3.110 \(\int \sin ^4(e+f x) (a+b \tan ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=222 \[ \frac {3 \left (a^2-8 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{8 f \sqrt {a-b}}-\frac {3 (a-4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 f}+\frac {3 (a-2 b) \sin ^2(e+f x) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 f}+\frac {3 \sqrt {b} (a-2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 f}-\frac {\sin ^3(e+f x) \cos (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{4 f} \]

[Out]

3/8*(a^2-8*a*b+8*b^2)*arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/f/(a-b)^(1/2)+3/2*(a-2*b)*arctan
h(b^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))*b^(1/2)/f-3/8*(a-4*b)*(a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/f+3/8
*(a-2*b)*sin(f*x+e)^2*(a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/f-1/4*cos(f*x+e)*sin(f*x+e)^3*(a+b*tan(f*x+e)^2)^(3/
2)/f

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Rubi [A]  time = 0.32, antiderivative size = 222, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3663, 467, 577, 582, 523, 217, 206, 377, 203} \[ \frac {3 \left (a^2-8 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{8 f \sqrt {a-b}}-\frac {3 (a-4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 f}+\frac {3 (a-2 b) \sin ^2(e+f x) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 f}+\frac {3 \sqrt {b} (a-2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 f}-\frac {\sin ^3(e+f x) \cos (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{4 f} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^4*(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

(3*(a^2 - 8*a*b + 8*b^2)*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(8*Sqrt[a - b]*f) + (3
*(a - 2*b)*Sqrt[b]*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(2*f) - (3*(a - 4*b)*Tan[e + f*
x]*Sqrt[a + b*Tan[e + f*x]^2])/(8*f) + (3*(a - 2*b)*Sin[e + f*x]^2*Tan[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(8
*f) - (Cos[e + f*x]*Sin[e + f*x]^3*(a + b*Tan[e + f*x]^2)^(3/2))/(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 577

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> -Simp[((b*e - a*f)*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(a*b*g*n*(p + 1)), x] + Dist[
1/(a*b*n*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(b*e*n*(p + 1) + (b*e - a*f)*(m
+ 1)) + d*(b*e*n*(p + 1) + (b*e - a*f)*(m + n*q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] &&
IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] &&  !(EqQ[q, 1] && SimplerQ[b*c - a*d, b*e - a*f])

Rule 582

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(f*g^(n - 1)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*d*(m + n*(p + q
+ 1) + 1)), x] - Dist[g^n/(b*d*(m + n*(p + q + 1) + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*
f*c*(m - n + 1) + (a*f*d*(m + n*q + 1) + b*(f*c*(m + n*p + 1) - e*d*(m + n*(p + q + 1) + 1)))*x^n, x], x], x]
/; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && IGtQ[n, 0] && GtQ[m, n - 1]

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) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^4 \left (a+b x^2\right )^{3/2}}{\left (1+x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {\cos (e+f x) \sin ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{4 f}+\frac {\operatorname {Subst}\left (\int \frac {x^2 \sqrt {a+b x^2} \left (3 a+6 b x^2\right )}{\left (1+x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=\frac {3 (a-2 b) \sin ^2(e+f x) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 f}-\frac {\cos (e+f x) \sin ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{4 f}-\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (3 a (a-6 b)+6 (a-4 b) b x^2\right )}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=-\frac {3 (a-4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 f}+\frac {3 (a-2 b) \sin ^2(e+f x) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 f}-\frac {\cos (e+f x) \sin ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{4 f}+\frac {\operatorname {Subst}\left (\int \frac {6 a (a-4 b) b+24 (a-2 b) b^2 x^2}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{16 b f}\\ &=-\frac {3 (a-4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 f}+\frac {3 (a-2 b) \sin ^2(e+f x) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 f}-\frac {\cos (e+f x) \sin ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{4 f}+\frac {(3 (a-2 b) b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac {\left (3 \left (a^2-8 a b+8 b^2\right )\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 f}\\ &=-\frac {3 (a-4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 f}+\frac {3 (a-2 b) \sin ^2(e+f x) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 f}-\frac {\cos (e+f x) \sin ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{4 f}+\frac {(3 (a-2 b) 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 )}{2 f}+\frac {\left (3 \left (a^2-8 a b+8 b^2\right )\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 f}\\ &=\frac {3 \left (a^2-8 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 \sqrt {a-b} f}+\frac {3 (a-2 b) \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 f}-\frac {3 (a-4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 f}+\frac {3 (a-2 b) \sin ^2(e+f x) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 f}-\frac {\cos (e+f x) \sin ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{4 f}\\ \end {align*}

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Mathematica [C]  time = 5.01, size = 278, normalized size = 1.25 \[ \frac {\sqrt {\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (\frac {3 a \sin (2 (e+f x)) \csc ^2(e+f x) \left (\left (a^2-5 a b+4 b^2\right ) F\left (\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right )-\left (a^2-8 a b+8 b^2\right ) \Pi \left (-\frac {b}{a-b};\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right )\right )}{\sqrt {2} b (a-b) \sqrt {\frac {\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}}}+\frac {1}{4} ((18 b-8 a) \sin (2 (e+f x))+(a-b) \sin (4 (e+f x))+16 b \tan (e+f x))\right )}{8 \sqrt {2} f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^4*(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

(Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]*((3*a*Csc[e + f*x]^2*((a^2 - 5*a*b + 4*b^2)*EllipticF
[ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1] - (a^2 - 8*a*b + 8*b^2)*Ellip
ticPi[-(b/(a - b)), ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1])*Sin[2*(e
+ f*x)])/(Sqrt[2]*(a - b)*b*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]) + ((-8*a + 18*b)*Sin[
2*(e + f*x)] + (a - b)*Sin[4*(e + f*x)] + 16*b*Tan[e + f*x])/4))/(8*Sqrt[2]*f)

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fricas [B]  time = 120.24, size = 2163, normalized size = 9.74 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/64*(3*(a^2 - 8*a*b + 8*b^2)*sqrt(-a + b)*cos(f*x + e)*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 - 1
1*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)) + 24*(a^2 - 3*a*b + 2*b^2)*sqrt(b)*cos(f*x + e)*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)*c
os(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)^4 - 5*(a^2 - 3*a*b + 2*b^2)*cos(f*x + e)^2 + 4*a*b - 4*b^2)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^
2)*sin(f*x + e))/((a - b)*f*cos(f*x + e)), -1/64*(48*(a^2 - 3*a*b + 2*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)))*cos(f*x + e) + 3*(a^2 - 8*a*b + 8*b^2)*sqrt(-a + b)*cos(f*x + e)*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)) - 8*(2*(a^2 - 2*a*b + b^2)*cos(f*x + e)^4 - 5*(a
^2 - 3*a*b + 2*b^2)*cos(f*x + e)^2 + 4*a*b - 4*b^2)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x
+ e))/((a - b)*f*cos(f*x + e)), 1/32*(3*(a^2 - 8*a*b + 8*b^2)*sqrt(a - b)*arctan(-1/4*(8*(a^2 - 2*a*b + b^2)*c
os(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)))*cos(f*x + e) - 12*(a^2 - 3*a*b
+ 2*b^2)*sqrt(b)*cos(f*x + e)*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)^4 - 5*(a^2 - 3*a*b + 2*b^2)*cos(f*x + e
)^2 + 4*a*b - 4*b^2)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/((a - b)*f*cos(f*x + e)),
1/32*(3*(a^2 - 8*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)/co
s(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)))*cos(f*x + e) - 24*(a^2 - 3*a*b + 2*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)))*cos(f*x + e) + 4*(2*(a^2 - 2*a*b + b^2)*cos(f*x + e)^4 - 5*(a^2 -
3*a*b + 2*b^2)*cos(f*x + e)^2 + 4*a*b - 4*b^2)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))
/((a - b)*f*cos(f*x + e))]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sin \left (f x + e\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e)^2 + a)^(3/2)*sin(f*x + e)^4, x)

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maple [C]  time = 1.17, size = 2630, normalized size = 11.85 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^4*(a+b*tan(f*x+e)^2)^(3/2),x)

[Out]

1/8/f*(-5*cos(f*x+e)^5*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*a^2+24*sin(f*x+e)*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))*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)*cos(f*x+e)^2*a*b+12*sin(f*x+e)*Ell
ipticF((-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))*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)*cos(f*x+e)^2*a*b-4*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/
a)^(1/2)*b^2-12*cos(f*x+e)^5*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*b^2-4*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/
a)^(1/2)*cos(f*x+e)^7*a*b+4*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^6*a*b+17*cos(f*x+e)^5*((2*I*(
a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*a*b-17*cos(f*x+e)^4*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*a*b-cos(f*x+e
)^3*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*a*b+cos(f*x+e)^2*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*a*b+5
*cos(f*x+e)^4*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*a^2+12*cos(f*x+e)^4*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a
)^(1/2)*b^2+6*cos(f*x+e)^3*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*b^2-6*cos(f*x+e)^2*((2*I*(a-b)^(1/2)*b^(1
/2)+a-2*b)/a)^(1/2)*b^2+4*cos(f*x+e)*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*b^2+2*((2*I*(a-b)^(1/2)*b^(1/2)
+a-2*b)/a)^(1/2)*cos(f*x+e)^7*a^2-2*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^6*a^2+2*b^2*((2*I*(a-
b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^7-2*b^2*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^6-3*c
os(f*x+e)^2*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^2+6*cos(f*x+e)^2*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)*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^
2+48*cos(f*x+e)^2*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*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)*EllipticPi((-1+cos(f*x+e))*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)/s
in(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^2-48*cos(f*x+e)^2*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)*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^2-48*cos(f*x+e)^2*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)*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)*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f
*x+e)^2*b+b)/cos(f*x+e)^2)^(3/2)*sin(f*x+e)/(-1+cos(f*x+e))/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^2/((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 \[ \int {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sin \left (f x + e\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e)^2 + a)^(3/2)*sin(f*x + e)^4, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\sin \left (e+f\,x\right )}^4\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^4*(a + b*tan(e + f*x)^2)^(3/2),x)

[Out]

int(sin(e + f*x)^4*(a + b*tan(e + f*x)^2)^(3/2), 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)**4*(a+b*tan(f*x+e)**2)**(3/2),x)

[Out]

Timed out

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