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]

-1/5*(5*a^2+10*a*b+b^2)*cos(f*x+e)/(a-b)^3/f/(a-b+b*sec(f*x+e)^2)^(3/2)+2/15*(5*a-b)*cos(f*x+e)^3/(a-b)^2/f/(a
-b+b*sec(f*x+e)^2)^(3/2)-1/5*cos(f*x+e)^5/(a-b)/f/(a-b+b*sec(f*x+e)^2)^(3/2)-4/15*b*(5*a^2+10*a*b+b^2)*sec(f*x
+e)/(a-b)^4/f/(a-b+b*sec(f*x+e)^2)^(3/2)-8/15*b*(5*a^2+10*a*b+b^2)*sec(f*x+e)/(a-b)^5/f/(a-b+b*sec(f*x+e)^2)^(
1/2)

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Rubi [A]  time = 0.23, antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, 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 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]

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 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 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 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 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]

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*}

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Mathematica [A]  time = 2.43, size = 294, normalized size = 1.19 \[ -\frac {\cos (e+f x) \left (-16 a^4 \cos (6 (e+f x))+3 a^4 \cos (8 (e+f x))+425 a^4+32 a^3 b \cos (6 (e+f x))-12 a^3 b \cos (8 (e+f x))+4700 a^3 b+18 a^2 b^2 \cos (8 (e+f x))+12 (a-b)^2 \left (7 a^2+50 a b+7 b^2\right ) \cos (4 (e+f x))+6134 a^2 b^2+48 \left (11 a^4+106 a^3 b-106 a b^3-11 b^4\right ) \cos (2 (e+f x))-32 a b^3 \cos (6 (e+f x))-12 a b^3 \cos (8 (e+f x))+4700 a b^3+16 b^4 \cos (6 (e+f x))+3 b^4 \cos (8 (e+f x))+425 b^4\right ) \sqrt {\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}}{480 \sqrt {2} f (a-b)^5 ((a-b) \cos (2 (e+f x))+a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/480*(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 - 10
6*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])/(Sqrt[2]*(a - b)^5*f*(a + b + (a - b)*Cos[2*(e
+ f*x)])^2)

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fricas [A]  time = 0.88, size = 370, normalized size = 1.49 \[ -\frac {{\left (3 \, {\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{9} - 2 \, {\left (5 \, a^{4} - 16 \, a^{3} b + 18 \, a^{2} b^{2} - 8 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{7} + 3 \, {\left (5 \, a^{4} - 14 \, a^{2} b^{2} + 8 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{5} + 12 \, {\left (5 \, a^{3} b + 5 \, a^{2} b^{2} - 9 \, a b^{3} - b^{4}\right )} \cos \left (f x + e\right )^{3} + 8 \, {\left (5 \, a^{2} b^{2} + 10 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, {\left ({\left (a^{7} - 7 \, a^{6} b + 21 \, a^{5} b^{2} - 35 \, a^{4} b^{3} + 35 \, a^{3} b^{4} - 21 \, a^{2} b^{5} + 7 \, a b^{6} - b^{7}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b - 6 \, a^{5} b^{2} + 15 \, a^{4} b^{3} - 20 \, a^{3} b^{4} + 15 \, a^{2} b^{5} - 6 \, a b^{6} + b^{7}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} b^{2} - 5 \, a^{4} b^{3} + 10 \, a^{3} b^{4} - 10 \, a^{2} b^{5} + 5 \, a b^{6} - b^{7}\right )} f\right )}} \]

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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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maple [A]  time = 4.52, size = 391, normalized size = 1.58 \[ \frac {\left (a -b \right )^{2} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right ) \left (3 \left (\cos ^{8}\left (f x +e \right )\right ) a^{4}-12 \left (\cos ^{8}\left (f x +e \right )\right ) a^{3} b +18 \left (\cos ^{8}\left (f x +e \right )\right ) a^{2} b^{2}-12 \left (\cos ^{8}\left (f x +e \right )\right ) a \,b^{3}+3 \left (\cos ^{8}\left (f x +e \right )\right ) b^{4}-10 \left (\cos ^{6}\left (f x +e \right )\right ) a^{4}+32 \left (\cos ^{6}\left (f x +e \right )\right ) a^{3} b -36 \left (\cos ^{6}\left (f x +e \right )\right ) a^{2} b^{2}+16 \left (\cos ^{6}\left (f x +e \right )\right ) a \,b^{3}-2 \left (\cos ^{6}\left (f x +e \right )\right ) b^{4}+15 \left (\cos ^{4}\left (f x +e \right )\right ) a^{4}-42 a^{2} b^{2} \left (\cos ^{4}\left (f x +e \right )\right )+24 \left (\cos ^{4}\left (f x +e \right )\right ) a \,b^{3}+3 \left (\cos ^{4}\left (f x +e \right )\right ) b^{4}+60 \left (\cos ^{2}\left (f x +e \right )\right ) a^{3} b +60 \left (\cos ^{2}\left (f x +e \right )\right ) a^{2} b^{2}-108 \left (\cos ^{2}\left (f x +e \right )\right ) a \,b^{3}-12 \left (\cos ^{2}\left (f x +e \right )\right ) b^{4}+40 a^{2} b^{2}+80 a \,b^{3}+8 b^{4}\right ) \sqrt {4}\, a^{7}}{30 f \left (\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\cos \left (f x +e \right )^{2}}\right )^{\frac {5}{2}} \cos \left (f x +e \right )^{5} \left (\sqrt {-\left (a -b \right ) b}+a -b \right )^{7} \left (\sqrt {-\left (a -b \right ) b}-a +b \right )^{7}} \]

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-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*a^2*b^2*cos(f*x+e)^4+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)*a^7/((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(5/2)/cos(f*x+e)^5/((-(a-
b)*b)^(1/2)+a-b)^7/((-(a-b)*b)^(1/2)-a+b)^7

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maxima [B]  time = 0.72, size = 534, normalized size = 2.15 \[ -\frac {\frac {15 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {3 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {5}{2}} \cos \left (f x + e\right )^{5} - 20 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} b \cos \left (f x + e\right )^{3} + 90 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} b^{2} \cos \left (f x + e\right )}{a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}} - \frac {10 \, {\left ({\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} - 9 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )\right )}}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} + \frac {5 \, {\left (12 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} b^{3} \cos \left (f x + e\right )^{2} - b^{4}\right )}}{{\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3}} + \frac {10 \, {\left (9 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} b^{2} \cos \left (f x + e\right )^{2} - b^{3}\right )}}{{\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3}} + \frac {5 \, {\left (6 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} b \cos \left (f x + e\right )^{2} - b^{2}\right )}}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3}}}{15 \, f} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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