\(\int \cos ^2(e+f x) (d \tan (e+f x))^{5/2} \, dx\) [251]

Optimal result

Integrand size = 21, antiderivative size = 225 \[ \int \cos ^2(e+f x) (d \tan (e+f x))^{5/2} \, dx=-\frac {3 d^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{4 \sqrt {2} f}+\frac {3 d^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{4 \sqrt {2} f}+\frac {3 d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{8 \sqrt {2} f}-\frac {3 d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{8 \sqrt {2} f}-\frac {d \cos ^2(e+f x) (d \tan (e+f x))^{3/2}}{2 f} \]

[Out]

-3/8*d^(5/2)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/f*2^(1/2)+3/8*d^(5/2)*arctan(1+2^(1/2)*(d*tan(f*x+
e))^(1/2)/d^(1/2))/f*2^(1/2)+3/16*d^(5/2)*ln(d^(1/2)-2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))/f*2^(1/2
)-3/16*d^(5/2)*ln(d^(1/2)+2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))/f*2^(1/2)-1/2*d*cos(f*x+e)^2*(d*tan
(f*x+e))^(3/2)/f

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2687, 294, 335, 303, 1176, 631, 210, 1179, 642} \[ \int \cos ^2(e+f x) (d \tan (e+f x))^{5/2} \, dx=-\frac {3 d^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{4 \sqrt {2} f}+\frac {3 d^{5/2} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{4 \sqrt {2} f}+\frac {3 d^{5/2} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{8 \sqrt {2} f}-\frac {3 d^{5/2} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{8 \sqrt {2} f}-\frac {d \cos ^2(e+f x) (d \tan (e+f x))^{3/2}}{2 f} \]

[In]

Int[Cos[e + f*x]^2*(d*Tan[e + f*x])^(5/2),x]

[Out]

(-3*d^(5/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(4*Sqrt[2]*f) + (3*d^(5/2)*ArcTan[1 + (Sqrt[2]
*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(4*Sqrt[2]*f) + (3*d^(5/2)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[
d*Tan[e + f*x]]])/(8*Sqrt[2]*f) - (3*d^(5/2)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]
])/(8*Sqrt[2]*f) - (d*Cos[e + f*x]^2*(d*Tan[e + f*x])^(3/2))/(2*f)

Rule 210

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

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

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

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 2687

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(d x)^{5/2}}{\left (1+x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {d \cos ^2(e+f x) (d \tan (e+f x))^{3/2}}{2 f}+\frac {\left (3 d^2\right ) \text {Subst}\left (\int \frac {\sqrt {d x}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{4 f} \\ & = -\frac {d \cos ^2(e+f x) (d \tan (e+f x))^{3/2}}{2 f}+\frac {(3 d) \text {Subst}\left (\int \frac {x^2}{1+\frac {x^4}{d^2}} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{2 f} \\ & = -\frac {d \cos ^2(e+f x) (d \tan (e+f x))^{3/2}}{2 f}-\frac {(3 d) \text {Subst}\left (\int \frac {d-x^2}{1+\frac {x^4}{d^2}} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{4 f}+\frac {(3 d) \text {Subst}\left (\int \frac {d+x^2}{1+\frac {x^4}{d^2}} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{4 f} \\ & = -\frac {d \cos ^2(e+f x) (d \tan (e+f x))^{3/2}}{2 f}+\frac {\left (3 d^{5/2}\right ) \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}+2 x}{-d-\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{8 \sqrt {2} f}+\frac {\left (3 d^{5/2}\right ) \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}-2 x}{-d+\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{8 \sqrt {2} f}+\frac {\left (3 d^3\right ) \text {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{8 f}+\frac {\left (3 d^3\right ) \text {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{8 f} \\ & = \frac {3 d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{8 \sqrt {2} f}-\frac {3 d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{8 \sqrt {2} f}-\frac {d \cos ^2(e+f x) (d \tan (e+f x))^{3/2}}{2 f}+\frac {\left (3 d^{5/2}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{4 \sqrt {2} f}-\frac {\left (3 d^{5/2}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{4 \sqrt {2} f} \\ & = -\frac {3 d^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{4 \sqrt {2} f}+\frac {3 d^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{4 \sqrt {2} f}+\frac {3 d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{8 \sqrt {2} f}-\frac {3 d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{8 \sqrt {2} f}-\frac {d \cos ^2(e+f x) (d \tan (e+f x))^{3/2}}{2 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.48 \[ \int \cos ^2(e+f x) (d \tan (e+f x))^{5/2} \, dx=-\frac {d^2 \left (3 \arcsin (\cos (e+f x)-\sin (e+f x)) \csc (e+f x)+3 \csc (e+f x) \log \left (\cos (e+f x)+\sin (e+f x)+\sqrt {\sin (2 (e+f x))}\right )+2 \sqrt {\sin (2 (e+f x))}\right ) \sqrt {\sin (2 (e+f x))} \sqrt {d \tan (e+f x)}}{8 f} \]

[In]

Integrate[Cos[e + f*x]^2*(d*Tan[e + f*x])^(5/2),x]

[Out]

-1/8*(d^2*(3*ArcSin[Cos[e + f*x] - Sin[e + f*x]]*Csc[e + f*x] + 3*Csc[e + f*x]*Log[Cos[e + f*x] + Sin[e + f*x]
+ Sqrt[Sin[2*(e + f*x)]]] + 2*Sqrt[Sin[2*(e + f*x)]])*Sqrt[Sin[2*(e + f*x)]]*Sqrt[d*Tan[e + f*x]])/f

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(550\) vs. \(2(169)=338\).

Time = 2.48 (sec) , antiderivative size = 551, normalized size of antiderivative = 2.45

method	result	size
default	\(\frac {\left (\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right ) \left (4 \sqrt {2}\, \cos \left (f x +e \right ) \sqrt {-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sin \left (f x +e \right )+4 \sqrt {2}\, \sqrt {-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sin \left (f x +e \right )-3 \ln \left (-\frac {\cot \left (f x +e \right ) \cos \left (f x +e \right )-2 \cot \left (f x +e \right )-2 \sin \left (f x +e \right ) \sqrt {-\left (\cot ^{3}\left (f x +e \right )\right )+3 \left (\cot ^{2}\left (f x +e \right )\right ) \csc \left (f x +e \right )-3 \left (\csc ^{2}\left (f x +e \right )\right ) \cot \left (f x +e \right )+\csc ^{3}\left (f x +e \right )+\cot \left (f x +e \right )-\csc \left (f x +e \right )}-2 \cos \left (f x +e \right )-\sin \left (f x +e \right )+\csc \left (f x +e \right )+2}{\cos \left (f x +e \right )-1}\right )+3 \ln \left (-\frac {\cot \left (f x +e \right ) \cos \left (f x +e \right )-2 \cot \left (f x +e \right )+2 \sin \left (f x +e \right ) \sqrt {-\left (\cot ^{3}\left (f x +e \right )\right )+3 \left (\cot ^{2}\left (f x +e \right )\right ) \csc \left (f x +e \right )-3 \left (\csc ^{2}\left (f x +e \right )\right ) \cot \left (f x +e \right )+\csc ^{3}\left (f x +e \right )+\cot \left (f x +e \right )-\csc \left (f x +e \right )}-2 \cos \left (f x +e \right )-\sin \left (f x +e \right )+\csc \left (f x +e \right )+2}{\cos \left (f x +e \right )-1}\right )-6 \arctan \left (\frac {\sqrt {2}\, \sqrt {-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sin \left (f x +e \right )-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )-1}\right )-6 \arctan \left (\frac {\sqrt {2}\, \sqrt {-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sin \left (f x +e \right )+\cos \left (f x +e \right )-1}{\cos \left (f x +e \right )-1}\right )\right ) \sqrt {d \tan \left (f x +e \right )}\, d^{2} \sqrt {2}}{16 f \left (\cos \left (f x +e \right )-1\right ) \left (\cos \left (f x +e \right )+1\right )^{2} \sqrt {-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\)	\(551\)

[In]

int(cos(f*x+e)^2*(d*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/16/f*sin(f*x+e)^2*cos(f*x+e)*(4*2^(1/2)*cos(f*x+e)*(-sin(f*x+e)*cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*sin(f*x+e
)+4*2^(1/2)*(-sin(f*x+e)*cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*sin(f*x+e)-3*ln(-(cot(f*x+e)*cos(f*x+e)-2*cot(f*x+
e)-2*sin(f*x+e)*(-cot(f*x+e)^3+3*cot(f*x+e)^2*csc(f*x+e)-3*csc(f*x+e)^2*cot(f*x+e)+csc(f*x+e)^3+cot(f*x+e)-csc
(f*x+e))^(1/2)-2*cos(f*x+e)-sin(f*x+e)+csc(f*x+e)+2)/(cos(f*x+e)-1))+3*ln(-(cot(f*x+e)*cos(f*x+e)-2*cot(f*x+e)
+2*sin(f*x+e)*(-cot(f*x+e)^3+3*cot(f*x+e)^2*csc(f*x+e)-3*csc(f*x+e)^2*cot(f*x+e)+csc(f*x+e)^3+cot(f*x+e)-csc(f
*x+e))^(1/2)-2*cos(f*x+e)-sin(f*x+e)+csc(f*x+e)+2)/(cos(f*x+e)-1))-6*arctan((2^(1/2)*(-sin(f*x+e)*cos(f*x+e)/(
cos(f*x+e)+1)^2)^(1/2)*sin(f*x+e)-cos(f*x+e)+1)/(cos(f*x+e)-1))-6*arctan((2^(1/2)*(-sin(f*x+e)*cos(f*x+e)/(cos
(f*x+e)+1)^2)^(1/2)*sin(f*x+e)+cos(f*x+e)-1)/(cos(f*x+e)-1)))*(d*tan(f*x+e))^(1/2)*d^2/(cos(f*x+e)-1)/(cos(f*x
+e)+1)^2/(-sin(f*x+e)*cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*2^(1/2)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.42 (sec) , antiderivative size = 969, normalized size of antiderivative = 4.31 \[ \int \cos ^2(e+f x) (d \tan (e+f x))^{5/2} \, dx=\text {Too large to display} \]

[In]

integrate(cos(f*x+e)^2*(d*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-1/32*(16*d^2*sqrt(d*sin(f*x + e)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) - 3*I*(-d^10/f^4)^(1/4)*f*log(27/2*d
^8*cos(f*x + e)*sin(f*x + e) + 27/4*(2*d^3*f^2*cos(f*x + e)^2 - d^3*f^2)*sqrt(-d^10/f^4) - 27/2*(I*(-d^10/f^4)
^(1/4)*d^5*f*cos(f*x + e)*sin(f*x + e) + I*(-d^10/f^4)^(3/4)*f^3*cos(f*x + e)^2)*sqrt(d*sin(f*x + e)/cos(f*x +
e))) + 3*I*(-d^10/f^4)^(1/4)*f*log(27/2*d^8*cos(f*x + e)*sin(f*x + e) + 27/4*(2*d^3*f^2*cos(f*x + e)^2 - d^3*
f^2)*sqrt(-d^10/f^4) - 27/2*(-I*(-d^10/f^4)^(1/4)*d^5*f*cos(f*x + e)*sin(f*x + e) - I*(-d^10/f^4)^(3/4)*f^3*co
s(f*x + e)^2)*sqrt(d*sin(f*x + e)/cos(f*x + e))) + 3*(-d^10/f^4)^(1/4)*f*log(27/2*d^8*cos(f*x + e)*sin(f*x + e
) - 27/4*(2*d^3*f^2*cos(f*x + e)^2 - d^3*f^2)*sqrt(-d^10/f^4) + 27/2*((-d^10/f^4)^(1/4)*d^5*f*cos(f*x + e)*sin
(f*x + e) - (-d^10/f^4)^(3/4)*f^3*cos(f*x + e)^2)*sqrt(d*sin(f*x + e)/cos(f*x + e))) - 3*(-d^10/f^4)^(1/4)*f*l
og(27/2*d^8*cos(f*x + e)*sin(f*x + e) - 27/4*(2*d^3*f^2*cos(f*x + e)^2 - d^3*f^2)*sqrt(-d^10/f^4) - 27/2*((-d^
10/f^4)^(1/4)*d^5*f*cos(f*x + e)*sin(f*x + e) - (-d^10/f^4)^(3/4)*f^3*cos(f*x + e)^2)*sqrt(d*sin(f*x + e)/cos(
f*x + e))) + 3*(-d^10/f^4)^(1/4)*f*log(27*d^8 + 54*((-d^10/f^4)^(1/4)*d^5*f*cos(f*x + e)^2 - (-d^10/f^4)^(3/4)
*f^3*cos(f*x + e)*sin(f*x + e))*sqrt(d*sin(f*x + e)/cos(f*x + e))) - 3*(-d^10/f^4)^(1/4)*f*log(27*d^8 - 54*((-
d^10/f^4)^(1/4)*d^5*f*cos(f*x + e)^2 - (-d^10/f^4)^(3/4)*f^3*cos(f*x + e)*sin(f*x + e))*sqrt(d*sin(f*x + e)/co
s(f*x + e))) - 3*I*(-d^10/f^4)^(1/4)*f*log(27*d^8 - 54*(I*(-d^10/f^4)^(1/4)*d^5*f*cos(f*x + e)^2 + I*(-d^10/f^
4)^(3/4)*f^3*cos(f*x + e)*sin(f*x + e))*sqrt(d*sin(f*x + e)/cos(f*x + e))) + 3*I*(-d^10/f^4)^(1/4)*f*log(27*d^
8 - 54*(-I*(-d^10/f^4)^(1/4)*d^5*f*cos(f*x + e)^2 - I*(-d^10/f^4)^(3/4)*f^3*cos(f*x + e)*sin(f*x + e))*sqrt(d*
sin(f*x + e)/cos(f*x + e))))/f

Sympy [F(-1)]

Timed out. \[ \int \cos ^2(e+f x) (d \tan (e+f x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate(cos(f*x+e)**2*(d*tan(f*x+e))**(5/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.86 \[ \int \cos ^2(e+f x) (d \tan (e+f x))^{5/2} \, dx=\frac {3 \, d^{4} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} - \frac {8 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} d^{4}}{d^{2} \tan \left (f x + e\right )^{2} + d^{2}}}{16 \, d f} \]

[In]

integrate(cos(f*x+e)^2*(d*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

1/16*(3*d^4*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) + 2*sqrt
(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) - sqrt(2)*log(d*tan(f*x +
e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d) + sqrt(2)*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x +
e))*sqrt(d) + d)/sqrt(d)) - 8*(d*tan(f*x + e))^(3/2)*d^4/(d^2*tan(f*x + e)^2 + d^2))/(d*f)

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.03 \[ \int \cos ^2(e+f x) (d \tan (e+f x))^{5/2} \, dx=-\frac {1}{16} \, {\left (\frac {8 \, \sqrt {d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right )}{{\left (d^{2} \tan \left (f x + e\right )^{2} + d^{2}\right )} f} - \frac {6 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{d f} - \frac {6 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{d f} + \frac {3 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{d f} - \frac {3 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{d f}\right )} d^{2} \]

[In]

integrate(cos(f*x+e)^2*(d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

-1/16*(8*sqrt(d*tan(f*x + e))*d^2*tan(f*x + e)/((d^2*tan(f*x + e)^2 + d^2)*f) - 6*sqrt(2)*abs(d)^(3/2)*arctan(
1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))/(d*f) - 6*sqrt(2)*abs(d)^(3/2)*arcta
n(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))/(d*f) + 3*sqrt(2)*abs(d)^(3/2)*lo
g(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)) + abs(d))/(d*f) - 3*sqrt(2)*abs(d)^(3/2)*log(d*ta
n(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)) + abs(d))/(d*f))*d^2

Mupad [F(-1)]

Timed out. \[ \int \cos ^2(e+f x) (d \tan (e+f x))^{5/2} \, dx=\int {\cos \left (e+f\,x\right )}^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2} \,d x \]

[In]

int(cos(e + f*x)^2*(d*tan(e + f*x))^(5/2),x)

[Out]

int(cos(e + f*x)^2*(d*tan(e + f*x))^(5/2), x)