3.99 \(\int \sin ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\)
Optimal. Leaf size=128 \[ \frac {(a-2 b) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 f \sqrt {a-b}}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}-\frac {\sin (e+f x) \cos (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 f} \]
[Out]
1/2*(a-2*b)*arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/f/(a-b)^(1/2)+arctanh(b^(1/2)*tan(f*x+e)/(
a+b*tan(f*x+e)^2)^(1/2))*b^(1/2)/f-1/2*cos(f*x+e)*sin(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2)/f
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Rubi [A] time = 0.13, antiderivative size = 128, normalized size of antiderivative = 1.00,
number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used =
{3663, 467, 523, 217, 206, 377, 203} \[ \frac {(a-2 b) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 f \sqrt {a-b}}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}-\frac {\sin (e+f x) \cos (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 f} \]
Antiderivative was successfully verified.
[In]
Int[Sin[e + f*x]^2*Sqrt[a + b*Tan[e + f*x]^2],x]
[Out]
((a - 2*b)*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(2*Sqrt[a - b]*f) + (Sqrt[b]*ArcTanh
[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/f - (Cos[e + f*x]*Sin[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2]
)/(2*f)
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 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 217
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] && !GtQ[a, 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 467
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)/(b*n*(p + 1)), x] - Dist[e^n/(b*n*(p + 1)), Int[(e*x)^
(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m - n + 1) + d*(m + n*(q - 1) + 1)*x^n, x], x], x] /;
FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] && GtQ[m - n + 1, 0] &
& IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
Rule 523
Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, 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 \sin ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^2 \sqrt {a+b x^2}}{\left (1+x^2\right )^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 f}+\frac {\operatorname {Subst}\left (\int \frac {a+2 b x^2}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=-\frac {\cos (e+f x) \sin (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 f}+\frac {(a-2 b) \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac {b \operatorname {Subst}\left (\int \frac {1}{\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 f}+\frac {(a-2 b) \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 f}+\frac {b \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}\\ &=\frac {(a-2 b) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 \sqrt {a-b} f}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}-\frac {\cos (e+f x) \sin (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 f}\\ \end {align*}
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Mathematica [C] time = 3.84, size = 273, normalized size = 2.13 \[ -\frac {\sin (2 (e+f x)) \sec ^2(e+f x) \left ((a-b) ((a-b) \cos (2 (e+f x))+a+b)+\sqrt {2} a (b-a) \sqrt {\frac {\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right )+\sqrt {2} a (a-2 b) \sqrt {\frac {\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} \Pi \left (-\frac {b}{a-b};\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right )\right )}{4 \sqrt {2} f (a-b) \sqrt {\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}} \]
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*(a - 2*b)*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*EllipticPi[-(b/(a - b)), Ar
cSin[Sqrt[((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)*f*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])
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fricas [B] time = 1.37, size = 1847, normalized size = 14.43 \[ \text {result too large to display} \]
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 - 2*b)*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)) - 4*(a - b)*sqrt(b)*log((
(a^2 - 8*a*b + 8*b^2)*cos(f*x + e)^4 + 8*(a*b - 2*b^2)*cos(f*x + e)^2 + 4*((a - 2*b)*cos(f*x + e)^3 + 2*b*cos(
f*x + e))*sqrt(b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e) + 8*b^2)/cos(f*x + e)^4))/((a
- b)*f), -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) + 8*(a
- b)*sqrt(-b)*arctan(1/2*((a - 2*b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*sqrt(((a - b)*cos(f*x + e)^2 +
b)/cos(f*x + e)^2)/(((a*b - b^2)*cos(f*x + e)^2 + b^2)*sin(f*x + e))) - (a - 2*b)*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)*sq
rt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)))/((a - b)*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 - 2*b)*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)*cos(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))) - 2*(a - b)*sqrt(b)*log
(((a^2 - 8*a*b + 8*b^2)*cos(f*x + e)^4 + 8*(a*b - 2*b^2)*cos(f*x + e)^2 + 4*((a - 2*b)*cos(f*x + e)^3 + 2*b*co
s(f*x + e))*sqrt(b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e) + 8*b^2)/cos(f*x + e)^4))/(
(a - b)*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 - 2*b)*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)*cos(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))) + 4*(a - b)*sqrt(-b)*arctan(1/2*((a - 2*b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*
sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(((a*b - b^2)*cos(f*x + e)^2 + b^2)*sin(f*x + e))))/((a - b)
*f)]
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \tan \left (f x + e\right )^{2} + a} \sin \left (f x + e\right )^{2}\,{d x} \]
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(sqrt(b*tan(f*x + e)^2 + a)*sin(f*x + e)^2, x)
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maple [C] time = 0.78, size = 1342, normalized size = 10.48 \[ \text {result too large to display} \]
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*(-4*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+co
s(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)*EllipticPi((-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))*b*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)*EllipticPi((-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)+4*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*co
s(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)*EllipticPi((-1+cos(f*x+e))*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)/si
n(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))*b*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)+c
os(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)*cos
(f*x+e)*sin(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(1/2)/(-1+cos(f*x+e))/(a*cos(f*x+e)^2-cos(
f*x+e)^2*b+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 \[ \int \sqrt {b \tan \left (f x + e\right )^{2} + a} \sin \left (f x + e\right )^{2}\,{d x} \]
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(sqrt(b*tan(f*x + e)^2 + a)*sin(f*x + e)^2, x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\sin \left (e+f\,x\right )}^2\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a} \,d x \]
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 \[ \int \sqrt {a + b \tan ^{2}{\left (e + f x \right )}} \sin ^{2}{\left (e + f x \right )}\, dx \]
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(sqrt(a + b*tan(e + f*x)**2)*sin(e + f*x)**2, x)
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