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

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

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.179872, antiderivative size = 253, normalized size of antiderivative = 1., 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 = {2607, 288, 290, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac{3 d^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{32 \sqrt{2} f}+\frac{3 d^{5/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{32 \sqrt{2} f}+\frac{3 d^{5/2} \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{64 \sqrt{2} f}-\frac{3 d^{5/2} \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{64 \sqrt{2} f}-\frac{d \cos ^4(e+f x) (d \tan (e+f x))^{3/2}}{4 f}+\frac{3 d \cos ^2(e+f x) (d \tan (e+f x))^{3/2}}{16 f} \]

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 2607

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])

Rule 288

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 290

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] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 329

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 297

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 1162

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 617

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 204

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

Rule 1165

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 628

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]

Rubi steps

\begin{align*} \int \cos ^4(e+f x) (d \tan (e+f x))^{5/2} \, dx &=\frac{\operatorname{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 ) \operatorname{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 ) \operatorname{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) \operatorname{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) \operatorname{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) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [A]  time = 0.188587, size = 125, normalized size = 0.49 \[ -\frac{d^2 \sqrt{d \tan (e+f x)} \left (-2 \sin (2 (e+f x))+2 \sin (4 (e+f x))+3 \sqrt{\sin (2 (e+f x))} \csc (e+f x) \sin ^{-1}(\cos (e+f x)-\sin (e+f x))+3 \sqrt{\sin (2 (e+f x))} \csc (e+f x) \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )\right )}{64 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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]])/(64*f)

________________________________________________________________________________________

Maple [C]  time = 0.145, size = 550, normalized size = 2.2 \begin{align*} \text{result too large to display} \end{align*}

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)

[Out]

1/64/f*2^(1/2)*(cos(f*x+e)-1)*(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)*((cos(f*x+e)-1+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*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)*((cos(f*x
+e)-1+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)^4*2^(1/2)+8*cos(f*x+e)^3*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)*((cos(f*x+e)-1+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)*((cos(f*x+e)-1+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))+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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

________________________________________________________________________________________

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

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

________________________________________________________________________________________

Giac [A]  time = 1.33293, size = 366, normalized size = 1.45 \begin{align*} \frac{1}{128} \, d^{5}{\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^{4} 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^{4} 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^{4} 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^{4} f} + \frac{8 \,{\left (3 \, \sqrt{d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right )^{3} - \sqrt{d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right )\right )}}{{\left (d^{2} \tan \left (f x + e\right )^{2} + d^{2}\right )}^{2} d^{2} f}\right )} \end{align*}

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^5*(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^4*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^4*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)) + ab
s(d))/(d^4*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^4*f) + 8*(3*sqrt(d*tan(f*x + e))*d^3*tan(f*x + e)^3 - sqrt(d*tan(f*x + e))*d^3*tan(f*x + e))/((d^2*tan(f*
x + e)^2 + d^2)^2*d^2*f))