3.3.52 \(\int \cos ^4(e+f x) (d \tan (e+f x))^{5/2} \, dx\) [252]

Optimal. Leaf size=253 \[ -\frac {3 d^{5/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{32 \sqrt {2} f}+\frac {3 d^{5/2} \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{32 \sqrt {2} f}+\frac {3 d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{64 \sqrt {2} f}-\frac {3 d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{64 \sqrt {2} f}+\frac {3 d \cos ^2(e+f x) (d \tan (e+f x))^{3/2}}{16 f}-\frac {d \cos ^4(e+f x) (d \tan (e+f x))^{3/2}}{4 f} \]

[Out]

-3/64*d^(5/2)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/f*2^(1/2)+3/64*d^(5/2)*arctan(1+2^(1/2)*(d*tan(f*
x+e))^(1/2)/d^(1/2))/f*2^(1/2)+3/128*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)-3/128*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*cos(f*x+e)^2*(
d*tan(f*x+e))^(3/2)/f-1/4*d*cos(f*x+e)^4*(d*tan(f*x+e))^(3/2)/f

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Rubi [A]
time = 0.13, antiderivative size = 253, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {2687, 294, 296, 335, 303, 1176, 631, 210, 1179, 642} \begin {gather*} -\frac {3 d^{5/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{32 \sqrt {2} f}+\frac {3 d^{5/2} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{32 \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 )}{64 \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 )}{64 \sqrt {2} f}-\frac {d \cos ^4(e+f x) (d \tan (e+f x))^{3/2}}{4 f}+\frac {3 d \cos ^2(e+f x) (d \tan (e+f x))^{3/2}}{16 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^4*(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]])/(32*Sqrt[2]*f) + (3*d^(5/2)*ArcTan[1 + (Sqrt[2
]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(32*Sqrt[2]*f) + (3*d^(5/2)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqr
t[d*Tan[e + f*x]]])/(64*Sqrt[2]*f) - (3*d^(5/2)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*
x]]])/(64*Sqrt[2]*f) + (3*d*Cos[e + f*x]^2*(d*Tan[e + f*x])^(3/2))/(16*f) - (d*Cos[e + f*x]^4*(d*Tan[e + f*x])
^(3/2))/(4*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 296

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(c*x)^(m + 1))*((a + b*x^n)^(p + 1)/
(a*c*n*(p + 1))), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && 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*} \int \cos ^4(e+f x) (d \tan (e+f x))^{5/2} \, dx &=\frac {\text {Subst}\left (\int \frac {(d x)^{5/2}}{\left (1+x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {d \cos ^4(e+f x) (d \tan (e+f x))^{3/2}}{4 f}+\frac {\left (3 d^2\right ) \text {Subst}\left (\int \frac {\sqrt {d x}}{\left (1+x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=\frac {3 d \cos ^2(e+f x) (d \tan (e+f x))^{3/2}}{16 f}-\frac {d \cos ^4(e+f x) (d \tan (e+f x))^{3/2}}{4 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 )}{32 f}\\ &=\frac {3 d \cos ^2(e+f x) (d \tan (e+f x))^{3/2}}{16 f}-\frac {d \cos ^4(e+f x) (d \tan (e+f x))^{3/2}}{4 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 )}{16 f}\\ &=\frac {3 d \cos ^2(e+f x) (d \tan (e+f x))^{3/2}}{16 f}-\frac {d \cos ^4(e+f x) (d \tan (e+f x))^{3/2}}{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 )}{32 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 )}{32 f}\\ &=\frac {3 d \cos ^2(e+f x) (d \tan (e+f x))^{3/2}}{16 f}-\frac {d \cos ^4(e+f x) (d \tan (e+f x))^{3/2}}{4 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 )}{64 \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 )}{64 \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 )}{64 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 )}{64 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 )}{64 \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 )}{64 \sqrt {2} f}+\frac {3 d \cos ^2(e+f x) (d \tan (e+f x))^{3/2}}{16 f}-\frac {d \cos ^4(e+f x) (d \tan (e+f x))^{3/2}}{4 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 )}{32 \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 )}{32 \sqrt {2} f}\\ &=-\frac {3 d^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{32 \sqrt {2} f}+\frac {3 d^{5/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{32 \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 )}{64 \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 )}{64 \sqrt {2} f}+\frac {3 d \cos ^2(e+f x) (d \tan (e+f x))^{3/2}}{16 f}-\frac {d \cos ^4(e+f x) (d \tan (e+f x))^{3/2}}{4 f}\\ \end {align*}

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Mathematica [A]
time = 0.21, size = 125, normalized size = 0.49 \begin {gather*} -\frac {d^2 \left (3 \text {ArcSin}(\cos (e+f x)-\sin (e+f x)) \csc (e+f x) \sqrt {\sin (2 (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 ) \sqrt {\sin (2 (e+f x))}-2 \sin (2 (e+f x))+2 \sin (4 (e+f x))\right ) \sqrt {d \tan (e+f x)}}{64 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^4*(d*Tan[e + f*x])^(5/2),x]

[Out]

-1/64*(d^2*(3*ArcSin[Cos[e + f*x] - Sin[e + f*x]]*Csc[e + f*x]*Sqrt[Sin[2*(e + f*x)]] + 3*Csc[e + f*x]*Log[Cos
[e + f*x] + Sin[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]*Sqrt[Sin[2*(e + f*x)]] - 2*Sin[2*(e + f*x)] + 2*Sin[4*(e +
f*x)])*Sqrt[d*Tan[e + f*x]])/f

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.30, size = 558, normalized size = 2.21

method	result	size
default	\(-\frac {\left (\cos \left (f x +e \right )-1\right ) \left (3 i \sqrt {\frac {\cos \left (f x +e \right )-1}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-3 i \sqrt {\frac {\cos \left (f x +e \right )-1}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+8 \sqrt {2}\, \left (\cos ^{4}\left (f x +e \right )\right )-3 \sqrt {\frac {\cos \left (f x +e \right )-1}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {\frac {\cos \left (f x +e \right )-1}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-8 \left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {2}-6 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {2}+6 \cos \left (f x +e \right ) \sqrt {2}\right ) \left (\cos ^{2}\left (f x +e \right )\right ) \left (\cos \left (f x +e \right )+1\right )^{2} \left (\frac {d \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \sqrt {2}}{64 f \sin \left (f x +e \right )^{5}}\)	\(558\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^4*(d*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/64/f*(cos(f*x+e)-1)*(3*I*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2-1/2*I,1/2*2^(1/2))*(
(cos(f*x+e)-1)/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(-1+cos(f*x+e)-sin(f*x+e))/si
n(f*x+e))^(1/2)-3*I*((cos(f*x+e)-1)/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(-1+cos(
f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2+1/2*I,1/2*2
^(1/2))+8*2^(1/2)*cos(f*x+e)^4-3*((cos(f*x+e)-1)/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/
2)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/
2-1/2*I,1/2*2^(1/2))-3*((cos(f*x+e)-1)/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(-1+c
os(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2+1/2*I,1/
2*2^(1/2))-8*cos(f*x+e)^3*2^(1/2)-6*cos(f*x+e)^2*2^(1/2)+6*cos(f*x+e)*2^(1/2))*cos(f*x+e)^2*(cos(f*x+e)+1)^2*(
d*sin(f*x+e)/cos(f*x+e))^(5/2)/sin(f*x+e)^5*2^(1/2)

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Maxima [A]
time = 0.49, size = 235, normalized size = 0.93 \begin {gather*} \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 (3 \, \left (d \tan \left (f x + e\right )\right )^{\frac {7}{2}} d^{4} - \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} d^{6}\right )}}{d^{4} \tan \left (f x + e\right )^{4} + 2 \, d^{4} \tan \left (f x + e\right )^{2} + d^{4}}}{128 \, d f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^4*(d*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

1/128*(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*sqr
t(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*(3*(d*tan(f*x + e))^(7/2)*d^4 - (d*tan(f*x + e))^(3/2)*d^6)/(d^4*tan(f*x + e)^
4 + 2*d^4*tan(f*x + e)^2 + d^4))/(d*f)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 2047 vs. \(2 (203) = 406\).
time = 64.32, size = 2047, normalized size = 8.09 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^4*(d*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/512*(12*sqrt(2)*(d^10/f^4)^(1/4)*f*arctan((sqrt(d^16 + 4*sqrt(d^10/f^4)*d^11*f^2*cos(f*x + e)*sin(f*x + e) -
2*(sqrt(2)*(d^10/f^4)^(1/4)*d^13*f*cos(f*x + e)^2 + sqrt(2)*(d^10/f^4)^(3/4)*d^8*f^3*cos(f*x + e)*sin(f*x + e
))*sqrt(d*sin(f*x + e)/cos(f*x + e)))*(2*d^8*cos(f*x + e)*sin(f*x + e) + sqrt(d^10/f^4)*d^3*f^2 + (sqrt(2)*(d^
10/f^4)^(1/4)*d^5*f*cos(f*x + e)*sin(f*x + e) + sqrt(2)*(d^10/f^4)^(3/4)*f^3*cos(f*x + e)^2)*sqrt(d*sin(f*x +
e)/cos(f*x + e))) + (sqrt(2)*(d^10/f^4)^(1/4)*d^13*f*cos(f*x + e)^2 + sqrt(2)*(d^10/f^4)^(3/4)*d^8*f^3*cos(f*x
+ e)*sin(f*x + e))*sqrt(d*sin(f*x + e)/cos(f*x + e)))/(2*d^16*cos(f*x + e)^2 - d^16)) + 12*sqrt(2)*(d^10/f^4)
^(1/4)*f*arctan(-(sqrt(d^16 + 4*sqrt(d^10/f^4)*d^11*f^2*cos(f*x + e)*sin(f*x + e) + 2*(sqrt(2)*(d^10/f^4)^(1/4
)*d^13*f*cos(f*x + e)^2 + sqrt(2)*(d^10/f^4)^(3/4)*d^8*f^3*cos(f*x + e)*sin(f*x + e))*sqrt(d*sin(f*x + e)/cos(
f*x + e)))*(2*d^8*cos(f*x + e)*sin(f*x + e) + sqrt(d^10/f^4)*d^3*f^2 - (sqrt(2)*(d^10/f^4)^(1/4)*d^5*f*cos(f*x
+ e)*sin(f*x + e) + sqrt(2)*(d^10/f^4)^(3/4)*f^3*cos(f*x + e)^2)*sqrt(d*sin(f*x + e)/cos(f*x + e))) - (sqrt(2
)*(d^10/f^4)^(1/4)*d^13*f*cos(f*x + e)^2 + sqrt(2)*(d^10/f^4)^(3/4)*d^8*f^3*cos(f*x + e)*sin(f*x + e))*sqrt(d*
sin(f*x + e)/cos(f*x + e)))/(2*d^16*cos(f*x + e)^2 - d^16)) + 12*sqrt(2)*(d^10/f^4)^(1/4)*f*arctan(-1/2*(2*d^1
6*sin(f*x + e) - sqrt(d^16 + 4*sqrt(d^10/f^4)*d^11*f^2*cos(f*x + e)*sin(f*x + e) + 2*(sqrt(2)*(d^10/f^4)^(1/4)
*d^13*f*cos(f*x + e)^2 + sqrt(2)*(d^10/f^4)^(3/4)*d^8*f^3*cos(f*x + e)*sin(f*x + e))*sqrt(d*sin(f*x + e)/cos(f
*x + e)))*(sqrt(2)*(d^10/f^4)^(1/4)*d^5*f*sin(f*x + e) + sqrt(2)*(d^10/f^4)^(3/4)*f^3*cos(f*x + e))*sqrt(d*sin
(f*x + e)/cos(f*x + e)) - 4*(d^11*f^2*cos(f*x + e)^3 - d^11*f^2*cos(f*x + e))*sqrt(d^10/f^4) + (sqrt(2)*(d^10/
f^4)^(1/4)*d^13*f*sin(f*x + e) + sqrt(2)*(d^10/f^4)^(3/4)*d^8*f^3*cos(f*x + e))*sqrt(d*sin(f*x + e)/cos(f*x +
e)))/((2*d^16*cos(f*x + e)^2 - d^16)*sin(f*x + e))) + 12*sqrt(2)*(d^10/f^4)^(1/4)*f*arctan(1/2*(2*d^16*sin(f*x
+ e) + sqrt(d^16 + 4*sqrt(d^10/f^4)*d^11*f^2*cos(f*x + e)*sin(f*x + e) - 2*(sqrt(2)*(d^10/f^4)^(1/4)*d^13*f*c
os(f*x + e)^2 + sqrt(2)*(d^10/f^4)^(3/4)*d^8*f^3*cos(f*x + e)*sin(f*x + e))*sqrt(d*sin(f*x + e)/cos(f*x + e)))
*(sqrt(2)*(d^10/f^4)^(1/4)*d^5*f*sin(f*x + e) + sqrt(2)*(d^10/f^4)^(3/4)*f^3*cos(f*x + e))*sqrt(d*sin(f*x + e)
/cos(f*x + e)) - 4*(d^11*f^2*cos(f*x + e)^3 - d^11*f^2*cos(f*x + e))*sqrt(d^10/f^4) - (sqrt(2)*(d^10/f^4)^(1/4
)*d^13*f*sin(f*x + e) + sqrt(2)*(d^10/f^4)^(3/4)*d^8*f^3*cos(f*x + e))*sqrt(d*sin(f*x + e)/cos(f*x + e)))/((2*
d^16*cos(f*x + e)^2 - d^16)*sin(f*x + e))) - 3*sqrt(2)*(d^10/f^4)^(1/4)*f*log(729*d^16 + 2916*sqrt(d^10/f^4)*d
^11*f^2*cos(f*x + e)*sin(f*x + e) + 1458*(sqrt(2)*(d^10/f^4)^(1/4)*d^13*f*cos(f*x + e)^2 + sqrt(2)*(d^10/f^4)^
(3/4)*d^8*f^3*cos(f*x + e)*sin(f*x + e))*sqrt(d*sin(f*x + e)/cos(f*x + e))) + 3*sqrt(2)*(d^10/f^4)^(1/4)*f*log
(729*d^16 + 2916*sqrt(d^10/f^4)*d^11*f^2*cos(f*x + e)*sin(f*x + e) - 1458*(sqrt(2)*(d^10/f^4)^(1/4)*d^13*f*cos
(f*x + e)^2 + sqrt(2)*(d^10/f^4)^(3/4)*d^8*f^3*cos(f*x + e)*sin(f*x + e))*sqrt(d*sin(f*x + e)/cos(f*x + e))) -
3*sqrt(2)*(d^10/f^4)^(1/4)*f*log(729/16*d^16 + 729/4*sqrt(d^10/f^4)*d^11*f^2*cos(f*x + e)*sin(f*x + e) + 729/
8*(sqrt(2)*(d^10/f^4)^(1/4)*d^13*f*cos(f*x + e)^2 + sqrt(2)*(d^10/f^4)^(3/4)*d^8*f^3*cos(f*x + e)*sin(f*x + e)
)*sqrt(d*sin(f*x + e)/cos(f*x + e))) + 3*sqrt(2)*(d^10/f^4)^(1/4)*f*log(729/16*d^16 + 729/4*sqrt(d^10/f^4)*d^1
1*f^2*cos(f*x + e)*sin(f*x + e) - 729/8*(sqrt(2)*(d^10/f^4)^(1/4)*d^13*f*cos(f*x + e)^2 + sqrt(2)*(d^10/f^4)^(
3/4)*d^8*f^3*cos(f*x + e)*sin(f*x + e))*sqrt(d*sin(f*x + e)/cos(f*x + e))) - 32*(4*d^2*cos(f*x + e)^3 - 3*d^2*
cos(f*x + e))*sqrt(d*sin(f*x + e)/cos(f*x + e))*sin(f*x + e))/f

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)**4*(d*tan(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [A]
time = 0.49, size = 268, normalized size = 1.06 \begin {gather*} \frac {1}{128} \, d^{2} {\left (\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} + \frac {8 \, {\left (3 \, \sqrt {d \tan \left (f x + e\right )} d^{4} \tan \left (f x + e\right )^{3} - \sqrt {d \tan \left (f x + e\right )} d^{4} \tan \left (f x + e\right )\right )}}{{\left (d^{2} \tan \left (f x + e\right )^{2} + d^{2}\right )}^{2} f}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^4*(d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

1/128*d^2*(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)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(f*x + e)))/sqrt(a
bs(d)))/(d*f) - 3*sqrt(2)*abs(d)^(3/2)*log(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*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)) + abs(d))/(d*f
) + 8*(3*sqrt(d*tan(f*x + e))*d^4*tan(f*x + e)^3 - sqrt(d*tan(f*x + e))*d^4*tan(f*x + e))/((d^2*tan(f*x + e)^2
+ d^2)^2*f))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\cos \left (e+f\,x\right )}^4\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(e + f*x)^4*(d*tan(e + f*x))^(5/2),x)

[Out]

int(cos(e + f*x)^4*(d*tan(e + f*x))^(5/2), x)

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