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]

-(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)

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Rubi [A]  time = 0.162636, antiderivative size = 225, normalized size of antiderivative = 1., 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 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 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 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 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]

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])

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*}

Mathematica [A]  time = 0.258561, 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]

-(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]])/(8*
b)

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Maple [C]  time = 0.143, size = 660, normalized size = 2.9 \begin{align*} \text{result too large to display} \end{align*}

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

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

[Out]

Exception raised: ValueError

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

[Out]

Timed out

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

[Out]

Timed out

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

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]

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