3.3.62 \(\int \frac {\cos ^2(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx\) [262]

Optimal. Leaf size=249 \[ \frac {5 \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b d^{3/2}}-\frac {5 \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b d^{3/2}}-\frac {5 \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b d^{3/2}}+\frac {5 \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b d^{3/2}}-\frac {5}{2 b d \sqrt {d \tan (a+b x)}}+\frac {\cos ^2(a+b x)}{2 b d \sqrt {d \tan (a+b x)}} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 249, 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, 296, 331, 335, 303, 1176, 631, 210, 1179, 642} \begin {gather*} \frac {5 \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b d^{3/2}}-\frac {5 \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{4 \sqrt {2} b d^{3/2}}-\frac {5 \log \left (\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{8 \sqrt {2} b d^{3/2}}+\frac {5 \log \left (\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{8 \sqrt {2} b d^{3/2}}-\frac {5}{2 b d \sqrt {d \tan (a+b x)}}+\frac {\cos ^2(a+b x)}{2 b d \sqrt {d \tan (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^2/(d*Tan[a + b*x])^(3/2),x]

[Out]

(5*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[a + b*x]])/Sqrt[d]])/(4*Sqrt[2]*b*d^(3/2)) - (5*ArcTan[1 + (Sqrt[2]*Sqrt[d*T
an[a + b*x]])/Sqrt[d]])/(4*Sqrt[2]*b*d^(3/2)) - (5*Log[Sqrt[d] + Sqrt[d]*Tan[a + b*x] - Sqrt[2]*Sqrt[d*Tan[a +
b*x]]])/(8*Sqrt[2]*b*d^(3/2)) + (5*Log[Sqrt[d] + Sqrt[d]*Tan[a + b*x] + Sqrt[2]*Sqrt[d*Tan[a + b*x]]])/(8*Sqr
t[2]*b*d^(3/2)) - 5/(2*b*d*Sqrt[d*Tan[a + b*x]]) + Cos[a + b*x]^2/(2*b*d*Sqrt[d*Tan[a + b*x]])

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 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 331

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

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 \frac {\cos ^2(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx &=\frac {\text {Subst}\left (\int \frac {1}{(d x)^{3/2} \left (1+x^2\right )^2} \, dx,x,\tan (a+b x)\right )}{b}\\ &=\frac {\cos ^2(a+b x)}{2 b d \sqrt {d \tan (a+b x)}}+\frac {5 \text {Subst}\left (\int \frac {1}{(d x)^{3/2} \left (1+x^2\right )} \, dx,x,\tan (a+b x)\right )}{4 b}\\ &=-\frac {5}{2 b d \sqrt {d \tan (a+b x)}}+\frac {\cos ^2(a+b x)}{2 b d \sqrt {d \tan (a+b x)}}-\frac {5 \text {Subst}\left (\int \frac {\sqrt {d x}}{1+x^2} \, dx,x,\tan (a+b x)\right )}{4 b d^2}\\ &=-\frac {5}{2 b d \sqrt {d \tan (a+b x)}}+\frac {\cos ^2(a+b x)}{2 b d \sqrt {d \tan (a+b x)}}-\frac {5 \text {Subst}\left (\int \frac {x^2}{1+\frac {x^4}{d^2}} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{2 b d^3}\\ &=-\frac {5}{2 b d \sqrt {d \tan (a+b x)}}+\frac {\cos ^2(a+b x)}{2 b d \sqrt {d \tan (a+b x)}}+\frac {5 \text {Subst}\left (\int \frac {d-x^2}{1+\frac {x^4}{d^2}} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{4 b d^3}-\frac {5 \text {Subst}\left (\int \frac {d+x^2}{1+\frac {x^4}{d^2}} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{4 b d^3}\\ &=-\frac {5}{2 b d \sqrt {d \tan (a+b x)}}+\frac {\cos ^2(a+b x)}{2 b d \sqrt {d \tan (a+b x)}}-\frac {5 \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}+2 x}{-d-\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b d^{3/2}}-\frac {5 \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}-2 x}{-d+\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b d^{3/2}}-\frac {5 \text {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{8 b d}-\frac {5 \text {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{8 b d}\\ &=-\frac {5 \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b d^{3/2}}+\frac {5 \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b d^{3/2}}-\frac {5}{2 b d \sqrt {d \tan (a+b x)}}+\frac {\cos ^2(a+b x)}{2 b d \sqrt {d \tan (a+b x)}}-\frac {5 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b d^{3/2}}+\frac {5 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b d^{3/2}}\\ &=\frac {5 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b d^{3/2}}-\frac {5 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b d^{3/2}}-\frac {5 \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b d^{3/2}}+\frac {5 \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b d^{3/2}}-\frac {5}{2 b d \sqrt {d \tan (a+b x)}}+\frac {\cos ^2(a+b x)}{2 b d \sqrt {d \tan (a+b x)}}\\ \end {align*}

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Mathematica [A]
time = 0.31, size = 115, normalized size = 0.46 \begin {gather*} \frac {\csc (a+b x) \left (-17 \cos (a+b x)+\cos (3 (a+b x))+5 \text {ArcSin}(\cos (a+b x)-\sin (a+b x)) \sqrt {\sin (2 (a+b x))}+5 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right ) \sqrt {\sin (2 (a+b x))}\right ) \sqrt {d \tan (a+b x)}}{8 b d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^2/(d*Tan[a + b*x])^(3/2),x]

[Out]

(Csc[a + b*x]*(-17*Cos[a + b*x] + Cos[3*(a + b*x)] + 5*ArcSin[Cos[a + b*x] - Sin[a + b*x]]*Sqrt[Sin[2*(a + b*x
)]] + 5*Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*(a + b*x)]]]*Sqrt[Sin[2*(a + b*x)]])*Sqrt[d*Tan[a + b*x]]
)/(8*b*d^2)

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.28, size = 982, normalized size = 3.94

method	result	size
default	\(\text {Expression too large to display}\)	\(982\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^2/(d*tan(b*x+a))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/8/b*(5*I*EllipticPi((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2-1/2*I,1/2*2^(1/2))*(-(cos(b*x+a)-1-si
n(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*co
s(b*x+a)-5*I*(-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+c
os(b*x+a))/sin(b*x+a))^(1/2)*EllipticPi((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))*c
os(b*x+a)+5*I*EllipticPi((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2-1/2*I,1/2*2^(1/2))*(-(cos(b*x+a)-1-
sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)-
5*I*(-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)
)/sin(b*x+a))^(1/2)*EllipticPi((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))-5*Elliptic
Pi((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2-1/2*I,1/2*2^(1/2))*(-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a)
)^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*cos(b*x+a)-5*(-(cos(b*
x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))
^(1/2)*EllipticPi((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(b*x+a)-5*EllipticPi
((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2-1/2*I,1/2*2^(1/2))*(-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^
(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)-5*(-(cos(b*x+a)-1-sin(b*
x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*Ellipt
icPi((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))-2*cos(b*x+a)^3*2^(1/2)+10*cos(b*x+a)
*2^(1/2))*sin(b*x+a)/cos(b*x+a)^2/(d*sin(b*x+a)/cos(b*x+a))^(3/2)*2^(1/2)

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Maxima [A]
time = 0.49, size = 204, normalized size = 0.82 \begin {gather*} -\frac {\frac {10 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {10 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {5 \, \sqrt {2} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {5 \, \sqrt {2} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {8 \, {\left (5 \, d^{2} \tan \left (b x + a\right )^{2} + 4 \, d^{2}\right )}}{\left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}} + \sqrt {d \tan \left (b x + a\right )} d^{2}}}{16 \, b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2/(d*tan(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

-1/16*(10*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(b*x + a)))/sqrt(d))/sqrt(d) + 10*sqrt(2)*
arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(b*x + a)))/sqrt(d))/sqrt(d) - 5*sqrt(2)*log(d*tan(b*x + a)
+ sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(d) + d)/sqrt(d) + 5*sqrt(2)*log(d*tan(b*x + a) - sqrt(2)*sqrt(d*tan(b*x +
a))*sqrt(d) + d)/sqrt(d) + 8*(5*d^2*tan(b*x + a)^2 + 4*d^2)/((d*tan(b*x + a))^(5/2) + sqrt(d*tan(b*x + a))*d^
2))/(b*d)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 2037 vs. \(2 (189) = 378\).
time = 62.54, size = 2037, normalized size = 8.18 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2/(d*tan(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

-1/64*(32*(cos(b*x + a)^3 - 5*cos(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a))*sin(b*x + a) - 20*(sqrt(2)*b*d^2
*cos(b*x + a)^2 - sqrt(2)*b*d^2)*(1/(b^4*d^6))^(1/4)*arctan((sqrt(4*b^2*d^3*sqrt(1/(b^4*d^6))*cos(b*x + a)*sin
(b*x + a) - 2*(sqrt(2)*b^3*d^4*(1/(b^4*d^6))^(3/4)*cos(b*x + a)*sin(b*x + a) + sqrt(2)*b*d*(1/(b^4*d^6))^(1/4)
*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 1)*(b^2*d^3*sqrt(1/(b^4*d^6)) + 2*cos(b*x + a)*sin(b*x +
a) + (sqrt(2)*b^3*d^4*(1/(b^4*d^6))^(3/4)*cos(b*x + a)^2 + sqrt(2)*b*d*(1/(b^4*d^6))^(1/4)*cos(b*x + a)*sin(b*
x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a))) - (sqrt(2)*b^3*d^4*(1/(b^4*d^6))^(3/4)*cos(b*x + a)*sin(b*x + a) +
sqrt(2)*b*d*(1/(b^4*d^6))^(1/4)*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a)))/(2*cos(b*x + a)^2 - 1)) - 2
0*(sqrt(2)*b*d^2*cos(b*x + a)^2 - sqrt(2)*b*d^2)*(1/(b^4*d^6))^(1/4)*arctan(-(sqrt(4*b^2*d^3*sqrt(1/(b^4*d^6))
*cos(b*x + a)*sin(b*x + a) + 2*(sqrt(2)*b^3*d^4*(1/(b^4*d^6))^(3/4)*cos(b*x + a)*sin(b*x + a) + sqrt(2)*b*d*(1
/(b^4*d^6))^(1/4)*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 1)*(b^2*d^3*sqrt(1/(b^4*d^6)) + 2*cos(b*
x + a)*sin(b*x + a) - (sqrt(2)*b^3*d^4*(1/(b^4*d^6))^(3/4)*cos(b*x + a)^2 + sqrt(2)*b*d*(1/(b^4*d^6))^(1/4)*co
s(b*x + a)*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a))) + (sqrt(2)*b^3*d^4*(1/(b^4*d^6))^(3/4)*cos(b*x + a
)*sin(b*x + a) + sqrt(2)*b*d*(1/(b^4*d^6))^(1/4)*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a)))/(2*cos(b*x
+ a)^2 - 1)) + 20*(sqrt(2)*b*d^2*cos(b*x + a)^2 - sqrt(2)*b*d^2)*(1/(b^4*d^6))^(1/4)*arctan(1/2*((sqrt(2)*b^3
*d^4*(1/(b^4*d^6))^(3/4)*cos(b*x + a) + sqrt(2)*b*d*(1/(b^4*d^6))^(1/4)*sin(b*x + a))*sqrt(4*b^2*d^3*sqrt(1/(b
^4*d^6))*cos(b*x + a)*sin(b*x + a) + 2*(sqrt(2)*b^3*d^4*(1/(b^4*d^6))^(3/4)*cos(b*x + a)*sin(b*x + a) + sqrt(2
)*b*d*(1/(b^4*d^6))^(1/4)*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 1)*sqrt(d*sin(b*x + a)/cos(b*x +
a)) - (sqrt(2)*b^3*d^4*(1/(b^4*d^6))^(3/4)*cos(b*x + a) + sqrt(2)*b*d*(1/(b^4*d^6))^(1/4)*sin(b*x + a))*sqrt(
d*sin(b*x + a)/cos(b*x + a)) + 4*(b^2*d^3*cos(b*x + a)^3 - b^2*d^3*cos(b*x + a))*sqrt(1/(b^4*d^6)) - 2*sin(b*x
+ a))/((2*cos(b*x + a)^2 - 1)*sin(b*x + a))) + 20*(sqrt(2)*b*d^2*cos(b*x + a)^2 - sqrt(2)*b*d^2)*(1/(b^4*d^6)
)^(1/4)*arctan(1/2*((sqrt(2)*b^3*d^4*(1/(b^4*d^6))^(3/4)*cos(b*x + a) + sqrt(2)*b*d*(1/(b^4*d^6))^(1/4)*sin(b*
x + a))*sqrt(4*b^2*d^3*sqrt(1/(b^4*d^6))*cos(b*x + a)*sin(b*x + a) - 2*(sqrt(2)*b^3*d^4*(1/(b^4*d^6))^(3/4)*co
s(b*x + a)*sin(b*x + a) + sqrt(2)*b*d*(1/(b^4*d^6))^(1/4)*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a)) +
1)*sqrt(d*sin(b*x + a)/cos(b*x + a)) - (sqrt(2)*b^3*d^4*(1/(b^4*d^6))^(3/4)*cos(b*x + a) + sqrt(2)*b*d*(1/(b^4
*d^6))^(1/4)*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) - 4*(b^2*d^3*cos(b*x + a)^3 - b^2*d^3*cos(b*x + a
))*sqrt(1/(b^4*d^6)) + 2*sin(b*x + a))/((2*cos(b*x + a)^2 - 1)*sin(b*x + a))) - 5*(sqrt(2)*b*d^2*cos(b*x + a)^
2 - sqrt(2)*b*d^2)*(1/(b^4*d^6))^(1/4)*log(4*b^2*d^3*sqrt(1/(b^4*d^6))*cos(b*x + a)*sin(b*x + a) + 2*(sqrt(2)*
b^3*d^4*(1/(b^4*d^6))^(3/4)*cos(b*x + a)*sin(b*x + a) + sqrt(2)*b*d*(1/(b^4*d^6))^(1/4)*cos(b*x + a)^2)*sqrt(d
*sin(b*x + a)/cos(b*x + a)) + 1) + 5*(sqrt(2)*b*d^2*cos(b*x + a)^2 - sqrt(2)*b*d^2)*(1/(b^4*d^6))^(1/4)*log(4*
b^2*d^3*sqrt(1/(b^4*d^6))*cos(b*x + a)*sin(b*x + a) - 2*(sqrt(2)*b^3*d^4*(1/(b^4*d^6))^(3/4)*cos(b*x + a)*sin(
b*x + a) + sqrt(2)*b*d*(1/(b^4*d^6))^(1/4)*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 1) - 5*(sqrt(2)
*b*d^2*cos(b*x + a)^2 - sqrt(2)*b*d^2)*(1/(b^4*d^6))^(1/4)*log(1/4*b^2*d^3*sqrt(1/(b^4*d^6))*cos(b*x + a)*sin(
b*x + a) + 1/8*(sqrt(2)*b^3*d^4*(1/(b^4*d^6))^(3/4)*cos(b*x + a)*sin(b*x + a) + sqrt(2)*b*d*(1/(b^4*d^6))^(1/4
)*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 1/16) + 5*(sqrt(2)*b*d^2*cos(b*x + a)^2 - sqrt(2)*b*d^2)
*(1/(b^4*d^6))^(1/4)*log(1/4*b^2*d^3*sqrt(1/(b^4*d^6))*cos(b*x + a)*sin(b*x + a) - 1/8*(sqrt(2)*b^3*d^4*(1/(b^
4*d^6))^(3/4)*cos(b*x + a)*sin(b*x + a) + sqrt(2)*b*d*(1/(b^4*d^6))^(1/4)*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/
cos(b*x + a)) + 1/16))/(b*d^2*cos(b*x + a)^2 - b*d^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cos ^{2}{\left (a + b x \right )}}{\left (d \tan {\left (a + b x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**2/(d*tan(b*x+a))**(3/2),x)

[Out]

Integral(cos(a + b*x)**2/(d*tan(a + b*x))**(3/2), x)

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Giac [A]
time = 0.51, size = 252, normalized size = 1.01 \begin {gather*} -\frac {\frac {10 \, \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 (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{b d^{2}} + \frac {10 \, \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 (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{b d^{2}} - \frac {5 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{b d^{2}} + \frac {5 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{b d^{2}} + \frac {8 \, {\left (5 \, d^{2} \tan \left (b x + a\right )^{2} + 4 \, d^{2}\right )}}{{\left (\sqrt {d \tan \left (b x + a\right )} d^{2} \tan \left (b x + a\right )^{2} + \sqrt {d \tan \left (b x + a\right )} d^{2}\right )} b}}{16 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2/(d*tan(b*x+a))^(3/2),x, algorithm="giac")

[Out]

-1/16*(10*sqrt(2)*abs(d)^(3/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(b*x + a)))/sqrt(abs(d))
)/(b*d^2) + 10*sqrt(2)*abs(d)^(3/2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(b*x + a)))/sqrt(a
bs(d)))/(b*d^2) - 5*sqrt(2)*abs(d)^(3/2)*log(d*tan(b*x + a) + sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(abs(d)) + abs(
d))/(b*d^2) + 5*sqrt(2)*abs(d)^(3/2)*log(d*tan(b*x + a) - sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(abs(d)) + abs(d))/
(b*d^2) + 8*(5*d^2*tan(b*x + a)^2 + 4*d^2)/((sqrt(d*tan(b*x + a))*d^2*tan(b*x + a)^2 + sqrt(d*tan(b*x + a))*d^
2)*b))/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\cos \left (a+b\,x\right )}^2}{{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^2/(d*tan(a + b*x))^(3/2),x)

[Out]

int(cos(a + b*x)^2/(d*tan(a + b*x))^(3/2), x)

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