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

Optimal. Leaf size=91 \[ -\frac {1}{f (a+b)^2 \sqrt {a+b \sin ^2(e+f x)}}-\frac {1}{3 f (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{f (a+b)^{5/2}} \]

[Out]

arctanh((a+b*sin(f*x+e)^2)^(1/2)/(a+b)^(1/2))/(a+b)^(5/2)/f-1/3/(a+b)/f/(a+b*sin(f*x+e)^2)^(3/2)-1/(a+b)^2/f/(
a+b*sin(f*x+e)^2)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3194, 51, 63, 208} \[ -\frac {1}{f (a+b)^2 \sqrt {a+b \sin ^2(e+f x)}}-\frac {1}{3 f (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{f (a+b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]]/((a + b)^(5/2)*f) - 1/(3*(a + b)*f*(a + b*Sin[e + f*x]^2)^(3/2
)) - 1/((a + b)^2*f*Sqrt[a + b*Sin[e + f*x]^2])

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 3194

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 0.08, size = 56, normalized size = 0.62 \[ -\frac {\, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};1-\frac {b \cos ^2(e+f x)}{a+b}\right )}{3 f (a+b) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*Hypergeometric2F1[-3/2, 1, -1/2, 1 - (b*Cos[e + f*x]^2)/(a + b)]/((a + b)*f*(a + b - b*Cos[e + f*x]^2)^(3
/2))

________________________________________________________________________________________

fricas [B]  time = 0.60, size = 521, normalized size = 5.73 \[ \left [\frac {3 \, {\left (b^{2} \cos \left (f x + e\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a + b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) + 2 \, {\left (3 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, a^{2} - 8 \, a b - 4 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{6 \, {\left ({\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 4 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} f\right )}}, -\frac {3 \, {\left (b^{2} \cos \left (f x + e\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{a + b}\right ) - {\left (3 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, a^{2} - 8 \, a b - 4 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{3 \, {\left ({\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 4 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*(b^2*cos(f*x + e)^4 - 2*(a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^2)*sqrt(a + b)*log((b*cos(f*x + e
)^2 - 2*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a + b) - 2*a - 2*b)/cos(f*x + e)^2) + 2*(3*(a*b + b^2)*cos(f*x +
e)^2 - 4*a^2 - 8*a*b - 4*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*f*cos(f*
x + e)^4 - 2*(a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 4*a*b^4 + b^5)*f*cos(f*x + e)^2 + (a^5 + 5*a^4*b + 10*a^3*b^2 +
10*a^2*b^3 + 5*a*b^4 + b^5)*f), -1/3*(3*(b^2*cos(f*x + e)^4 - 2*(a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^2
)*sqrt(-a - b)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a - b)/(a + b)) - (3*(a*b + b^2)*cos(f*x + e)^2 -
4*a^2 - 8*a*b - 4*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*f*cos(f*x + e)^
4 - 2*(a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 4*a*b^4 + b^5)*f*cos(f*x + e)^2 + (a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*
b^3 + 5*a*b^4 + b^5)*f)]

________________________________________________________________________________________

giac [B]  time = 1.22, size = 842, normalized size = 9.25 \[ -\frac {\frac {{\left ({\left (\frac {{\left (4 \, a^{9} b^{2} + 33 \, a^{8} b^{3} + 120 \, a^{7} b^{4} + 252 \, a^{6} b^{5} + 336 \, a^{5} b^{6} + 294 \, a^{4} b^{7} + 168 \, a^{3} b^{8} + 60 \, a^{2} b^{9} + 12 \, a b^{10} + b^{11}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{a^{10} b^{2} + 10 \, a^{9} b^{3} + 45 \, a^{8} b^{4} + 120 \, a^{7} b^{5} + 210 \, a^{6} b^{6} + 252 \, a^{5} b^{7} + 210 \, a^{4} b^{8} + 120 \, a^{3} b^{9} + 45 \, a^{2} b^{10} + 10 \, a b^{11} + b^{12}} + \frac {3 \, {\left (4 \, a^{9} b^{2} + 37 \, a^{8} b^{3} + 152 \, a^{7} b^{4} + 364 \, a^{6} b^{5} + 560 \, a^{5} b^{6} + 574 \, a^{4} b^{7} + 392 \, a^{3} b^{8} + 172 \, a^{2} b^{9} + 44 \, a b^{10} + 5 \, b^{11}\right )}}{a^{10} b^{2} + 10 \, a^{9} b^{3} + 45 \, a^{8} b^{4} + 120 \, a^{7} b^{5} + 210 \, a^{6} b^{6} + 252 \, a^{5} b^{7} + 210 \, a^{4} b^{8} + 120 \, a^{3} b^{9} + 45 \, a^{2} b^{10} + 10 \, a b^{11} + b^{12}}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + \frac {3 \, {\left (4 \, a^{9} b^{2} + 37 \, a^{8} b^{3} + 152 \, a^{7} b^{4} + 364 \, a^{6} b^{5} + 560 \, a^{5} b^{6} + 574 \, a^{4} b^{7} + 392 \, a^{3} b^{8} + 172 \, a^{2} b^{9} + 44 \, a b^{10} + 5 \, b^{11}\right )}}{a^{10} b^{2} + 10 \, a^{9} b^{3} + 45 \, a^{8} b^{4} + 120 \, a^{7} b^{5} + 210 \, a^{6} b^{6} + 252 \, a^{5} b^{7} + 210 \, a^{4} b^{8} + 120 \, a^{3} b^{9} + 45 \, a^{2} b^{10} + 10 \, a b^{11} + b^{12}}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + \frac {4 \, a^{9} b^{2} + 33 \, a^{8} b^{3} + 120 \, a^{7} b^{4} + 252 \, a^{6} b^{5} + 336 \, a^{5} b^{6} + 294 \, a^{4} b^{7} + 168 \, a^{3} b^{8} + 60 \, a^{2} b^{9} + 12 \, a b^{10} + b^{11}}{a^{10} b^{2} + 10 \, a^{9} b^{3} + 45 \, a^{8} b^{4} + 120 \, a^{7} b^{5} + 210 \, a^{6} b^{6} + 252 \, a^{5} b^{7} + 210 \, a^{4} b^{8} + 120 \, a^{3} b^{9} + 45 \, a^{2} b^{10} + 10 \, a b^{11} + b^{12}}}{{\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a\right )}^{\frac {3}{2}}} + \frac {6 \, \arctan \left (-\frac {\sqrt {a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a} - \sqrt {a}}{2 \, \sqrt {-a - b}}\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-a - b}}}{3 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(((((4*a^9*b^2 + 33*a^8*b^3 + 120*a^7*b^4 + 252*a^6*b^5 + 336*a^5*b^6 + 294*a^4*b^7 + 168*a^3*b^8 + 60*a^
2*b^9 + 12*a*b^10 + b^11)*tan(1/2*f*x + 1/2*e)^2/(a^10*b^2 + 10*a^9*b^3 + 45*a^8*b^4 + 120*a^7*b^5 + 210*a^6*b
^6 + 252*a^5*b^7 + 210*a^4*b^8 + 120*a^3*b^9 + 45*a^2*b^10 + 10*a*b^11 + b^12) + 3*(4*a^9*b^2 + 37*a^8*b^3 + 1
52*a^7*b^4 + 364*a^6*b^5 + 560*a^5*b^6 + 574*a^4*b^7 + 392*a^3*b^8 + 172*a^2*b^9 + 44*a*b^10 + 5*b^11)/(a^10*b
^2 + 10*a^9*b^3 + 45*a^8*b^4 + 120*a^7*b^5 + 210*a^6*b^6 + 252*a^5*b^7 + 210*a^4*b^8 + 120*a^3*b^9 + 45*a^2*b^
10 + 10*a*b^11 + b^12))*tan(1/2*f*x + 1/2*e)^2 + 3*(4*a^9*b^2 + 37*a^8*b^3 + 152*a^7*b^4 + 364*a^6*b^5 + 560*a
^5*b^6 + 574*a^4*b^7 + 392*a^3*b^8 + 172*a^2*b^9 + 44*a*b^10 + 5*b^11)/(a^10*b^2 + 10*a^9*b^3 + 45*a^8*b^4 + 1
20*a^7*b^5 + 210*a^6*b^6 + 252*a^5*b^7 + 210*a^4*b^8 + 120*a^3*b^9 + 45*a^2*b^10 + 10*a*b^11 + b^12))*tan(1/2*
f*x + 1/2*e)^2 + (4*a^9*b^2 + 33*a^8*b^3 + 120*a^7*b^4 + 252*a^6*b^5 + 336*a^5*b^6 + 294*a^4*b^7 + 168*a^3*b^8
+ 60*a^2*b^9 + 12*a*b^10 + b^11)/(a^10*b^2 + 10*a^9*b^3 + 45*a^8*b^4 + 120*a^7*b^5 + 210*a^6*b^6 + 252*a^5*b^
7 + 210*a^4*b^8 + 120*a^3*b^9 + 45*a^2*b^10 + 10*a*b^11 + b^12))/(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x +
1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a)^(3/2) + 6*arctan(-1/2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan
(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a) - sqrt(a))/sqrt(-a - b))/((
a^2 + 2*a*b + b^2)*sqrt(-a - b)))/f

________________________________________________________________________________________

maple [B]  time = 5.35, size = 898, normalized size = 9.87 \[ \frac {-8 a^{3} b^{3} \sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a \,b^{2}+b^{3}}{b^{2}}}\, \sqrt {a +b}-8 a^{2} b^{4} \sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a \,b^{2}+b^{3}}{b^{2}}}\, \sqrt {a +b}+3 a^{4} b^{3} \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right )+3 a^{4} b^{3} \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right )+3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{2} b^{5}+3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{2} b^{5}+6 a^{3} b^{4} \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right )+6 a^{3} b^{4} \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right )+3 a^{2} b^{5} \left (\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right )+\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right )\right ) \left (\cos ^{4}\left (f x +e \right )\right )+6 \left (\cos ^{2}\left (f x +e \right )\right ) a^{2} b^{4} \left (\sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a \,b^{2}+b^{3}}{b^{2}}}\, \sqrt {a +b}-\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a -\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) b -\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a -\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) b \right )}{6 b^{3} \sqrt {a +b}\, a^{2} \left (a^{2} b^{2} \left (\cos ^{4}\left (f x +e \right )\right )+2 a \,b^{3} \left (\cos ^{4}\left (f x +e \right )\right )+b^{4} \left (\cos ^{4}\left (f x +e \right )\right )-2 a^{3} b \left (\cos ^{2}\left (f x +e \right )\right )-6 a^{2} b^{2} \left (\cos ^{2}\left (f x +e \right )\right )-6 a \,b^{3} \left (\cos ^{2}\left (f x +e \right )\right )-2 b^{4} \left (\cos ^{2}\left (f x +e \right )\right )+a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right ) f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/b^3/(a+b)^(1/2)/a^2/(a^2*b^2*cos(f*x+e)^4+2*a*b^3*cos(f*x+e)^4+b^4*cos(f*x+e)^4-2*a^3*b*cos(f*x+e)^2-6*a^2
*b^2*cos(f*x+e)^2-6*a*b^3*cos(f*x+e)^2-2*b^4*cos(f*x+e)^2+a^4+4*a^3*b+6*a^2*b^2+4*a*b^3+b^4)*(-8*a^3*b^3*(-b*c
os(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*(a+b)^(1/2)-8*a^2*b^4*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*(a+b)^(1/2)+3
*a^4*b^3*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))+3*a^4*b^3*ln(2/(1+sin(f*
x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))+3*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f
*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^2*b^5+3*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*
x+e)+a))*a^2*b^5+6*a^3*b^4*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))+6*a^3*
b^4*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))+3*a^2*b^5*(ln(2/(sin(f*x+e)-1
)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))+ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^
2)^(1/2)-b*sin(f*x+e)+a)))*cos(f*x+e)^4+6*cos(f*x+e)^2*a^2*b^4*((-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*(a+b)^
(1/2)-ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a-ln(2/(1+sin(f*x+e))*((a+b
)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*b-ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1
/2)+b*sin(f*x+e)+a))*a-ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*b))/f

________________________________________________________________________________________

maxima [B]  time = 0.77, size = 203, normalized size = 2.23 \[ -\frac {\frac {2}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a + {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b} + \frac {6}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{2} + 2 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} a b + \sqrt {b \sin \left (f x + e\right )^{2} + a} b^{2}} + \frac {3 \, \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}}\right )}{{\left (a + b\right )}^{\frac {5}{2}}} - \frac {3 \, \operatorname {arsinh}\left (-\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}}\right )}{{\left (a + b\right )}^{\frac {5}{2}}}}{6 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2/((b*sin(f*x + e)^2 + a)^(3/2)*a + (b*sin(f*x + e)^2 + a)^(3/2)*b) + 6/(sqrt(b*sin(f*x + e)^2 + a)*a^2
+ 2*sqrt(b*sin(f*x + e)^2 + a)*a*b + sqrt(b*sin(f*x + e)^2 + a)*b^2) + 3*arcsinh(b*sin(f*x + e)/(sqrt(a*b)*(si
n(f*x + e) + 1)) - a/(sqrt(a*b)*(sin(f*x + e) + 1)))/(a + b)^(5/2) - 3*arcsinh(-b*sin(f*x + e)/(sqrt(a*b)*(sin
(f*x + e) - 1)) - a/(sqrt(a*b)*(sin(f*x + e) - 1)))/(a + b)^(5/2))/f

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {tan}\left (e+f\,x\right )}{{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________