3.98 \(\int \sin ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\)
Optimal. Leaf size=189 \[ \frac {\left (3 a^2-12 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{8 f (a-b)^{3/2}}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}-\frac {\sin ^3(e+f x) \cos (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 f}-\frac {(3 a-4 b) \sin (e+f x) \cos (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 f (a-b)} \]
[Out]
1/8*(3*a^2-12*a*b+8*b^2)*arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/(a-b)^(3/2)/f+arctanh(b^(1/2)
*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))*b^(1/2)/f-1/8*(3*a-4*b)*cos(f*x+e)*sin(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2)/(
a-b)/f-1/4*cos(f*x+e)*sin(f*x+e)^3*(a+b*tan(f*x+e)^2)^(1/2)/f
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Rubi [A] time = 0.24, antiderivative size = 189, normalized size of antiderivative = 1.00,
number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used =
{3663, 467, 578, 523, 217, 206, 377, 203} \[ \frac {\left (3 a^2-12 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{8 f (a-b)^{3/2}}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}-\frac {\sin ^3(e+f x) \cos (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 f}-\frac {(3 a-4 b) \sin (e+f x) \cos (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 f (a-b)} \]
Antiderivative was successfully verified.
[In]
Int[Sin[e + f*x]^4*Sqrt[a + b*Tan[e + f*x]^2],x]
[Out]
((3*a^2 - 12*a*b + 8*b^2)*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(8*(a - b)^(3/2)*f) +
(Sqrt[b]*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/f - ((3*a - 4*b)*Cos[e + f*x]*Sin[e + f*
x]*Sqrt[a + b*Tan[e + f*x]^2])/(8*(a - b)*f) - (Cos[e + f*x]*Sin[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/(4*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 578
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[(g^(n - 1)*(b*e - a*f)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c -
a*d)*(p + 1)), x] - Dist[g^n/(b*n*(b*c - a*d)*(p + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*S
imp[c*(b*e - a*f)*(m - n + 1) + (d*(b*e - a*f)*(m + n*q + 1) - b*n*(c*f - d*e)*(p + 1))*x^n, x], x], x] /; Fre
eQ[{a, b, c, d, e, f, g, q}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, 0]
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 ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^4 \sqrt {a+b x^2}}{\left (1+x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {\cos (e+f x) \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 f}+\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (3 a+4 b x^2\right )}{\left (1+x^2\right )^2 \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=-\frac {(3 a-4 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 (a-b) f}-\frac {\cos (e+f x) \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 f}+\frac {\operatorname {Subst}\left (\int \frac {a (3 a-4 b)+8 (a-b) b x^2}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{8 (a-b) f}\\ &=-\frac {(3 a-4 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 (a-b) f}-\frac {\cos (e+f x) \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 f}+\frac {b \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}+\frac {\left (3 a^2-12 a b+8 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{8 (a-b) f}\\ &=-\frac {(3 a-4 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 (a-b) f}-\frac {\cos (e+f x) \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 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 {\left (3 a^2-12 a b+8 b^2\right ) \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 )}{8 (a-b) f}\\ &=\frac {\left (3 a^2-12 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{8 (a-b)^{3/2} 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 {(3 a-4 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 (a-b) f}-\frac {\cos (e+f x) \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 f}\\ \end {align*}
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Mathematica [C] time = 3.80, size = 330, normalized size = 1.75 \[ \frac {\sin (2 (e+f x)) \sec ^2(e+f x) \left (-\left ((a-b) \left (6 \left (a^2-3 a b+2 b^2\right ) \cos (2 (e+f x))+7 a^2-(a-b)^2 \cos (4 (e+f x))-11 b^2\right )\right )+2 \sqrt {2} a \left (3 a^2-7 a b+4 b^2\right ) \sqrt {\frac {\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right )-2 \sqrt {2} a \left (3 a^2-12 a b+8 b^2\right ) \sqrt {\frac {\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} \Pi \left (-\frac {b}{a-b};\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right )\right )}{32 \sqrt {2} f (a-b)^2 \sqrt {\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}} \]
Antiderivative was successfully verified.
[In]
Integrate[Sin[e + f*x]^4*Sqrt[a + b*Tan[e + f*x]^2],x]
[Out]
((-((a - b)*(7*a^2 - 11*b^2 + 6*(a^2 - 3*a*b + 2*b^2)*Cos[2*(e + f*x)] - (a - b)^2*Cos[4*(e + f*x)])) + 2*Sqrt
[2]*a*(3*a^2 - 7*a*b + 4*b^2)*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*EllipticF[ArcSin[Sqr
t[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1] - 2*Sqrt[2]*a*(3*a^2 - 12*a*b + 8*b^2)*S
qrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*EllipticPi[-(b/(a - b)), ArcSin[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)])/(32*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 = 13.42, size = 2068, normalized size = 10.94 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sin(f*x+e)^4*(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")
[Out]
[1/64*((3*a^2 - 12*a*b + 8*b^2)*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)) + 16*(a^2 - 2*a*b + b^2)*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) + 8*(2*(a^2 - 2*a*b + b^2)*cos(f*x + e)^3 - (5*a^2 - 11*a*b + 6*b
^2)*cos(f*x + e))*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/
64*(32*(a^2 - 2*a*b + b^2)*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))) - (3*a^2 - 12*a*b
+ 8*b^2)*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 + 1
6*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)) - 8*(2*(a^2
- 2*a*b + b^2)*cos(f*x + e)^3 - (5*a^2 - 11*a*b + 6*b^2)*cos(f*x + e))*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), 1/32*((3*a^2 - 12*a*b + 8*b^2)*sqrt(a - 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))) + 8*(a^2 -
2*a*b + b^2)*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) + 4*(2*(a^2 - 2*a*b + b^2)*cos(f*x + e)^3 - (5*a^2 - 11*a*b + 6*b^2)*cos(f*x + e))*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), 1/32*((3*a^2 - 12*a*b + 8*
b^2)*sqrt(a - 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))) - 16*(a^2 - 2*a*b + b^2)*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)
)) + 4*(2*(a^2 - 2*a*b + b^2)*cos(f*x + e)^3 - (5*a^2 - 11*a*b + 6*b^2)*cos(f*x + e))*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)]
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \tan \left (f x + e\right )^{2} + a} \sin \left (f x + e\right )^{4}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sin(f*x+e)^4*(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)^4, x)
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maple [C] time = 1.99, size = 2498, normalized size = 13.22 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sin(f*x+e)^4*(a+b*tan(f*x+e)^2)^(1/2),x)
[Out]
1/8/f*(2*cos(f*x+e)^5*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*a^2-4*cos(f*x+e)^5*((2*I*(a-b)^(1/2)*b^(1/2)+a
-2*b)/a)^(1/2)*a*b+2*cos(f*x+e)^5*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*b^2+6*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+co
s(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^2*sin(f*x+e)-24*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*c
os(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)*Ellip
ticPi((-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*b*sin(f*x+e)+16*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))*b^2*sin(f*
x+e)+16*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+co
s(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*b*sin(f*x+e)-16*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))*b^2*sin(f*x+e)-3*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*c
os(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)/si
n(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^2*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*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*b*sin(f*x+e)-2*cos(f*x+e)^4*((2*I*(a-b)^(1/2)*b^(1
/2)+a-2*b)/a)^(1/2)*a^2+4*cos(f*x+e)^4*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*a*b-2*cos(f*x+e)^4*((2*I*(a-b
)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*b^2-5*cos(f*x+e)^3*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*a^2+13*cos(f*x+e)
^3*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*a*b-8*cos(f*x+e)^3*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*b^2+
5*cos(f*x+e)^2*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*a^2-13*cos(f*x+e)^2*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/
a)^(1/2)*a*b+8*cos(f*x+e)^2*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*b^2-5*cos(f*x+e)*((2*I*(a-b)^(1/2)*b^(1/
2)+a-2*b)/a)^(1/2)*a*b+6*cos(f*x+e)*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*b^2+5*((2*I*(a-b)^(1/2)*b^(1/2)+
a-2*b)/a)^(1/2)*a*b-6*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*b^2)*cos(f*x+e)*sin(f*x+e)*((a*cos(f*x+e)^2-co
s(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)/(a-b)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \tan \left (f x + e\right )^{2} + a} \sin \left (f x + e\right )^{4}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sin(f*x+e)^4*(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)^4, x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\sin \left (e+f\,x\right )}^4\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sin(e + f*x)^4*(a + b*tan(e + f*x)^2)^(1/2),x)
[Out]
int(sin(e + f*x)^4*(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 \[ \int \sqrt {a + b \tan ^{2}{\left (e + f x \right )}} \sin ^{4}{\left (e + f x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sin(f*x+e)**4*(a+b*tan(f*x+e)**2)**(1/2),x)
[Out]
Integral(sqrt(a + b*tan(e + f*x)**2)*sin(e + f*x)**4, x)
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