3.533 \(\int \sec ^4(c+d x) (a+b \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=75 \[ \frac{\left (a^2+b^2\right ) (a+b \tan (c+d x))^4}{4 b^3 d}+\frac{(a+b \tan (c+d x))^6}{6 b^3 d}-\frac{2 a (a+b \tan (c+d x))^5}{5 b^3 d} \]

[Out]

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

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Rubi [A]  time = 0.0708607, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {3506, 697} \[ \frac{\left (a^2+b^2\right ) (a+b \tan (c+d x))^4}{4 b^3 d}+\frac{(a+b \tan (c+d x))^6}{6 b^3 d}-\frac{2 a (a+b \tan (c+d x))^5}{5 b^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + b*Tan[c + d*x])^3,x]

[Out]

((a^2 + b^2)*(a + b*Tan[c + d*x])^4)/(4*b^3*d) - (2*a*(a + b*Tan[c + d*x])^5)/(5*b^3*d) + (a + b*Tan[c + d*x])
^6/(6*b^3*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 697

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \sec ^4(c+d x) (a+b \tan (c+d x))^3 \, dx &=\frac{\operatorname{Subst}\left (\int (a+x)^3 \left (1+\frac{x^2}{b^2}\right ) \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{\left (a^2+b^2\right ) (a+x)^3}{b^2}-\frac{2 a (a+x)^4}{b^2}+\frac{(a+x)^5}{b^2}\right ) \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac{\left (a^2+b^2\right ) (a+b \tan (c+d x))^4}{4 b^3 d}-\frac{2 a (a+b \tan (c+d x))^5}{5 b^3 d}+\frac{(a+b \tan (c+d x))^6}{6 b^3 d}\\ \end{align*}

Mathematica [A]  time = 0.346668, size = 54, normalized size = 0.72 \[ \frac{(a+b \tan (c+d x))^4 \left (a^2-4 a b \tan (c+d x)+10 b^2 \tan ^2(c+d x)+15 b^2\right )}{60 b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(a + b*Tan[c + d*x])^3,x]

[Out]

((a + b*Tan[c + d*x])^4*(a^2 + 15*b^2 - 4*a*b*Tan[c + d*x] + 10*b^2*Tan[c + d*x]^2))/(60*b^3*d)

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Maple [A]  time = 0.063, size = 127, normalized size = 1.7 \begin{align*}{\frac{1}{d} \left ({b}^{3} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{6\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{12\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}}} \right ) +3\,a{b}^{2} \left ( 1/5\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+2/15\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) +{\frac{3\,b{a}^{2}}{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{a}^{3} \left ( -{\frac{2}{3}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) \tan \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(a+b*tan(d*x+c))^3,x)

[Out]

1/d*(b^3*(1/6*sin(d*x+c)^4/cos(d*x+c)^6+1/12*sin(d*x+c)^4/cos(d*x+c)^4)+3*a*b^2*(1/5*sin(d*x+c)^3/cos(d*x+c)^5
+2/15*sin(d*x+c)^3/cos(d*x+c)^3)+3/4*b*a^2/cos(d*x+c)^4-a^3*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c))

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Maxima [A]  time = 1.15361, size = 132, normalized size = 1.76 \begin{align*} \frac{10 \, b^{3} \tan \left (d x + c\right )^{6} + 36 \, a b^{2} \tan \left (d x + c\right )^{5} + 90 \, a^{2} b \tan \left (d x + c\right )^{2} + 15 \,{\left (3 \, a^{2} b + b^{3}\right )} \tan \left (d x + c\right )^{4} + 60 \, a^{3} \tan \left (d x + c\right ) + 20 \,{\left (a^{3} + 3 \, a b^{2}\right )} \tan \left (d x + c\right )^{3}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(10*b^3*tan(d*x + c)^6 + 36*a*b^2*tan(d*x + c)^5 + 90*a^2*b*tan(d*x + c)^2 + 15*(3*a^2*b + b^3)*tan(d*x +
c)^4 + 60*a^3*tan(d*x + c) + 20*(a^3 + 3*a*b^2)*tan(d*x + c)^3)/d

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Fricas [A]  time = 1.88605, size = 246, normalized size = 3.28 \begin{align*} \frac{10 \, b^{3} + 15 \,{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left (2 \,{\left (5 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} + 9 \, a b^{2} \cos \left (d x + c\right ) +{\left (5 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{60 \, d \cos \left (d x + c\right )^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/60*(10*b^3 + 15*(3*a^2*b - b^3)*cos(d*x + c)^2 + 4*(2*(5*a^3 - 3*a*b^2)*cos(d*x + c)^5 + 9*a*b^2*cos(d*x + c
) + (5*a^3 - 3*a*b^2)*cos(d*x + c)^3)*sin(d*x + c))/(d*cos(d*x + c)^6)

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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} \sec ^{4}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(a+b*tan(d*x+c))**3,x)

[Out]

Integral((a + b*tan(c + d*x))**3*sec(c + d*x)**4, x)

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Giac [A]  time = 1.75557, size = 151, normalized size = 2.01 \begin{align*} \frac{10 \, b^{3} \tan \left (d x + c\right )^{6} + 36 \, a b^{2} \tan \left (d x + c\right )^{5} + 45 \, a^{2} b \tan \left (d x + c\right )^{4} + 15 \, b^{3} \tan \left (d x + c\right )^{4} + 20 \, a^{3} \tan \left (d x + c\right )^{3} + 60 \, a b^{2} \tan \left (d x + c\right )^{3} + 90 \, a^{2} b \tan \left (d x + c\right )^{2} + 60 \, a^{3} \tan \left (d x + c\right )}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(10*b^3*tan(d*x + c)^6 + 36*a*b^2*tan(d*x + c)^5 + 45*a^2*b*tan(d*x + c)^4 + 15*b^3*tan(d*x + c)^4 + 20*a
^3*tan(d*x + c)^3 + 60*a*b^2*tan(d*x + c)^3 + 90*a^2*b*tan(d*x + c)^2 + 60*a^3*tan(d*x + c))/d