3.123 \(\int \frac {\sin ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\)

Optimal. Leaf size=93 \[ \frac {a \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 f (a-b)^{3/2}}-\frac {\sin (e+f x) \cos (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 f (a-b)} \]

[Out]

1/2*a*arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/(a-b)^(3/2)/f-1/2*cos(f*x+e)*sin(f*x+e)*(a+b*tan
(f*x+e)^2)^(1/2)/(a-b)/f

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Rubi [A]  time = 0.11, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3663, 471, 12, 377, 203} \[ \frac {a \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 f (a-b)^{3/2}}-\frac {\sin (e+f x) \cos (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 f (a-b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(2*(a - b)^(3/2)*f) - (Cos[e + f*x]*Sin[e +
f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(2*(a - b)*f)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 203

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

Rule 377

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 471

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 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] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 3663

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

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Mathematica [C]  time = 3.26, size = 270, normalized size = 2.90 \[ -\frac {\sin (2 (e+f x)) \sec ^2(e+f x) \left (\sqrt {2} a^2 \sqrt {\frac {\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} \Pi \left (-\frac {b}{a-b};\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right )+(a-b) ((a-b) \cos (2 (e+f x))+a+b)+\sqrt {2} a (b-a) \sqrt {\frac {\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right )\right )}{4 \sqrt {2} f (a-b)^2 \sqrt {\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(((a - b)*(a + b + (a - b)*Cos[2*(e + f*x)]) + Sqrt[2]*a*(-a + b)*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)]
)*Csc[e + f*x]^2)/b]*EllipticF[ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1]
+ Sqrt[2]*a^2*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*EllipticPi[-(b/(a - b)), ArcSin[Sqr
t[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1])*Sec[e + f*x]^2*Sin[2*(e + f*x)])/(Sqrt[
2]*(a - b)^2*f*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])

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fricas [B]  time = 0.70, size = 696, normalized size = 7.48 \[ \left [-\frac {8 \, {\left (a - b\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - a \sqrt {-a + b} \log \left (128 \, {\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 )^{8} - 256 \, {\left (a^{4} - 5 \, a^{3} b + 9 \, a^{2} b^{2} - 7 \, a b^{3} + 2 \, b^{4}\right )} \cos \left (f x + e\right )^{6} + 32 \, {\left (5 \, a^{4} - 34 \, a^{3} b + 77 \, a^{2} b^{2} - 72 \, a b^{3} + 24 \, b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{4} - 32 \, a^{3} b + 160 \, a^{2} b^{2} - 256 \, a b^{3} + 128 \, b^{4} - 32 \, {\left (a^{4} - 11 \, a^{3} b + 34 \, a^{2} b^{2} - 40 \, a b^{3} + 16 \, b^{4}\right )} \cos \left (f x + e\right )^{2} - 8 \, {\left (16 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{7} - 24 \, {\left (a^{3} - 4 \, a^{2} b + 5 \, a b^{2} - 2 \, b^{3}\right )} \cos \left (f x + e\right )^{5} + 2 \, {\left (5 \, a^{3} - 29 \, a^{2} b + 48 \, a b^{2} - 24 \, b^{3}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} - 10 \, a^{2} b + 24 \, a b^{2} - 16 \, b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-a + b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )\right )}{16 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} f}, -\frac {4 \, {\left (a - b\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - \sqrt {a - b} a \arctan \left (-\frac {{\left (8 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - 8 \, {\left (a^{2} - 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {a - b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{4} - a^{2} b + 3 \, a b^{2} - 2 \, b^{3} - {\left (a^{3} - 6 \, a^{2} b + 9 \, a b^{2} - 4 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right )}{8 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^2/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[-1/16*(8*(a - b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)*sin(f*x + e) - a*sqrt(-a + b)
*log(128*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^8 - 256*(a^4 - 5*a^3*b + 9*a^2*b^2 - 7*a*b^3
+ 2*b^4)*cos(f*x + e)^6 + 32*(5*a^4 - 34*a^3*b + 77*a^2*b^2 - 72*a*b^3 + 24*b^4)*cos(f*x + e)^4 + a^4 - 32*a^
3*b + 160*a^2*b^2 - 256*a*b^3 + 128*b^4 - 32*(a^4 - 11*a^3*b + 34*a^2*b^2 - 40*a*b^3 + 16*b^4)*cos(f*x + e)^2
- 8*(16*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^7 - 24*(a^3 - 4*a^2*b + 5*a*b^2 - 2*b^3)*cos(f*x + e)^5 +
2*(5*a^3 - 29*a^2*b + 48*a*b^2 - 24*b^3)*cos(f*x + e)^3 - (a^3 - 10*a^2*b + 24*a*b^2 - 16*b^3)*cos(f*x + e))*
sqrt(-a + b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)))/((a^2 - 2*a*b + b^2)*f), -1/8*(4
*(a - b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)*sin(f*x + e) - sqrt(a - b)*a*arctan(-1
/4*(8*(a^2 - 2*a*b + b^2)*cos(f*x + e)^5 - 8*(a^2 - 3*a*b + 2*b^2)*cos(f*x + e)^3 + (a^2 - 8*a*b + 8*b^2)*cos(
f*x + e))*sqrt(a - b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((2*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*co
s(f*x + e)^4 - a^2*b + 3*a*b^2 - 2*b^3 - (a^3 - 6*a^2*b + 9*a*b^2 - 4*b^3)*cos(f*x + e)^2)*sin(f*x + e))))/((a
^2 - 2*a*b + b^2)*f)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^2/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.84, size = 795, normalized size = 8.55 \[ -\frac {\sin \left (f x +e \right ) \left (\sqrt {2}\, \sqrt {\frac {i \cos \left (f x +e \right ) \sqrt {a -b}\, \sqrt {b}-i \sqrt {a -b}\, \sqrt {b}+a \cos \left (f x +e \right )-b \cos \left (f x +e \right )+b}{\left (1+\cos \left (f x +e \right )\right ) a}}\, \sqrt {-\frac {2 \left (i \cos \left (f x +e \right ) \sqrt {a -b}\, \sqrt {b}-i \sqrt {a -b}\, \sqrt {b}-a \cos \left (f x +e \right )+b \cos \left (f x +e \right )-b \right )}{\left (1+\cos \left (f x +e \right )\right ) a}}\, \EllipticF \left (\frac {\left (-1+\cos \left (f x +e \right )\right ) \sqrt {\frac {2 i \sqrt {a -b}\, \sqrt {b}+a -2 b}{a}}}{\sin \left (f x +e \right )}, \sqrt {\frac {8 i \sqrt {a -b}\, b^{\frac {3}{2}}-4 i \sqrt {a -b}\, \sqrt {b}\, a +a^{2}-8 a b +8 b^{2}}{a^{2}}}\right ) a \sin \left (f x +e \right )-2 \sqrt {2}\, \sqrt {\frac {i \cos \left (f x +e \right ) \sqrt {a -b}\, \sqrt {b}-i \sqrt {a -b}\, \sqrt {b}+a \cos \left (f x +e \right )-b \cos \left (f x +e \right )+b}{\left (1+\cos \left (f x +e \right )\right ) a}}\, \sqrt {-\frac {2 \left (i \cos \left (f x +e \right ) \sqrt {a -b}\, \sqrt {b}-i \sqrt {a -b}\, \sqrt {b}-a \cos \left (f x +e \right )+b \cos \left (f x +e \right )-b \right )}{\left (1+\cos \left (f x +e \right )\right ) a}}\, \EllipticPi \left (\frac {\left (-1+\cos \left (f x +e \right )\right ) \sqrt {\frac {2 i \sqrt {a -b}\, \sqrt {b}+a -2 b}{a}}}{\sin \left (f x +e \right )}, -\frac {a}{2 i \sqrt {a -b}\, \sqrt {b}+a -2 b}, \frac {\sqrt {-\frac {2 i \sqrt {a -b}\, \sqrt {b}-a +2 b}{a}}}{\sqrt {\frac {2 i \sqrt {a -b}\, \sqrt {b}+a -2 b}{a}}}\right ) a \sin \left (f x +e \right )+\left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {\frac {2 i \sqrt {a -b}\, \sqrt {b}+a -2 b}{a}}\, a -\left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {\frac {2 i \sqrt {a -b}\, \sqrt {b}+a -2 b}{a}}\, b -\left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {\frac {2 i \sqrt {a -b}\, \sqrt {b}+a -2 b}{a}}\, a +\left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {\frac {2 i \sqrt {a -b}\, \sqrt {b}+a -2 b}{a}}\, b +\cos \left (f x +e \right ) \sqrt {\frac {2 i \sqrt {a -b}\, \sqrt {b}+a -2 b}{a}}\, b -\sqrt {\frac {2 i \sqrt {a -b}\, \sqrt {b}+a -2 b}{a}}\, b \right )}{2 f \left (-1+\cos \left (f x +e \right )\right ) \sqrt {\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}}}\, \cos \left (f x +e \right ) \left (a -b \right ) \sqrt {\frac {2 i \sqrt {a -b}\, \sqrt {b}+a -2 b}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^2/(a+b*tan(f*x+e)^2)^(1/2),x)

[Out]

-1/2/f*sin(f*x+e)*(2^(1/2)*((I*cos(f*x+e)*(a-b)^(1/2)*b^(1/2)-I*(a-b)^(1/2)*b^(1/2)+a*cos(f*x+e)-b*cos(f*x+e)+
b)/(1+cos(f*x+e))/a)^(1/2)*(-2*(I*cos(f*x+e)*(a-b)^(1/2)*b^(1/2)-I*(a-b)^(1/2)*b^(1/2)-a*cos(f*x+e)+b*cos(f*x+
e)-b)/(1+cos(f*x+e))/a)^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),(
(8*I*(a-b)^(1/2)*b^(3/2)-4*I*(a-b)^(1/2)*b^(1/2)*a+a^2-8*a*b+8*b^2)/a^2)^(1/2))*a*sin(f*x+e)-2*2^(1/2)*((I*cos
(f*x+e)*(a-b)^(1/2)*b^(1/2)-I*(a-b)^(1/2)*b^(1/2)+a*cos(f*x+e)-b*cos(f*x+e)+b)/(1+cos(f*x+e))/a)^(1/2)*(-2*(I*
cos(f*x+e)*(a-b)^(1/2)*b^(1/2)-I*(a-b)^(1/2)*b^(1/2)-a*cos(f*x+e)+b*cos(f*x+e)-b)/(1+cos(f*x+e))/a)^(1/2)*Elli
pticPi((-1+cos(f*x+e))*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),-1/(2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)
*a,(-(2*I*(a-b)^(1/2)*b^(1/2)-a+2*b)/a)^(1/2)/((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2))*a*sin(f*x+e)+cos(f*x+
e)^3*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*a-cos(f*x+e)^3*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*b-cos(
f*x+e)^2*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*a+cos(f*x+e)^2*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*b+
cos(f*x+e)*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*b-((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*b)/(-1+cos(f*
x+e))/((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(1/2)/cos(f*x+e)/(a-b)/((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)
/a)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^2/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^2/(a + b*tan(e + f*x)^2)^(1/2),x)

[Out]

int(sin(e + f*x)^2/(a + b*tan(e + f*x)^2)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**2/(a+b*tan(f*x+e)**2)**(1/2),x)

[Out]

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

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