[e
+ f*x])/(4*(a - b)*f*(a + b*Tan[e + f*x]^2)^(3/2)) - (b*(23*a + 12*b)*Tan[e + f*x])/(24*(a - b)^3*f*(a + b*Tan
[e + f*x]^2)^(3/2)) - (5*b*(11*a + 10*b)*Tan[e + f*x])/(24*(a - b)^4*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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 385

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 481

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 541

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 3744

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 )^{5/2}} \, dx &=\frac {\text {Subst}\left (\int \frac {x^4}{\left (1+x^2\right )^3 \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\text {Subst}\left (\int \frac {a-2 (2 a+b) x^2}{\left (1+x^2\right )^2 \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{4 (a-b) f}\\ &=-\frac {(5 a+2 b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\text {Subst}\left (\int \frac {a (3 a+4 b)-4 b (5 a+2 b) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{8 (a-b)^2 f}\\ &=-\frac {(5 a+2 b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {b (23 a+12 b) \tan (e+f x)}{24 (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\text {Subst}\left (\int \frac {a^2 (9 a+26 b)-2 a b (23 a+12 b) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{24 a (a-b)^3 f}\\ &=-\frac {(5 a+2 b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {b (23 a+12 b) \tan (e+f x)}{24 (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {5 b (11 a+10 b) \tan (e+f x)}{24 (a-b)^4 f \sqrt {a+b \tan ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {3 a^2 \left (3 a^2+24 a b+8 b^2\right )}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{24 a^2 (a-b)^4 f}\\ &=-\frac {(5 a+2 b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {b (23 a+12 b) \tan (e+f x)}{24 (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {5 b (11 a+10 b) \tan (e+f x)}{24 (a-b)^4 f \sqrt {a+b \tan ^2(e+f x)}}+\frac {\left (3 a^2+24 a b+8 b^2\right ) \text {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)^4 f}\\ &=-\frac {(5 a+2 b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {b (23 a+12 b) \tan (e+f x)}{24 (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {5 b (11 a+10 b) \tan (e+f x)}{24 (a-b)^4 f \sqrt {a+b \tan ^2(e+f x)}}+\frac {\left (3 a^2+24 a b+8 b^2\right ) \text {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)^4 f}\\ &=\frac {\left (3 a^2+24 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{8 (a-b)^{9/2} f}-\frac {(5 a+2 b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {b (23 a+12 b) \tan (e+f x)}{24 (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {5 b (11 a+10 b) \tan (e+f x)}{24 (a-b)^4 f \sqrt {a+b \tan ^2(e+f x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
time = 3.75, size = 378, normalized size = 1.54 \begin {gather*} -\frac {\sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)} \left (-3 \sqrt {2} a b \left (3 a^2+24 a b+8 b^2\right ) \left (\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}\right )^{3/2} \left (2 (a-b) F\left (\left .\text {ArcSin}\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 )-2 a \Pi \left (-\frac {b}{a-b};\left .\text {ArcSin}\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 ) \sin ^2(e+f x) \sin (2 (e+f x))-a (a-b) \left (64 a b^2 \sin (2 (e+f x))-64 b (3 a+2 b) (a+b+(a-b) \cos (2 (e+f x))) \sin (2 (e+f x))-6 (4 a+7 b) (a+b+(a-b) \cos (2 (e+f x)))^2 \sin (2 (e+f x))+3 (a-b) (a+b+(a-b) \cos (2 (e+f x)))^2 \sin (4 (e+f x))\right )\right )}{96 \sqrt {2} a (a-b)^5 f (a+b+(a-b) \cos (2 (e+f x)))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^4/(a + b*Tan[e + f*x]^2)^(5/2),x]

[Out]

-1/96*(Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]*(-3*Sqrt[2]*a*b*(3*a^2 + 24*a*b + 8*b^2)*(((a +
b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b)^(3/2)*(2*(a - b)*EllipticF[ArcSin[Sqrt[((a + b + (a - b)*Cos
[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1] - 2*a*EllipticPi[-(b/(a - b)), ArcSin[Sqrt[((a + b + (a - b)*Co
s[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1])*Sin[e + f*x]^2*Sin[2*(e + f*x)] - a*(a - b)*(64*a*b^2*Sin[2*(
e + f*x)] - 64*b*(3*a + 2*b)*(a + b + (a - b)*Cos[2*(e + f*x)])*Sin[2*(e + f*x)] - 6*(4*a + 7*b)*(a + b + (a -
b)*Cos[2*(e + f*x)])^2*Sin[2*(e + f*x)] + 3*(a - b)*(a + b + (a - b)*Cos[2*(e + f*x)])^2*Sin[4*(e + f*x)])))/
(Sqrt[2]*a*(a - b)^5*f*(a + b + (a - b)*Cos[2*(e + f*x)])^2)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(7942\) vs. \(2(222)=444\).
time = 2.51, size = 7943, normalized size = 32.29

method	result	size
default	\(\text {Expression too large to display}\)	\(7943\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^4/(a+b*tan(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: sign: argument cannot be imaginary
; found %i

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sin ^{4}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**4/(a+b*tan(f*x+e)**2)**(5/2),x)

[Out]

Integral(sin(e + f*x)**4/(a + b*tan(e + f*x)**2)**(5/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

integrate(sin(f*x + e)^4/(b*tan(f*x + e)^2 + a)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\sin \left (e+f\,x\right )}^4}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^4/(a + b*tan(e + f*x)^2)^(5/2),x)

[Out]

int(sin(e + f*x)^4/(a + b*tan(e + f*x)^2)^(5/2), x)

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