3.6.38 \(\int \frac {\tan ^4(e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\) [538]

Optimal. Leaf size=348 \[ \frac {(5 a-3 b) b \cos (e+f x) \sin (e+f x)}{3 (a+b)^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {8 (a-b) b \cos (e+f x) \sin (e+f x)}{3 (a+b)^4 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {8 (a-b) \sqrt {\cos ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 (a+b)^4 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}-\frac {(5 a-3 b) \sqrt {\cos ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{3 (a+b)^3 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 (2 a-b) \tan (e+f x)}{3 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\sec ^2(e+f x) \tan (e+f x)}{3 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

[Out]

1/3*(5*a-3*b)*b*cos(f*x+e)*sin(f*x+e)/(a+b)^3/f/(a+b*sin(f*x+e)^2)^(3/2)+8/3*(a-b)*b*cos(f*x+e)*sin(f*x+e)/(a+
b)^4/f/(a+b*sin(f*x+e)^2)^(1/2)+8/3*(a-b)*EllipticE(sin(f*x+e),(-b/a)^(1/2))*sec(f*x+e)*(cos(f*x+e)^2)^(1/2)*(
a+b*sin(f*x+e)^2)^(1/2)/(a+b)^4/f/(1+b*sin(f*x+e)^2/a)^(1/2)-1/3*(5*a-3*b)*EllipticF(sin(f*x+e),(-b/a)^(1/2))*
sec(f*x+e)*(cos(f*x+e)^2)^(1/2)*(1+b*sin(f*x+e)^2/a)^(1/2)/(a+b)^3/f/(a+b*sin(f*x+e)^2)^(1/2)-2/3*(2*a-b)*tan(
f*x+e)/(a+b)^2/f/(a+b*sin(f*x+e)^2)^(3/2)+1/3*sec(f*x+e)^2*tan(f*x+e)/(a+b)/f/(a+b*sin(f*x+e)^2)^(3/2)

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Rubi [A]
time = 0.29, antiderivative size = 348, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3275, 481, 541, 538, 437, 435, 432, 430} \begin {gather*} -\frac {(5 a-3 b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} F\left (\text {ArcSin}(\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{3 f (a+b)^3 \sqrt {a+b \sin ^2(e+f x)}}+\frac {8 (a-b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} E\left (\text {ArcSin}(\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{3 f (a+b)^4 \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}-\frac {2 (2 a-b) \tan (e+f x)}{3 f (a+b)^2 \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {8 b (a-b) \sin (e+f x) \cos (e+f x)}{3 f (a+b)^4 \sqrt {a+b \sin ^2(e+f x)}}+\frac {b (5 a-3 b) \sin (e+f x) \cos (e+f x)}{3 f (a+b)^3 \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\tan (e+f x) \sec ^2(e+f x)}{3 f (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((5*a - 3*b)*b*Cos[e + f*x]*Sin[e + f*x])/(3*(a + b)^3*f*(a + b*Sin[e + f*x]^2)^(3/2)) + (8*(a - b)*b*Cos[e +
f*x]*Sin[e + f*x])/(3*(a + b)^4*f*Sqrt[a + b*Sin[e + f*x]^2]) + (8*(a - b)*Sqrt[Cos[e + f*x]^2]*EllipticE[ArcS
in[Sin[e + f*x]], -(b/a)]*Sec[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(3*(a + b)^4*f*Sqrt[1 + (b*Sin[e + f*x]^2)/
a]) - ((5*a - 3*b)*Sqrt[Cos[e + f*x]^2]*EllipticF[ArcSin[Sin[e + f*x]], -(b/a)]*Sec[e + f*x]*Sqrt[1 + (b*Sin[e
+ f*x]^2)/a])/(3*(a + b)^3*f*Sqrt[a + b*Sin[e + f*x]^2]) - (2*(2*a - b)*Tan[e + f*x])/(3*(a + b)^2*f*(a + b*S
in[e + f*x]^2)^(3/2)) + (Sec[e + f*x]^2*Tan[e + f*x])/(3*(a + b)*f*(a + b*Sin[e + f*x]^2)^(3/2))

Rule 430

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]
))*EllipticF[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && Gt
Q[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])

Rule 432

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 + (d/c)*x^2]/Sqrt[c + d*
x^2], Int[1/(Sqrt[a + b*x^2]*Sqrt[1 + (d/c)*x^2]), x], x] /; FreeQ[{a, b, c, d}, x] &&  !GtQ[c, 0]

Rule 435

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*Ell
ipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0
]

Rule 437

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[a + b*x^2]/Sqrt[1 + (b/a)*x^2]
, Int[Sqrt[1 + (b/a)*x^2]/Sqrt[c + d*x^2], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &&  !GtQ
[a, 0]

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 538

Int[((e_) + (f_.)*(x_)^(n_))/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/
b, Int[Sqrt[a + b*x^n]/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]),
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&  !(EqQ[n, 2] && ((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[
d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-b/a, -d/c]))))))

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 3275

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = FreeF
actors[Sin[e + f*x], x]}, Dist[ff^(m + 1)*(Sqrt[Cos[e + f*x]^2]/(f*Cos[e + f*x])), Subst[Int[x^m*((a + b*ff^2*
x^2)^p/(1 - ff^2*x^2)^((m + 1)/2)), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2]
&&  !IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {\tan ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx &=\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {x^4}{\left (1-x^2\right )^{5/2} \left (a+b x^2\right )^{5/2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\sec ^2(e+f x) \tan (e+f x)}{3 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {a+(3 a-2 b) x^2}{\left (1-x^2\right )^{3/2} \left (a+b x^2\right )^{5/2}} \, dx,x,\sin (e+f x)\right )}{3 (a+b) f}\\ &=-\frac {2 (2 a-b) \tan (e+f x)}{3 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\sec ^2(e+f x) \tan (e+f x)}{3 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {-3 a (a-b)+6 (2 a-b) b x^2}{\sqrt {1-x^2} \left (a+b x^2\right )^{5/2}} \, dx,x,\sin (e+f x)\right )}{3 (a+b)^2 f}\\ &=\frac {(5 a-3 b) b \cos (e+f x) \sin (e+f x)}{3 (a+b)^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {2 (2 a-b) \tan (e+f x)}{3 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\sec ^2(e+f x) \tan (e+f x)}{3 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {3 a^2 (3 a-5 b)-3 a (5 a-3 b) b x^2}{\sqrt {1-x^2} \left (a+b x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{9 a (a+b)^3 f}\\ &=\frac {(5 a-3 b) b \cos (e+f x) \sin (e+f x)}{3 (a+b)^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {8 (a-b) b \cos (e+f x) \sin (e+f x)}{3 (a+b)^4 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 (2 a-b) \tan (e+f x)}{3 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\sec ^2(e+f x) \tan (e+f x)}{3 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {-3 a^2 (a-3 b) (3 a-b)-24 a^2 (a-b) b x^2}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{9 a^2 (a+b)^4 f}\\ &=\frac {(5 a-3 b) b \cos (e+f x) \sin (e+f x)}{3 (a+b)^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {8 (a-b) b \cos (e+f x) \sin (e+f x)}{3 (a+b)^4 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 (2 a-b) \tan (e+f x)}{3 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\sec ^2(e+f x) \tan (e+f x)}{3 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\left (8 (a-b) \sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {\sqrt {a+b x^2}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{3 (a+b)^4 f}-\frac {\left ((5 a-3 b) \sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{3 (a+b)^3 f}\\ &=\frac {(5 a-3 b) b \cos (e+f x) \sin (e+f x)}{3 (a+b)^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {8 (a-b) b \cos (e+f x) \sin (e+f x)}{3 (a+b)^4 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 (2 a-b) \tan (e+f x)}{3 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\sec ^2(e+f x) \tan (e+f x)}{3 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\left (8 (a-b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {\sqrt {1+\frac {b x^2}{a}}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{3 (a+b)^4 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}-\frac {\left ((5 a-3 b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+\frac {b x^2}{a}}} \, dx,x,\sin (e+f x)\right )}{3 (a+b)^3 f \sqrt {a+b \sin ^2(e+f x)}}\\ &=\frac {(5 a-3 b) b \cos (e+f x) \sin (e+f x)}{3 (a+b)^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {8 (a-b) b \cos (e+f x) \sin (e+f x)}{3 (a+b)^4 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {8 (a-b) \sqrt {\cos ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 (a+b)^4 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}-\frac {(5 a-3 b) \sqrt {\cos ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{3 (a+b)^3 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 (2 a-b) \tan (e+f x)}{3 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\sec ^2(e+f x) \tan (e+f x)}{3 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 2.33, size = 235, normalized size = 0.68 \begin {gather*} \frac {2 a b \left (\frac {2 a+b-b \cos (2 (e+f x))}{a}\right )^{3/2} \left (8 a (a-b) E\left (e+f x\left |-\frac {b}{a}\right .\right )+\left (-5 a^2-2 a b+3 b^2\right ) F\left (e+f x\left |-\frac {b}{a}\right .\right )\right )+\sqrt {2} b \left (2 a b (a+b) \sin (2 (e+f x))+4 (a-b) b (2 a+b-b \cos (2 (e+f x))) \sin (2 (e+f x))-4 (a-b) (2 a+b-b \cos (2 (e+f x)))^2 \tan (e+f x)+(a+b) (2 a+b-b \cos (2 (e+f x)))^2 \sec ^2(e+f x) \tan (e+f x)\right )}{6 b (a+b)^4 f (2 a+b-b \cos (2 (e+f x)))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*b*((2*a + b - b*Cos[2*(e + f*x)])/a)^(3/2)*(8*a*(a - b)*EllipticE[e + f*x, -(b/a)] + (-5*a^2 - 2*a*b + 3*
b^2)*EllipticF[e + f*x, -(b/a)]) + Sqrt[2]*b*(2*a*b*(a + b)*Sin[2*(e + f*x)] + 4*(a - b)*b*(2*a + b - b*Cos[2*
(e + f*x)])*Sin[2*(e + f*x)] - 4*(a - b)*(2*a + b - b*Cos[2*(e + f*x)])^2*Tan[e + f*x] + (a + b)*(2*a + b - b*
Cos[2*(e + f*x)])^2*Sec[e + f*x]^2*Tan[e + f*x]))/(6*b*(a + b)^4*f*(2*a + b - b*Cos[2*(e + f*x)])^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(666\) vs. \(2(318)=636\).
time = 26.05, size = 667, normalized size = 1.92

method	result	size
default	\(-\frac {-8 \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, b^{2} \left (a -b \right ) \sin \left (f x +e \right ) \left (\cos ^{6}\left (f x +e \right )\right )+\sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, b \left (13 a^{2}+2 a b -11 b^{2}\right ) \left (\cos ^{4}\left (f x +e \right )\right ) \sin \left (f x +e \right )-2 \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (2 a^{3}+3 a^{2} b -b^{3}\right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+\sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \sin \left (f x +e \right )+\sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, b \left (5 \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2}+2 \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a b -3 \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b^{2}-8 \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2}+8 \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a b \right ) \left (\cos ^{4}\left (f x +e \right )\right )-\sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, \left (5 \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{3}+7 \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2} b -\EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a \,b^{2}-3 \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b^{3}-8 \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{3}+8 \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a \,b^{2}\right ) \left (\cos ^{2}\left (f x +e \right )\right )}{3 \left (1+\sin \left (f x +e \right )\right ) \sqrt {-\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right ) \left (\sin \left (f x +e \right )-1\right ) \left (1+\sin \left (f x +e \right )\right )}\, \left (\sin \left (f x +e \right )-1\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}} \left (a +b \right )^{4} \cos \left (f x +e \right ) f}\)	\(667\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(-8*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*b^2*(a-b)*sin(f*x+e)*cos(f*x+e)^6+(-b*cos(f*x+e)^4+(a+b)*c
os(f*x+e)^2)^(1/2)*b*(13*a^2+2*a*b-11*b^2)*cos(f*x+e)^4*sin(f*x+e)-2*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2
)*(2*a^3+3*a^2*b-b^3)*cos(f*x+e)^2*sin(f*x+e)+(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*(a^3+3*a^2*b+3*a*b^2+
b^3)*sin(f*x+e)+(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1
/2)*b*(5*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a^2+2*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a*b-3*EllipticF(sin(f
*x+e),(-1/a*b)^(1/2))*b^2-8*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a^2+8*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a*
b)*cos(f*x+e)^4-(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1
/2)*(5*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a^3+7*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a^2*b-EllipticF(sin(f*x
+e),(-1/a*b)^(1/2))*a*b^2-3*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*b^3-8*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a^
3+8*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a*b^2)*cos(f*x+e)^2)/(1+sin(f*x+e))/(-(a+b*sin(f*x+e)^2)*(sin(f*x+e)-
1)*(1+sin(f*x+e)))^(1/2)/(sin(f*x+e)-1)/(a+b*sin(f*x+e)^2)^(3/2)/(a+b)^4/cos(f*x+e)/f

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

[Out]

Timed out

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Fricas [C] Result contains complex when optimal does not.
time = 0.36, size = 1730, normalized size = 4.97 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(4*(2*((-I*a*b^4 + I*b^5)*cos(f*x + e)^7 + 2*(I*a^2*b^3 - I*b^5)*cos(f*x + e)^5 + (-I*a^3*b^2 - I*a^2*b^3
+ I*a*b^4 + I*b^5)*cos(f*x + e)^3)*sqrt(-b)*sqrt((a^2 + a*b)/b^2) + ((-2*I*a^2*b^3 + I*a*b^4 + I*b^5)*cos(f*x
+ e)^7 + 2*(2*I*a^3*b^2 + I*a^2*b^3 - 2*I*a*b^4 - I*b^5)*cos(f*x + e)^5 + (-2*I*a^4*b - 3*I*a^3*b^2 + I*a^2*b
^3 + 3*I*a*b^4 + I*b^5)*cos(f*x + e)^3)*sqrt(-b))*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_e(arc
sin(sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*(cos(f*x + e) + I*sin(f*x + e))), (8*a^2 + 8*a*b + b^2 - 4*(
2*a*b + b^2)*sqrt((a^2 + a*b)/b^2))/b^2) + 4*(2*((I*a*b^4 - I*b^5)*cos(f*x + e)^7 + 2*(-I*a^2*b^3 + I*b^5)*cos
(f*x + e)^5 + (I*a^3*b^2 + I*a^2*b^3 - I*a*b^4 - I*b^5)*cos(f*x + e)^3)*sqrt(-b)*sqrt((a^2 + a*b)/b^2) + ((2*I
*a^2*b^3 - I*a*b^4 - I*b^5)*cos(f*x + e)^7 + 2*(-2*I*a^3*b^2 - I*a^2*b^3 + 2*I*a*b^4 + I*b^5)*cos(f*x + e)^5 +
(2*I*a^4*b + 3*I*a^3*b^2 - I*a^2*b^3 - 3*I*a*b^4 - I*b^5)*cos(f*x + e)^3)*sqrt(-b))*sqrt((2*b*sqrt((a^2 + a*b
)/b^2) + 2*a + b)/b)*elliptic_e(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*(cos(f*x + e) - I*sin(f*x
+ e))), (8*a^2 + 8*a*b + b^2 - 4*(2*a*b + b^2)*sqrt((a^2 + a*b)/b^2))/b^2) - (2*((-3*I*a^2*b^3 + 2*I*a*b^4 +
5*I*b^5)*cos(f*x + e)^7 - 2*(-3*I*a^3*b^2 - I*a^2*b^3 + 7*I*a*b^4 + 5*I*b^5)*cos(f*x + e)^5 + (-3*I*a^4*b - 4*
I*a^3*b^2 + 6*I*a^2*b^3 + 12*I*a*b^4 + 5*I*b^5)*cos(f*x + e)^3)*sqrt(-b)*sqrt((a^2 + a*b)/b^2) - ((-6*I*a^3*b^
2 + 17*I*a^2*b^3 + 4*I*a*b^4 - 3*I*b^5)*cos(f*x + e)^7 + 2*(6*I*a^4*b - 11*I*a^3*b^2 - 21*I*a^2*b^3 - I*a*b^4
+ 3*I*b^5)*cos(f*x + e)^5 + (-6*I*a^5 + 5*I*a^4*b + 32*I*a^3*b^2 + 22*I*a^2*b^3 - 2*I*a*b^4 - 3*I*b^5)*cos(f*x
+ e)^3)*sqrt(-b))*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_f(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/
b^2) + 2*a + b)/b)*(cos(f*x + e) + I*sin(f*x + e))), (8*a^2 + 8*a*b + b^2 - 4*(2*a*b + b^2)*sqrt((a^2 + a*b)/b
^2))/b^2) - (2*((3*I*a^2*b^3 - 2*I*a*b^4 - 5*I*b^5)*cos(f*x + e)^7 - 2*(3*I*a^3*b^2 + I*a^2*b^3 - 7*I*a*b^4 -
5*I*b^5)*cos(f*x + e)^5 + (3*I*a^4*b + 4*I*a^3*b^2 - 6*I*a^2*b^3 - 12*I*a*b^4 - 5*I*b^5)*cos(f*x + e)^3)*sqrt(
-b)*sqrt((a^2 + a*b)/b^2) - ((6*I*a^3*b^2 - 17*I*a^2*b^3 - 4*I*a*b^4 + 3*I*b^5)*cos(f*x + e)^7 + 2*(-6*I*a^4*b
+ 11*I*a^3*b^2 + 21*I*a^2*b^3 + I*a*b^4 - 3*I*b^5)*cos(f*x + e)^5 + (6*I*a^5 - 5*I*a^4*b - 32*I*a^3*b^2 - 22*
I*a^2*b^3 + 2*I*a*b^4 + 3*I*b^5)*cos(f*x + e)^3)*sqrt(-b))*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*ellip
tic_f(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*(cos(f*x + e) - I*sin(f*x + e))), (8*a^2 + 8*a*b +
b^2 - 4*(2*a*b + b^2)*sqrt((a^2 + a*b)/b^2))/b^2) + (8*(a*b^4 - b^5)*cos(f*x + e)^6 - a^3*b^2 - 3*a^2*b^3 - 3*
a*b^4 - b^5 - (13*a^2*b^3 + 2*a*b^4 - 11*b^5)*cos(f*x + e)^4 + 2*(2*a^3*b^2 + 3*a^2*b^3 - b^5)*cos(f*x + e)^2)
*sqrt(-b*cos(f*x + e)^2 + a + b)*sin(f*x + e))/((a^4*b^4 + 4*a^3*b^5 + 6*a^2*b^6 + 4*a*b^7 + b^8)*f*cos(f*x +
e)^7 - 2*(a^5*b^3 + 5*a^4*b^4 + 10*a^3*b^5 + 10*a^2*b^6 + 5*a*b^7 + b^8)*f*cos(f*x + e)^5 + (a^6*b^2 + 6*a^5*b
^3 + 15*a^4*b^4 + 20*a^3*b^5 + 15*a^2*b^6 + 6*a*b^7 + b^8)*f*cos(f*x + e)^3)

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

[Out]

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

[Out]

integrate(tan(f*x + e)^4/(b*sin(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 {{\mathrm {tan}\left (e+f\,x\right )}^4}{{\left (b\,{\sin \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(tan(e + f*x)^4/(a + b*sin(e + f*x)^2)^(5/2),x)

[Out]

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

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