3.91 \(\int \sqrt {a+a \sin (e+f x)} \tan ^4(e+f x) \, dx\)

Optimal. Leaf size=162 \[ \frac {5 \tan ^3(e+f x) \sqrt {a (\sin (e+f x)+1)}}{12 f}+\frac {29 \tan (e+f x) \sqrt {a \sin (e+f x)+a}}{12 f}-\frac {\sec ^3(e+f x) \sqrt {a (\sin (e+f x)+1)}}{12 f}-\frac {27 \sec (e+f x) \sqrt {a (\sin (e+f x)+1)}}{8 f}+\frac {11 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{8 \sqrt {2} f} \]

[Out]

11/16*arctanh(1/2*cos(f*x+e)*a^(1/2)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))*a^(1/2)/f*2^(1/2)-27/8*sec(f*x+e)*(a*(1+s
in(f*x+e)))^(1/2)/f-1/12*sec(f*x+e)^3*(a*(1+sin(f*x+e)))^(1/2)/f+29/12*(a+a*sin(f*x+e))^(1/2)*tan(f*x+e)/f+5/1
2*(a*(1+sin(f*x+e)))^(1/2)*tan(f*x+e)^3/f

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Rubi [A]  time = 0.92, antiderivative size = 195, normalized size of antiderivative = 1.20, number of steps used = 15, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {2714, 2646, 4401, 2675, 2687, 2650, 2649, 206, 2878, 2855} \[ \frac {11 a^2 \cos (e+f x)}{8 f (a \sin (e+f x)+a)^{3/2}}-\frac {2 a \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}+\frac {4 \sec ^3(e+f x) (a \sin (e+f x)+a)^{3/2}}{3 a f}-\frac {7 \sec ^3(e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}-\frac {11 a \sec (e+f x)}{6 f \sqrt {a \sin (e+f x)+a}}+\frac {11 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{8 \sqrt {2} f} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Sin[e + f*x]]*Tan[e + f*x]^4,x]

[Out]

(11*Sqrt[a]*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(8*Sqrt[2]*f) + (11*a^2*Cos[e
+ f*x])/(8*f*(a + a*Sin[e + f*x])^(3/2)) - (2*a*Cos[e + f*x])/(f*Sqrt[a + a*Sin[e + f*x]]) - (11*a*Sec[e + f*x
])/(6*f*Sqrt[a + a*Sin[e + f*x]]) - (7*Sec[e + f*x]^3*Sqrt[a + a*Sin[e + f*x]])/(3*f) + (4*Sec[e + f*x]^3*(a +
a*Sin[e + f*x])^(3/2))/(3*a*f)

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 2646

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(-2*b*Cos[c + d*x])/(d*Sqrt[a + b*Sin[c + d*
x]]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2650

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

Rule 2675

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(p + 1)), x] + Dist[(a*(m + p + 1))/(g^2*(p + 1)), Int[(
g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& GtQ[m, 0] && LeQ[p, -2*m] && IntegersQ[m + 1/2, 2*p]

Rule 2687

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> -Simp[(b*(g*
Cos[e + f*x])^(p + 1))/(a*f*g*(p + 1)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(a*(2*p + 1))/(2*g^2*(p + 1)), Int[
(g*Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[p, -1] && IntegerQ[2*p]

Rule 2714

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^4, x_Symbol] :> Int[(a + b*Sin[e + f*x
])^m, x] - Int[((a + b*Sin[e + f*x])^m*(1 - 2*Sin[e + f*x]^2))/Cos[e + f*x]^4, x] /; FreeQ[{a, b, e, f, m}, x]
&& EqQ[a^2 - b^2, 0] && IntegerQ[m - 1/2]

Rule 2855

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[((b*c + a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(p +
1)), x] + Dist[(b*(a*d*m + b*c*(m + p + 1)))/(a*g^2*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x]
)^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]

Rule 2878

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> -Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*g*(m + p + 2)), x] + Dist[1/
(b*(m + p + 2)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m*(b*(m + 1) - a*(p + 1)*Sin[e + f*x]), x], x] /;
FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[m + p + 2, 0]

Rule 4401

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rubi steps

\begin {align*} \int \sqrt {a+a \sin (e+f x)} \tan ^4(e+f x) \, dx &=\int \sqrt {a+a \sin (e+f x)} \, dx-\int \sec ^4(e+f x) \sqrt {a+a \sin (e+f x)} \left (1-2 \sin ^2(e+f x)\right ) \, dx\\ &=-\frac {2 a \cos (e+f x)}{f \sqrt {a+a \sin (e+f x)}}-\int \left (\sec ^4(e+f x) \sqrt {a (1+\sin (e+f x))}-2 \sec ^2(e+f x) \sqrt {a (1+\sin (e+f x))} \tan ^2(e+f x)\right ) \, dx\\ &=-\frac {2 a \cos (e+f x)}{f \sqrt {a+a \sin (e+f x)}}+2 \int \sec ^2(e+f x) \sqrt {a (1+\sin (e+f x))} \tan ^2(e+f x) \, dx-\int \sec ^4(e+f x) \sqrt {a (1+\sin (e+f x))} \, dx\\ &=-\frac {2 a \cos (e+f x)}{f \sqrt {a+a \sin (e+f x)}}-\frac {\sec ^3(e+f x) \sqrt {a+a \sin (e+f x)}}{3 f}+\frac {4 \sec ^3(e+f x) (a+a \sin (e+f x))^{3/2}}{3 a f}-\frac {4 \int \sec ^4(e+f x) \sqrt {a+a \sin (e+f x)} \left (\frac {3 a}{2}+3 a \sin (e+f x)\right ) \, dx}{3 a}-\frac {1}{6} (5 a) \int \frac {\sec ^2(e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx\\ &=-\frac {2 a \cos (e+f x)}{f \sqrt {a+a \sin (e+f x)}}-\frac {5 a \sec (e+f x)}{6 f \sqrt {a+a \sin (e+f x)}}-\frac {7 \sec ^3(e+f x) \sqrt {a+a \sin (e+f x)}}{3 f}+\frac {4 \sec ^3(e+f x) (a+a \sin (e+f x))^{3/2}}{3 a f}-a \int \frac {\sec ^2(e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx-\frac {1}{4} \left (5 a^2\right ) \int \frac {1}{(a+a \sin (e+f x))^{3/2}} \, dx\\ &=\frac {5 a^2 \cos (e+f x)}{8 f (a+a \sin (e+f x))^{3/2}}-\frac {2 a \cos (e+f x)}{f \sqrt {a+a \sin (e+f x)}}-\frac {11 a \sec (e+f x)}{6 f \sqrt {a+a \sin (e+f x)}}-\frac {7 \sec ^3(e+f x) \sqrt {a+a \sin (e+f x)}}{3 f}+\frac {4 \sec ^3(e+f x) (a+a \sin (e+f x))^{3/2}}{3 a f}-\frac {1}{16} (5 a) \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx-\frac {1}{2} \left (3 a^2\right ) \int \frac {1}{(a+a \sin (e+f x))^{3/2}} \, dx\\ &=\frac {11 a^2 \cos (e+f x)}{8 f (a+a \sin (e+f x))^{3/2}}-\frac {2 a \cos (e+f x)}{f \sqrt {a+a \sin (e+f x)}}-\frac {11 a \sec (e+f x)}{6 f \sqrt {a+a \sin (e+f x)}}-\frac {7 \sec ^3(e+f x) \sqrt {a+a \sin (e+f x)}}{3 f}+\frac {4 \sec ^3(e+f x) (a+a \sin (e+f x))^{3/2}}{3 a f}-\frac {1}{8} (3 a) \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx+\frac {(5 a) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{8 f}\\ &=\frac {5 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{8 \sqrt {2} f}+\frac {11 a^2 \cos (e+f x)}{8 f (a+a \sin (e+f x))^{3/2}}-\frac {2 a \cos (e+f x)}{f \sqrt {a+a \sin (e+f x)}}-\frac {11 a \sec (e+f x)}{6 f \sqrt {a+a \sin (e+f x)}}-\frac {7 \sec ^3(e+f x) \sqrt {a+a \sin (e+f x)}}{3 f}+\frac {4 \sec ^3(e+f x) (a+a \sin (e+f x))^{3/2}}{3 a f}+\frac {(3 a) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{4 f}\\ &=\frac {11 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{8 \sqrt {2} f}+\frac {11 a^2 \cos (e+f x)}{8 f (a+a \sin (e+f x))^{3/2}}-\frac {2 a \cos (e+f x)}{f \sqrt {a+a \sin (e+f x)}}-\frac {11 a \sec (e+f x)}{6 f \sqrt {a+a \sin (e+f x)}}-\frac {7 \sec ^3(e+f x) \sqrt {a+a \sin (e+f x)}}{3 f}+\frac {4 \sec ^3(e+f x) (a+a \sin (e+f x))^{3/2}}{3 a f}\\ \end {align*}

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Mathematica [C]  time = 5.57, size = 394, normalized size = 2.43 \[ \frac {\sqrt {a (\sin (e+f x)+1)} \left (-48 \left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \cos \left (\frac {f x}{2}\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2+48 \left (\sin \left (\frac {e}{2}\right )+\cos \left (\frac {e}{2}\right )\right ) \sin \left (\frac {f x}{2}\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2-\frac {36 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2}{\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )}+\frac {4 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2}{\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}-\frac {3 \left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )}{\sin \left (\frac {e}{2}\right )+\cos \left (\frac {e}{2}\right )}+\frac {6 \sin \left (\frac {f x}{2}\right )}{\sin \left (\frac {e}{2}\right )+\cos \left (\frac {e}{2}\right )}+(33+33 i) (-1)^{3/4} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2 \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \sec \left (\frac {f x}{4}\right ) \left (\cos \left (\frac {1}{4} (2 e+f x)\right )-\sin \left (\frac {1}{4} (2 e+f x)\right )\right )\right )\right )}{24 f \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + a*Sin[e + f*x]]*Tan[e + f*x]^4,x]

[Out]

(((6*Sin[(f*x)/2])/(Cos[e/2] + Sin[e/2]) - (3*(Cos[e/2] - Sin[e/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))/(Co
s[e/2] + Sin[e/2]) + (33 + 33*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*Sec[(f*x)/4]*(Cos[(2*e + f*x)/4] -
Sin[(2*e + f*x)/4])]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - 48*Cos[(f*x)/2]*(Cos[e/2] - Sin[e/2])*(Cos[(e +
f*x)/2] + Sin[(e + f*x)/2])^2 + 48*(Cos[e/2] + Sin[e/2])*Sin[(f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2
+ (4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2)/(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3 - (36*(Cos[(e + f*x)/2]
+ Sin[(e + f*x)/2])^2)/(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))*Sqrt[a*(1 + Sin[e + f*x])])/(24*f*(Cos[(e + f*x
)/2] + Sin[(e + f*x)/2])^3)

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fricas [A]  time = 0.47, size = 200, normalized size = 1.23 \[ \frac {33 \, \sqrt {2} \sqrt {a} \cos \left (f x + e\right )^{3} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {a \sin \left (f x + e\right ) + a} {\left (\sqrt {2} \cos \left (f x + e\right ) - \sqrt {2} \sin \left (f x + e\right ) + \sqrt {2}\right )} \sqrt {a} + 3 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (81 \, \cos \left (f x + e\right )^{2} - 2 \, {\left (24 \, \cos \left (f x + e\right )^{2} + 5\right )} \sin \left (f x + e\right ) + 2\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{96 \, f \cos \left (f x + e\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(1/2)*tan(f*x+e)^4,x, algorithm="fricas")

[Out]

1/96*(33*sqrt(2)*sqrt(a)*cos(f*x + e)^3*log(-(a*cos(f*x + e)^2 + 2*sqrt(a*sin(f*x + e) + a)*(sqrt(2)*cos(f*x +
e) - sqrt(2)*sin(f*x + e) + sqrt(2))*sqrt(a) + 3*a*cos(f*x + e) - (a*cos(f*x + e) - 2*a)*sin(f*x + e) + 2*a)/
(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) - 4*(81*cos(f*x + e)^2 - 2*(24*cos(f*x
+ e)^2 + 5)*sin(f*x + e) + 2)*sqrt(a*sin(f*x + e) + a))/(f*cos(f*x + e)^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(1/2)*tan(f*x+e)^4,x, algorithm="giac")

[Out]

integrate(sqrt(a*sin(f*x + e) + a)*tan(f*x + e)^4, x)

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maple [A]  time = 0.70, size = 172, normalized size = 1.06 \[ -\frac {96 a^{\frac {5}{2}} \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )+\left (33 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a +20 a^{\frac {5}{2}}\right ) \sin \left (f x +e \right )-162 a^{\frac {5}{2}} \left (\cos ^{2}\left (f x +e \right )\right )+33 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a -4 a^{\frac {5}{2}}}{48 a^{\frac {3}{2}} \left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^(1/2)*tan(f*x+e)^4,x)

[Out]

-1/48/a^(3/2)*(96*a^(5/2)*sin(f*x+e)*cos(f*x+e)^2+(33*(a-a*sin(f*x+e))^(3/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+
e))^(1/2)*2^(1/2)/a^(1/2))*a+20*a^(5/2))*sin(f*x+e)-162*a^(5/2)*cos(f*x+e)^2+33*(a-a*sin(f*x+e))^(3/2)*2^(1/2)
*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a-4*a^(5/2))/(sin(f*x+e)-1)/cos(f*x+e)/(a+a*sin(f*x+e))^(
1/2)/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(1/2)*tan(f*x+e)^4,x, algorithm="maxima")

[Out]

integrate(sqrt(a*sin(f*x + e) + a)*tan(f*x + e)^4, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {tan}\left (e+f\,x\right )}^4\,\sqrt {a+a\,\sin \left (e+f\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)^4*(a + a*sin(e + f*x))^(1/2),x)

[Out]

int(tan(e + f*x)^4*(a + a*sin(e + f*x))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )} \tan ^{4}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**(1/2)*tan(f*x+e)**4,x)

[Out]

Integral(sqrt(a*(sin(e + f*x) + 1))*tan(e + f*x)**4, x)

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