3.240 \(\int \cos ^2(a+b x) (d \tan (a+b x))^{3/2} \, dx\)

Optimal. Leaf size=225 \[ -\frac {d^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b}+\frac {d^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{4 \sqrt {2} b}-\frac {d^{3/2} \log \left (\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{8 \sqrt {2} b}+\frac {d^{3/2} \log \left (\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{8 \sqrt {2} b}-\frac {d \cos ^2(a+b x) \sqrt {d \tan (a+b x)}}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 225, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2607, 288, 329, 211, 1165, 628, 1162, 617, 204} \[ -\frac {d^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b}+\frac {d^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{4 \sqrt {2} b}-\frac {d^{3/2} \log \left (\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{8 \sqrt {2} b}+\frac {d^{3/2} \log \left (\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{8 \sqrt {2} b}-\frac {d \cos ^2(a+b x) \sqrt {d \tan (a+b x)}}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 211

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

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 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 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 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]

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 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 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])

Rubi steps

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

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Mathematica [A]  time = 0.28, size = 110, normalized size = 0.49 \[ -\frac {d \csc (a+b x) \sqrt {d \tan (a+b x)} \left (\sin (a+b x)+\sin (3 (a+b x))+\sqrt {\sin (2 (a+b x))} \sin ^{-1}(\cos (a+b x)-\sin (a+b x))-\sqrt {\sin (2 (a+b x))} \log \left (\sin (a+b x)+\sqrt {\sin (2 (a+b x))}+\cos (a+b x)\right )\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 83.68, size = 1558, normalized size = 6.92 \[ \text {result too large to display} \]

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]

-1/32*(16*d*sqrt(d*sin(b*x + a)/cos(b*x + a))*cos(b*x + a)^2 + 2*sqrt(2)*(d^6/b^4)^(1/4)*b*arctan(1/2*(2*d^10*
sin(b*x + a) + sqrt(4*sqrt(d^6/b^4)*b^2*d^7*cos(b*x + a)*sin(b*x + a) + d^10 + 2*(sqrt(2)*(d^6/b^4)^(1/4)*b*d^
8*cos(b*x + a)*sin(b*x + a) + sqrt(2)*(d^6/b^4)^(3/4)*b^3*d^5*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a)
))*(sqrt(2)*(d^6/b^4)^(1/4)*b*d^3*cos(b*x + a) + sqrt(2)*(d^6/b^4)^(3/4)*b^3*sin(b*x + a))*sqrt(d*sin(b*x + a)
/cos(b*x + a)) - 4*(b^2*d^7*cos(b*x + a)^3 - b^2*d^7*cos(b*x + a))*sqrt(d^6/b^4) + (sqrt(2)*(d^6/b^4)^(1/4)*b*
d^8*cos(b*x + a) + sqrt(2)*(d^6/b^4)^(3/4)*b^3*d^5*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)))/((2*d^10*c
os(b*x + a)^2 - d^10)*sin(b*x + a))) + 2*sqrt(2)*(d^6/b^4)^(1/4)*b*arctan(-1/2*(2*d^10*sin(b*x + a) - sqrt(4*s
qrt(d^6/b^4)*b^2*d^7*cos(b*x + a)*sin(b*x + a) + d^10 - 2*(sqrt(2)*(d^6/b^4)^(1/4)*b*d^8*cos(b*x + a)*sin(b*x
+ a) + sqrt(2)*(d^6/b^4)^(3/4)*b^3*d^5*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a)))*(sqrt(2)*(d^6/b^4)^(
1/4)*b*d^3*cos(b*x + a) + sqrt(2)*(d^6/b^4)^(3/4)*b^3*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) - 4*(b^2
*d^7*cos(b*x + a)^3 - b^2*d^7*cos(b*x + a))*sqrt(d^6/b^4) - (sqrt(2)*(d^6/b^4)^(1/4)*b*d^8*cos(b*x + a) + sqrt
(2)*(d^6/b^4)^(3/4)*b^3*d^5*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)))/((2*d^10*cos(b*x + a)^2 - d^10)*s
in(b*x + a))) + 2*sqrt(2)*(d^6/b^4)^(1/4)*b*arctan(1/2*(sqrt(4*sqrt(d^6/b^4)*b^2*d^7*cos(b*x + a)*sin(b*x + a)
+ d^10 - 2*(sqrt(2)*(d^6/b^4)^(1/4)*b*d^8*cos(b*x + a)*sin(b*x + a) + sqrt(2)*(d^6/b^4)^(3/4)*b^3*d^5*cos(b*x
+ a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a)))*(2*d^5*sin(b*x + a) + (sqrt(2)*(d^6/b^4)^(1/4)*b*d^3*cos(b*x + a)
+ sqrt(2)*(d^6/b^4)^(3/4)*b^3*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a))) + (sqrt(2)*(d^6/b^4)^(1/4)*b*d^
8*cos(b*x + a) - sqrt(2)*(d^6/b^4)^(3/4)*b^3*d^5*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)))/(d^10*sin(b*
x + a))) + 2*sqrt(2)*(d^6/b^4)^(1/4)*b*arctan(-1/2*(sqrt(4*sqrt(d^6/b^4)*b^2*d^7*cos(b*x + a)*sin(b*x + a) + d
^10 + 2*(sqrt(2)*(d^6/b^4)^(1/4)*b*d^8*cos(b*x + a)*sin(b*x + a) + sqrt(2)*(d^6/b^4)^(3/4)*b^3*d^5*cos(b*x + a
)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a)))*(2*d^5*sin(b*x + a) - (sqrt(2)*(d^6/b^4)^(1/4)*b*d^3*cos(b*x + a) + sq
rt(2)*(d^6/b^4)^(3/4)*b^3*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a))) - (sqrt(2)*(d^6/b^4)^(1/4)*b*d^8*co
s(b*x + a) - sqrt(2)*(d^6/b^4)^(3/4)*b^3*d^5*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)))/(d^10*sin(b*x +
a))) - sqrt(2)*(d^6/b^4)^(1/4)*b*log(4*sqrt(d^6/b^4)*b^2*d^7*cos(b*x + a)*sin(b*x + a) + d^10 + 2*(sqrt(2)*(d^
6/b^4)^(1/4)*b*d^8*cos(b*x + a)*sin(b*x + a) + sqrt(2)*(d^6/b^4)^(3/4)*b^3*d^5*cos(b*x + a)^2)*sqrt(d*sin(b*x
+ a)/cos(b*x + a))) + sqrt(2)*(d^6/b^4)^(1/4)*b*log(4*sqrt(d^6/b^4)*b^2*d^7*cos(b*x + a)*sin(b*x + a) + d^10 -
2*(sqrt(2)*(d^6/b^4)^(1/4)*b*d^8*cos(b*x + a)*sin(b*x + a) + sqrt(2)*(d^6/b^4)^(3/4)*b^3*d^5*cos(b*x + a)^2)*
sqrt(d*sin(b*x + a)/cos(b*x + a))))/b

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giac [A]  time = 0.59, size = 210, normalized size = 0.93 \[ \frac {1}{16} \, d {\left (\frac {2 \, \sqrt {2} \sqrt {{\left | d \right |}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{b} + \frac {2 \, \sqrt {2} \sqrt {{\left | d \right |}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{b} + \frac {\sqrt {2} \sqrt {{\left | d \right |}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{b} - \frac {\sqrt {2} \sqrt {{\left | d \right |}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{b} - \frac {8 \, \sqrt {d \tan \left (b x + a\right )} d^{2}}{{\left (d^{2} \tan \left (b x + a\right )^{2} + d^{2}\right )} b}\right )} \]

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*d*(2*sqrt(2)*sqrt(abs(d))*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(b*x + a)))/sqrt(abs(d))
)/b + 2*sqrt(2)*sqrt(abs(d))*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(b*x + a)))/sqrt(abs(d)))
/b + sqrt(2)*sqrt(abs(d))*log(d*tan(b*x + a) + sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(abs(d)) + abs(d))/b - sqrt(2)
*sqrt(abs(d))*log(d*tan(b*x + a) - sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(abs(d)) + abs(d))/b - 8*sqrt(d*tan(b*x +
a))*d^2/((d^2*tan(b*x + a)^2 + d^2)*b))

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maple [C]  time = 0.47, size = 670, normalized size = 2.98 \[ \frac {\left (-1+\cos \left (b x +a \right )\right ) \left (i \sin \left (b x +a \right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticPi \left (\sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-i \sin \left (b x +a \right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticPi \left (\sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-\sin \left (b x +a \right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticPi \left (\sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-\sin \left (b x +a \right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticPi \left (\sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+2 \sin \left (b x +a \right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticF \left (\sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right )-2 \left (\cos ^{3}\left (b x +a \right )\right ) \sqrt {2}+2 \left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}\right ) \left (\cos \left (b x +a \right )+1\right )^{2} \cos \left (b x +a \right ) \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {3}{2}} \sqrt {2}}{8 b \sin \left (b x +a \right )^{5}} \]

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)

[Out]

1/8/b*(-1+cos(b*x+a))*(I*sin(b*x+a)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))
^(1/2)*(-(-sin(b*x+a)-1+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticPi((-(-sin(b*x+a)-1+cos(b*x+a))/sin(b*x+a))^(1/2
),1/2+1/2*I,1/2*2^(1/2))-I*sin(b*x+a)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a
))^(1/2)*(-(-sin(b*x+a)-1+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticPi((-(-sin(b*x+a)-1+cos(b*x+a))/sin(b*x+a))^(1
/2),1/2-1/2*I,1/2*2^(1/2))-sin(b*x+a)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a
))^(1/2)*(-(-sin(b*x+a)-1+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticPi((-(-sin(b*x+a)-1+cos(b*x+a))/sin(b*x+a))^(1
/2),1/2+1/2*I,1/2*2^(1/2))-sin(b*x+a)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a
))^(1/2)*(-(-sin(b*x+a)-1+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticPi((-(-sin(b*x+a)-1+cos(b*x+a))/sin(b*x+a))^(1
/2),1/2-1/2*I,1/2*2^(1/2))+2*sin(b*x+a)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x
+a))^(1/2)*(-(-sin(b*x+a)-1+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticF((-(-sin(b*x+a)-1+cos(b*x+a))/sin(b*x+a))^(
1/2),1/2*2^(1/2))-2*cos(b*x+a)^3*2^(1/2)+2*cos(b*x+a)^2*2^(1/2))*(cos(b*x+a)+1)^2*cos(b*x+a)*(d*sin(b*x+a)/cos
(b*x+a))^(3/2)/sin(b*x+a)^5*2^(1/2)

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maxima [A]  time = 0.61, size = 188, normalized size = 0.84 \[ \frac {2 \, \sqrt {2} d^{\frac {5}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right ) + 2 \, \sqrt {2} d^{\frac {5}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right ) + \sqrt {2} d^{\frac {5}{2}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right ) - \sqrt {2} d^{\frac {5}{2}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right ) - \frac {8 \, \sqrt {d \tan \left (b x + a\right )} d^{4}}{d^{2} \tan \left (b x + a\right )^{2} + d^{2}}}{16 \, b d} \]

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

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

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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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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