3.535 \(\int \cos ^2(c+d x) (a+b \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=86 \[ \frac{1}{2} a x \left (a^2+3 b^2\right )-\frac{a b^2 \tan (c+d x)}{2 d}-\frac{\cos ^2(c+d x) (b-a \tan (c+d x)) (a+b \tan (c+d x))^2}{2 d}-\frac{b^3 \log (\cos (c+d x))}{d} \]

[Out]

(a*(a^2 + 3*b^2)*x)/2 - (b^3*Log[Cos[c + d*x]])/d - (a*b^2*Tan[c + d*x])/(2*d) - (Cos[c + d*x]^2*(b - a*Tan[c
+ d*x])*(a + b*Tan[c + d*x])^2)/(2*d)

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Rubi [A]  time = 0.0932581, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3506, 739, 774, 635, 203, 260} \[ \frac{1}{2} a x \left (a^2+3 b^2\right )-\frac{a b^2 \tan (c+d x)}{2 d}-\frac{\cos ^2(c+d x) (b-a \tan (c+d x)) (a+b \tan (c+d x))^2}{2 d}-\frac{b^3 \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(a^2 + 3*b^2)*x)/2 - (b^3*Log[Cos[c + d*x]])/d - (a*b^2*Tan[c + d*x])/(2*d) - (Cos[c + d*x]^2*(b - a*Tan[c
+ d*x])*(a + b*Tan[c + d*x])^2)/(2*d)

Rule 3506

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(b*f), Subst
[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && NeQ[a^2 + b
^2, 0] && IntegerQ[m/2]

Rule 739

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(a*e - c*d*x)*(a
+ c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] + Dist[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^
2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 774

Int[(((d_.) + (e_.)*(x_))*((f_) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*g*x)/c, x] + Dist[1
/c, Int[(c*d*f - a*e*g + c*(e*f + d*g)*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 203

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

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

Mathematica [B]  time = 0.746237, size = 401, normalized size = 4.66 \[ \frac{a b \left (-2 a^2 b^2+a^4-3 b^4\right ) \sin (2 (c+d x))+\left (-2 a^2 b^4-3 a^4 b^2+b^6\right ) \cos (2 (c+d x))+2 a^2 b^4 \log \left (\sqrt{-b^2}-b \tan (c+d x)\right )+2 a^2 b^4 \log \left (\sqrt{-b^2}+b \tan (c+d x)\right )+4 a^3 \left (-b^2\right )^{3/2} \log \left (\sqrt{-b^2}-b \tan (c+d x)\right )-a^5 \sqrt{-b^2} \log \left (\sqrt{-b^2}-b \tan (c+d x)\right )-4 a^3 \left (-b^2\right )^{3/2} \log \left (\sqrt{-b^2}+b \tan (c+d x)\right )+a^5 \sqrt{-b^2} \log \left (\sqrt{-b^2}+b \tan (c+d x)\right )+2 a^2 b^4+5 a^4 b^2+3 a \sqrt{-b^2} b^4 \log \left (\sqrt{-b^2}+b \tan (c+d x)\right )-3 a \left (-b^2\right )^{5/2} \log \left (\sqrt{-b^2}-b \tan (c+d x)\right )+2 b^6 \log \left (\sqrt{-b^2}-b \tan (c+d x)\right )+2 b^6 \log \left (\sqrt{-b^2}+b \tan (c+d x)\right )-b^6}{4 b d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a^4*b^2 + 2*a^2*b^4 - b^6 + (-3*a^4*b^2 - 2*a^2*b^4 + b^6)*Cos[2*(c + d*x)] + 2*a^2*b^4*Log[Sqrt[-b^2] - b*
Tan[c + d*x]] + 2*b^6*Log[Sqrt[-b^2] - b*Tan[c + d*x]] - a^5*Sqrt[-b^2]*Log[Sqrt[-b^2] - b*Tan[c + d*x]] + 4*a
^3*(-b^2)^(3/2)*Log[Sqrt[-b^2] - b*Tan[c + d*x]] - 3*a*(-b^2)^(5/2)*Log[Sqrt[-b^2] - b*Tan[c + d*x]] + 2*a^2*b
^4*Log[Sqrt[-b^2] + b*Tan[c + d*x]] + 2*b^6*Log[Sqrt[-b^2] + b*Tan[c + d*x]] + a^5*Sqrt[-b^2]*Log[Sqrt[-b^2] +
b*Tan[c + d*x]] + 3*a*b^4*Sqrt[-b^2]*Log[Sqrt[-b^2] + b*Tan[c + d*x]] - 4*a^3*(-b^2)^(3/2)*Log[Sqrt[-b^2] + b
*Tan[c + d*x]] + a*b*(a^4 - 2*a^2*b^2 - 3*b^4)*Sin[2*(c + d*x)])/(4*b*(a^2 + b^2)*d)

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Maple [A]  time = 0.059, size = 123, normalized size = 1.4 \begin{align*} -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}{b}^{3}}{2\,d}}-{\frac{{b}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}-{\frac{3\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) a{b}^{2}}{2\,d}}+{\frac{3\,a{b}^{2}x}{2}}+{\frac{3\,a{b}^{2}c}{2\,d}}-{\frac{3\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}b}{2\,d}}+{\frac{{a}^{3}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{3}x}{2}}+{\frac{{a}^{3}c}{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d*sin(d*x+c)^2*b^3-b^3*ln(cos(d*x+c))/d-3/2/d*sin(d*x+c)*cos(d*x+c)*a*b^2+3/2*a*b^2*x+3/2/d*a*b^2*c-3/2/d
*a^2*cos(d*x+c)^2*b+1/2/d*a^3*sin(d*x+c)*cos(d*x+c)+1/2*a^3*x+1/2/d*a^3*c

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Maxima [A]  time = 1.73724, size = 109, normalized size = 1.27 \begin{align*} \frac{b^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) +{\left (a^{3} + 3 \, a b^{2}\right )}{\left (d x + c\right )} - \frac{3 \, a^{2} b - b^{3} -{\left (a^{3} - 3 \, a b^{2}\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^3*log(tan(d*x + c)^2 + 1) + (a^3 + 3*a*b^2)*(d*x + c) - (3*a^2*b - b^3 - (a^3 - 3*a*b^2)*tan(d*x + c))/
(tan(d*x + c)^2 + 1))/d

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Fricas [A]  time = 1.98182, size = 181, normalized size = 2.1 \begin{align*} -\frac{2 \, b^{3} \log \left (-\cos \left (d x + c\right )\right ) -{\left (a^{3} + 3 \, a b^{2}\right )} d x +{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} -{\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*b^3*log(-cos(d*x + c)) - (a^3 + 3*a*b^2)*d*x + (3*a^2*b - b^3)*cos(d*x + c)^2 - (a^3 - 3*a*b^2)*cos(d*
x + c)*sin(d*x + c))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.91175, size = 811, normalized size = 9.43 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a^3*d*x*tan(d*x)^2*tan(c)^2 + 6*a*b^2*d*x*tan(d*x)^2*tan(c)^2 - 2*b^3*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*
tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)^2*tan(c)^
2 + 2*a^3*d*x*tan(d*x)^2 + 6*a*b^2*d*x*tan(d*x)^2 + 2*a^3*d*x*tan(c)^2 + 6*a*b^2*d*x*tan(c)^2 - 3*a^2*b*tan(d*
x)^2*tan(c)^2 + b^3*tan(d*x)^2*tan(c)^2 - 2*b^3*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c
) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)^2 - 2*a^3*tan(d*x)^2*tan(c) + 6*a*b^2*
tan(d*x)^2*tan(c) - 2*b^3*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^
2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(c)^2 - 2*a^3*tan(d*x)*tan(c)^2 + 6*a*b^2*tan(d*x)*tan(c)^2 + 2*a^
3*d*x + 6*a*b^2*d*x + 3*a^2*b*tan(d*x)^2 - b^3*tan(d*x)^2 + 12*a^2*b*tan(d*x)*tan(c) - 4*b^3*tan(d*x)*tan(c) +
3*a^2*b*tan(c)^2 - b^3*tan(c)^2 - 2*b^3*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan
(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)) + 2*a^3*tan(d*x) - 6*a*b^2*tan(d*x) + 2*a^3*tan(c) - 6
*a*b^2*tan(c) - 3*a^2*b + b^3)/(d*tan(d*x)^2*tan(c)^2 + d*tan(d*x)^2 + d*tan(c)^2 + d)