3.140 \(\int \frac{\sin ^5(e+f x)}{(a+b \tan ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=248 \[ -\frac{8 b \left (5 a^2+10 a b+b^2\right ) \sec (e+f x)}{15 f (a-b)^5 \sqrt{a+b \sec ^2(e+f x)-b}}-\frac{4 b \left (5 a^2+10 a b+b^2\right ) \sec (e+f x)}{15 f (a-b)^4 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac{\left (5 a^2+10 a b+b^2\right ) \cos (e+f x)}{5 f (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac{\cos ^5(e+f x)}{5 f (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}+\frac{2 (5 a-b) \cos ^3(e+f x)}{15 f (a-b)^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}} \]

[Out]

-((5*a^2 + 10*a*b + b^2)*Cos[e + f*x])/(5*(a - b)^3*f*(a - b + b*Sec[e + f*x]^2)^(3/2)) + (2*(5*a - b)*Cos[e +
f*x]^3)/(15*(a - b)^2*f*(a - b + b*Sec[e + f*x]^2)^(3/2)) - Cos[e + f*x]^5/(5*(a - b)*f*(a - b + b*Sec[e + f*
x]^2)^(3/2)) - (4*b*(5*a^2 + 10*a*b + b^2)*Sec[e + f*x])/(15*(a - b)^4*f*(a - b + b*Sec[e + f*x]^2)^(3/2)) - (
8*b*(5*a^2 + 10*a*b + b^2)*Sec[e + f*x])/(15*(a - b)^5*f*Sqrt[a - b + b*Sec[e + f*x]^2])

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Rubi [A]  time = 0.231493, antiderivative size = 248, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3664, 462, 453, 271, 192, 191} \[ -\frac{8 b \left (5 a^2+10 a b+b^2\right ) \sec (e+f x)}{15 f (a-b)^5 \sqrt{a+b \sec ^2(e+f x)-b}}-\frac{4 b \left (5 a^2+10 a b+b^2\right ) \sec (e+f x)}{15 f (a-b)^4 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac{\left (5 a^2+10 a b+b^2\right ) \cos (e+f x)}{5 f (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac{\cos ^5(e+f x)}{5 f (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}+\frac{2 (5 a-b) \cos ^3(e+f x)}{15 f (a-b)^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((5*a^2 + 10*a*b + b^2)*Cos[e + f*x])/(5*(a - b)^3*f*(a - b + b*Sec[e + f*x]^2)^(3/2)) + (2*(5*a - b)*Cos[e +
f*x]^3)/(15*(a - b)^2*f*(a - b + b*Sec[e + f*x]^2)^(3/2)) - Cos[e + f*x]^5/(5*(a - b)*f*(a - b + b*Sec[e + f*
x]^2)^(3/2)) - (4*b*(5*a^2 + 10*a*b + b^2)*Sec[e + f*x])/(15*(a - b)^4*f*(a - b + b*Sec[e + f*x]^2)^(3/2)) - (
8*b*(5*a^2 + 10*a*b + b^2)*Sec[e + f*x])/(15*(a - b)^5*f*Sqrt[a - b + b*Sec[e + f*x]^2])

Rule 3664

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

Rule 462

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

Rule 453

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

Rule 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

\begin{align*} \int \frac{\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right )^2}{x^6 \left (a-b+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{-2 (5 a-b)+5 (a-b) x^2}{x^4 \left (a-b+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{5 (a-b) f}\\ &=\frac{2 (5 a-b) \cos ^3(e+f x)}{15 (a-b)^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}+\frac{\left (5 a^2+10 a b+b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a-b+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{5 (a-b)^2 f}\\ &=-\frac{\left (5 a^2+10 a b+b^2\right ) \cos (e+f x)}{5 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}+\frac{2 (5 a-b) \cos ^3(e+f x)}{15 (a-b)^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\left (4 b \left (5 a^2+10 a b+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\left (a-b+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{5 (a-b)^3 f}\\ &=-\frac{\left (5 a^2+10 a b+b^2\right ) \cos (e+f x)}{5 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}+\frac{2 (5 a-b) \cos ^3(e+f x)}{15 (a-b)^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{4 b \left (5 a^2+10 a b+b^2\right ) \sec (e+f x)}{15 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\left (8 b \left (5 a^2+10 a b+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\left (a-b+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{15 (a-b)^4 f}\\ &=-\frac{\left (5 a^2+10 a b+b^2\right ) \cos (e+f x)}{5 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}+\frac{2 (5 a-b) \cos ^3(e+f x)}{15 (a-b)^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{4 b \left (5 a^2+10 a b+b^2\right ) \sec (e+f x)}{15 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{8 b \left (5 a^2+10 a b+b^2\right ) \sec (e+f x)}{15 (a-b)^5 f \sqrt{a-b+b \sec ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 2.2824, size = 294, normalized size = 1.19 \[ -\frac{\cos (e+f x) \left (18 a^2 b^2 \cos (8 (e+f x))+48 \left (106 a^3 b+11 a^4-106 a b^3-11 b^4\right ) \cos (2 (e+f x))+12 (a-b)^2 \left (7 a^2+50 a b+7 b^2\right ) \cos (4 (e+f x))+6134 a^2 b^2+32 a^3 b \cos (6 (e+f x))-12 a^3 b \cos (8 (e+f x))+4700 a^3 b-16 a^4 \cos (6 (e+f x))+3 a^4 \cos (8 (e+f x))+425 a^4-32 a b^3 \cos (6 (e+f x))-12 a b^3 \cos (8 (e+f x))+4700 a b^3+16 b^4 \cos (6 (e+f x))+3 b^4 \cos (8 (e+f x))+425 b^4\right ) \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}}{480 \sqrt{2} f (a-b)^5 ((a-b) \cos (2 (e+f x))+a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cos[e + f*x]*(425*a^4 + 4700*a^3*b + 6134*a^2*b^2 + 4700*a*b^3 + 425*b^4 + 48*(11*a^4 + 106*a^3*b - 106*a*b^
3 - 11*b^4)*Cos[2*(e + f*x)] + 12*(a - b)^2*(7*a^2 + 50*a*b + 7*b^2)*Cos[4*(e + f*x)] - 16*a^4*Cos[6*(e + f*x)
] + 32*a^3*b*Cos[6*(e + f*x)] - 32*a*b^3*Cos[6*(e + f*x)] + 16*b^4*Cos[6*(e + f*x)] + 3*a^4*Cos[8*(e + f*x)] -
12*a^3*b*Cos[8*(e + f*x)] + 18*a^2*b^2*Cos[8*(e + f*x)] - 12*a*b^3*Cos[8*(e + f*x)] + 3*b^4*Cos[8*(e + f*x)])
*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(480*Sqrt[2]*(a - b)^5*f*(a + b + (a - b)*Cos[2*(e +
f*x)])^2)

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Maple [A]  time = 1.559, size = 391, normalized size = 1.6 \begin{align*}{\frac{{a}^{7} \left ( a-b \right ) ^{2} \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) \left ( 3\, \left ( \cos \left ( fx+e \right ) \right ) ^{8}{a}^{4}-12\, \left ( \cos \left ( fx+e \right ) \right ) ^{8}{a}^{3}b+18\, \left ( \cos \left ( fx+e \right ) \right ) ^{8}{a}^{2}{b}^{2}-12\, \left ( \cos \left ( fx+e \right ) \right ) ^{8}a{b}^{3}+3\, \left ( \cos \left ( fx+e \right ) \right ) ^{8}{b}^{4}-10\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}{a}^{4}+32\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}{a}^{3}b-36\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}{a}^{2}{b}^{2}+16\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}a{b}^{3}-2\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}{b}^{4}+15\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{4}-42\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{2}{b}^{2}+24\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}a{b}^{3}+3\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{b}^{4}+60\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{a}^{3}b+60\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{a}^{2}{b}^{2}-108\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}a{b}^{3}-12\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{b}^{4}+40\,{a}^{2}{b}^{2}+80\,a{b}^{3}+8\,{b}^{4} \right ) \sqrt{4}}{30\,f \left ( \cos \left ( fx+e \right ) \right ) ^{5}} \left ( \sqrt{-b \left ( a-b \right ) }+a-b \right ) ^{-7} \left ( \sqrt{-b \left ( a-b \right ) }-a+b \right ) ^{-7} \left ({\frac{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b}{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30/f*a^7/((-b*(a-b))^(1/2)+a-b)^7/((-b*(a-b))^(1/2)-a+b)^7*(a-b)^2*(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)*(3*cos(
f*x+e)^8*a^4-12*cos(f*x+e)^8*a^3*b+18*cos(f*x+e)^8*a^2*b^2-12*cos(f*x+e)^8*a*b^3+3*cos(f*x+e)^8*b^4-10*cos(f*x
+e)^6*a^4+32*cos(f*x+e)^6*a^3*b-36*cos(f*x+e)^6*a^2*b^2+16*cos(f*x+e)^6*a*b^3-2*cos(f*x+e)^6*b^4+15*cos(f*x+e)
^4*a^4-42*cos(f*x+e)^4*a^2*b^2+24*cos(f*x+e)^4*a*b^3+3*cos(f*x+e)^4*b^4+60*cos(f*x+e)^2*a^3*b+60*cos(f*x+e)^2*
a^2*b^2-108*cos(f*x+e)^2*a*b^3-12*cos(f*x+e)^2*b^4+40*a^2*b^2+80*a*b^3+8*b^4)*4^(1/2)/cos(f*x+e)^5/((a*cos(f*x
+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(5/2)

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Maxima [B]  time = 1.07457, size = 721, normalized size = 2.91 \begin{align*} -\frac{\frac{15 \, \sqrt{a - b + \frac{b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac{3 \,{\left (a - b + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{5}{2}} \cos \left (f x + e\right )^{5} - 20 \,{\left (a - b + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{3}{2}} b \cos \left (f x + e\right )^{3} + 90 \, \sqrt{a - b + \frac{b}{\cos \left (f x + e\right )^{2}}} b^{2} \cos \left (f x + e\right )}{a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}} - \frac{10 \,{\left ({\left (a - b + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{3}{2}} \cos \left (f x + e\right )^{3} - 9 \, \sqrt{a - b + \frac{b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )\right )}}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} + \frac{5 \,{\left (12 \,{\left (a - b + \frac{b}{\cos \left (f x + e\right )^{2}}\right )} b^{3} \cos \left (f x + e\right )^{2} - b^{4}\right )}}{{\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )}{\left (a - b + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{3}{2}} \cos \left (f x + e\right )^{3}} + \frac{10 \,{\left (9 \,{\left (a - b + \frac{b}{\cos \left (f x + e\right )^{2}}\right )} b^{2} \cos \left (f x + e\right )^{2} - b^{3}\right )}}{{\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )}{\left (a - b + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{3}{2}} \cos \left (f x + e\right )^{3}} + \frac{5 \,{\left (6 \,{\left (a - b + \frac{b}{\cos \left (f x + e\right )^{2}}\right )} b \cos \left (f x + e\right )^{2} - b^{2}\right )}}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}{\left (a - b + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{3}{2}} \cos \left (f x + e\right )^{3}}}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(15*sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (3*(a - b + b/cos(f*x
+ e)^2)^(5/2)*cos(f*x + e)^5 - 20*(a - b + b/cos(f*x + e)^2)^(3/2)*b*cos(f*x + e)^3 + 90*sqrt(a - b + b/cos(f*
x + e)^2)*b^2*cos(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) - 10*((a - b + b/cos(f*x
+ e)^2)^(3/2)*cos(f*x + e)^3 - 9*sqrt(a - b + b/cos(f*x + e)^2)*b*cos(f*x + e))/(a^4 - 4*a^3*b + 6*a^2*b^2 -
4*a*b^3 + b^4) + 5*(12*(a - b + b/cos(f*x + e)^2)*b^3*cos(f*x + e)^2 - b^4)/((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*
a^2*b^3 + 5*a*b^4 - b^5)*(a - b + b/cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3) + 10*(9*(a - b + b/cos(f*x + e)^2)*b
^2*cos(f*x + e)^2 - b^3)/((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*(a - b + b/cos(f*x + e)^2)^(3/2)*cos(f*x
+ e)^3) + 5*(6*(a - b + b/cos(f*x + e)^2)*b*cos(f*x + e)^2 - b^2)/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*(a - b + b
/cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3))/f

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Fricas [A]  time = 6.25133, size = 829, normalized size = 3.34 \begin{align*} -\frac{{\left (3 \,{\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{9} - 2 \,{\left (5 \, a^{4} - 16 \, a^{3} b + 18 \, a^{2} b^{2} - 8 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{7} + 3 \,{\left (5 \, a^{4} - 14 \, a^{2} b^{2} + 8 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{5} + 12 \,{\left (5 \, a^{3} b + 5 \, a^{2} b^{2} - 9 \, a b^{3} - b^{4}\right )} \cos \left (f x + e\right )^{3} + 8 \,{\left (5 \, a^{2} b^{2} + 10 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \,{\left ({\left (a^{7} - 7 \, a^{6} b + 21 \, a^{5} b^{2} - 35 \, a^{4} b^{3} + 35 \, a^{3} b^{4} - 21 \, a^{2} b^{5} + 7 \, a b^{6} - b^{7}\right )} f \cos \left (f x + e\right )^{4} + 2 \,{\left (a^{6} b - 6 \, a^{5} b^{2} + 15 \, a^{4} b^{3} - 20 \, a^{3} b^{4} + 15 \, a^{2} b^{5} - 6 \, a b^{6} + b^{7}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{5} b^{2} - 5 \, a^{4} b^{3} + 10 \, a^{3} b^{4} - 10 \, a^{2} b^{5} + 5 \, a b^{6} - b^{7}\right )} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(3*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^9 - 2*(5*a^4 - 16*a^3*b + 18*a^2*b^2 - 8*a*b
^3 + b^4)*cos(f*x + e)^7 + 3*(5*a^4 - 14*a^2*b^2 + 8*a*b^3 + b^4)*cos(f*x + e)^5 + 12*(5*a^3*b + 5*a^2*b^2 - 9
*a*b^3 - b^4)*cos(f*x + e)^3 + 8*(5*a^2*b^2 + 10*a*b^3 + b^4)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/
cos(f*x + e)^2)/((a^7 - 7*a^6*b + 21*a^5*b^2 - 35*a^4*b^3 + 35*a^3*b^4 - 21*a^2*b^5 + 7*a*b^6 - b^7)*f*cos(f*x
+ e)^4 + 2*(a^6*b - 6*a^5*b^2 + 15*a^4*b^3 - 20*a^3*b^4 + 15*a^2*b^5 - 6*a*b^6 + b^7)*f*cos(f*x + e)^2 + (a^5
*b^2 - 5*a^4*b^3 + 10*a^3*b^4 - 10*a^2*b^5 + 5*a*b^6 - b^7)*f)

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Sympy [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)**5/(a+b*tan(f*x+e)**2)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )^{5}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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