\(\int \cos ^2(a+b x) (d \tan (a+b x))^{3/2} \, dx\) [240]

Optimal result

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

[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

Rubi [A] (verified)

Time = 0.19 (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, 217, 1179, 642, 1176, 631, 210} \[ \int \cos ^2(a+b x) (d \tan (a+b x))^{3/2} \, dx=-\frac {d^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b}+\frac {d^{3/2} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{4 \sqrt {2} b}-\frac {d^{3/2} \log \left (\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{8 \sqrt {2} b}+\frac {d^{3/2} \log \left (\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{8 \sqrt {2} b}-\frac {d \cos ^2(a+b x) \sqrt {d \tan (a+b x)}}{2 b} \]

[In]

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

[Out]

-1/4*(d^(3/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[a + b*x]])/Sqrt[d]])/(Sqrt[2]*b) + (d^(3/2)*ArcTan[1 + (Sqrt[2]*S
qrt[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*Ta
n[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*
Sqrt[2]*b) - (d*Cos[a + b*x]^2*Sqrt[d*Tan[a + b*x]])/(2*b)

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 217

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 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 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)^{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 \text {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 \text {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 {\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}+\frac {\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} \\ & = -\frac {d \cos ^2(a+b x) \sqrt {d \tan (a+b x)}}{2 b}-\frac {d^{3/2} \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}-\frac {d^{3/2} \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}+\frac {d^2 \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}+\frac {d^2 \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} \\ & = -\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} \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}-\frac {d^{3/2} \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} \\ & = -\frac {d^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b}+\frac {d^{3/2} \arctan \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*}

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.49 \[ \int \cos ^2(a+b x) (d \tan (a+b x))^{3/2} \, dx=-\frac {d \csc (a+b x) \left (\sin (a+b x)+\arcsin (\cos (a+b x)-\sin (a+b x)) \sqrt {\sin (2 (a+b x))}-\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right ) \sqrt {\sin (2 (a+b x))}+\sin (3 (a+b x))\right ) \sqrt {d \tan (a+b x)}}{8 b} \]

[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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(524\) vs. \(2(169)=338\).

Time = 2.17 (sec) , antiderivative size = 525, normalized size of antiderivative = 2.33

method	result	size
default	\(-\frac {\cos \left (b x +a \right ) \left (4 \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sqrt {2}\, \left (\cos ^{2}\left (b x +a \right )\right )+4 \cos \left (b x +a \right ) \sqrt {2}\, \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}-\ln \left (\frac {2 \sin \left (b x +a \right ) \sqrt {-\left (\cot ^{3}\left (b x +a \right )\right )+3 \left (\cot ^{2}\left (b x +a \right )\right ) \csc \left (b x +a \right )-3 \cot \left (b x +a \right ) \left (\csc ^{2}\left (b x +a \right )\right )+\csc ^{3}\left (b x +a \right )+\cot \left (b x +a \right )-\csc \left (b x +a \right )}-\cot \left (b x +a \right ) \cos \left (b x +a \right )+2 \cot \left (b x +a \right )+2 \cos \left (b x +a \right )+\sin \left (b x +a \right )-\csc \left (b x +a \right )-2}{-1+\cos \left (b x +a \right )}\right )+\ln \left (-\frac {\cot \left (b x +a \right ) \cos \left (b x +a \right )-2 \cot \left (b x +a \right )+2 \sin \left (b x +a \right ) \sqrt {-\left (\cot ^{3}\left (b x +a \right )\right )+3 \left (\cot ^{2}\left (b x +a \right )\right ) \csc \left (b x +a \right )-3 \cot \left (b x +a \right ) \left (\csc ^{2}\left (b x +a \right )\right )+\csc ^{3}\left (b x +a \right )+\cot \left (b x +a \right )-\csc \left (b x +a \right )}-2 \cos \left (b x +a \right )-\sin \left (b x +a \right )+\csc \left (b x +a \right )+2}{-1+\cos \left (b x +a \right )}\right )+2 \arctan \left (\frac {\sin \left (b x +a \right ) \sqrt {2}\, \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}+\cos \left (b x +a \right )-1}{-1+\cos \left (b x +a \right )}\right )+2 \arctan \left (\frac {\sin \left (b x +a \right ) \sqrt {2}\, \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}-\cos \left (b x +a \right )+1}{-1+\cos \left (b x +a \right )}\right )\right ) d \sqrt {d \tan \left (b x +a \right )}\, \sqrt {2}}{16 b \left (\cos \left (b x +a \right )+1\right ) \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\)	\(525\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.38 (sec) , antiderivative size = 927, normalized size of antiderivative = 4.12 \[ \int \cos ^2(a+b x) (d \tan (a+b x))^{3/2} \, dx=\text {Too large to display} \]

[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 - (-d^6/b^4)^(1/4)*b*log(-2*d^5*cos(b*x + a)^2 +
2*sqrt(-d^6/b^4)*b^2*d^2*cos(b*x + a)*sin(b*x + a) + d^5 + 2*((-d^6/b^4)^(1/4)*b*d^3*cos(b*x + a)*sin(b*x + a)
+ (-d^6/b^4)^(3/4)*b^3*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a))) + (-d^6/b^4)^(1/4)*b*log(-2*d^5*cos
(b*x + a)^2 + 2*sqrt(-d^6/b^4)*b^2*d^2*cos(b*x + a)*sin(b*x + a) + d^5 - 2*((-d^6/b^4)^(1/4)*b*d^3*cos(b*x + a
)*sin(b*x + a) + (-d^6/b^4)^(3/4)*b^3*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a))) + I*(-d^6/b^4)^(1/4)*
b*log(-2*d^5*cos(b*x + a)^2 - 2*sqrt(-d^6/b^4)*b^2*d^2*cos(b*x + a)*sin(b*x + a) + d^5 - 2*(I*(-d^6/b^4)^(1/4)
*b*d^3*cos(b*x + a)*sin(b*x + a) - I*(-d^6/b^4)^(3/4)*b^3*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a))) -
I*(-d^6/b^4)^(1/4)*b*log(-2*d^5*cos(b*x + a)^2 - 2*sqrt(-d^6/b^4)*b^2*d^2*cos(b*x + a)*sin(b*x + a) + d^5 - 2
*(-I*(-d^6/b^4)^(1/4)*b*d^3*cos(b*x + a)*sin(b*x + a) + I*(-d^6/b^4)^(3/4)*b^3*cos(b*x + a)^2)*sqrt(d*sin(b*x
+ a)/cos(b*x + a))) + (-d^6/b^4)^(1/4)*b*log(-d^5 + 2*((-d^6/b^4)^(1/4)*b*d^3*cos(b*x + a)*sin(b*x + a) - (-d^
6/b^4)^(3/4)*b^3*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a))) - (-d^6/b^4)^(1/4)*b*log(-d^5 - 2*((-d^6/b
^4)^(1/4)*b*d^3*cos(b*x + a)*sin(b*x + a) - (-d^6/b^4)^(3/4)*b^3*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x +
a))) - I*(-d^6/b^4)^(1/4)*b*log(-d^5 - 2*(I*(-d^6/b^4)^(1/4)*b*d^3*cos(b*x + a)*sin(b*x + a) + I*(-d^6/b^4)^(
3/4)*b^3*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a))) + I*(-d^6/b^4)^(1/4)*b*log(-d^5 - 2*(-I*(-d^6/b^4)
^(1/4)*b*d^3*cos(b*x + a)*sin(b*x + a) - I*(-d^6/b^4)^(3/4)*b^3*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x +
a))))/b

Sympy [F(-1)]

Timed out. \[ \int \cos ^2(a+b x) (d \tan (a+b x))^{3/2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.84 \[ \int \cos ^2(a+b x) (d \tan (a+b x))^{3/2} \, dx=\frac {2 \, \sqrt {2} d^{\frac {5}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right ) + 2 \, \sqrt {2} d^{\frac {5}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right ) + \sqrt {2} d^{\frac {5}{2}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right ) - \sqrt {2} d^{\frac {5}{2}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right ) - \frac {8 \, \sqrt {d \tan \left (b x + a\right )} d^{4}}{d^{2} \tan \left (b x + a\right )^{2} + d^{2}}}{16 \, b d} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.93 \[ \int \cos ^2(a+b x) (d \tan (a+b x))^{3/2} \, dx=\frac {1}{16} \, d {\left (\frac {2 \, \sqrt {2} \sqrt {{\left | d \right |}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{b} + \frac {2 \, \sqrt {2} \sqrt {{\left | d \right |}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{b} + \frac {\sqrt {2} \sqrt {{\left | d \right |}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{b} - \frac {\sqrt {2} \sqrt {{\left | d \right |}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{b} - \frac {8 \, \sqrt {d \tan \left (b x + a\right )} d^{2}}{{\left (d^{2} \tan \left (b x + a\right )^{2} + d^{2}\right )} b}\right )} \]

[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))

Mupad [F(-1)]

Timed out. \[ \int \cos ^2(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int {\cos \left (a+b\,x\right )}^2\,{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2} \,d x \]

[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)