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

Optimal. Leaf size=87 \[ \frac{\left (a^2-b^2\right ) \sec ^2(c+d x)}{2 d}+\frac{a^2 \log (\cos (c+d x))}{d}+\frac{2 a b \sec ^3(c+d x)}{3 d}-\frac{2 a b \sec (c+d x)}{d}+\frac{b^2 \sec ^4(c+d x)}{4 d} \]

[Out]

(a^2*Log[Cos[c + d*x]])/d - (2*a*b*Sec[c + d*x])/d + ((a^2 - b^2)*Sec[c + d*x]^2)/(2*d) + (2*a*b*Sec[c + d*x]^
3)/(3*d) + (b^2*Sec[c + d*x]^4)/(4*d)

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Rubi [A]  time = 0.069152, antiderivative size = 87, 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 = {3885, 894} \[ \frac{\left (a^2-b^2\right ) \sec ^2(c+d x)}{2 d}+\frac{a^2 \log (\cos (c+d x))}{d}+\frac{2 a b \sec ^3(c+d x)}{3 d}-\frac{2 a b \sec (c+d x)}{d}+\frac{b^2 \sec ^4(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Log[Cos[c + d*x]])/d - (2*a*b*Sec[c + d*x])/d + ((a^2 - b^2)*Sec[c + d*x]^2)/(2*d) + (2*a*b*Sec[c + d*x]^
3)/(3*d) + (b^2*Sec[c + d*x]^4)/(4*d)

Rule 3885

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[((b^2 - x^2)^((m - 1)/2)*(a + x)^n)/x, x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

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

Mathematica [A]  time = 0.435898, size = 74, normalized size = 0.85 \[ \frac{6 \left (a^2-b^2\right ) \sec ^2(c+d x)+12 a^2 \log (\cos (c+d x))+8 a b \sec ^3(c+d x)-24 a b \sec (c+d x)+3 b^2 \sec ^4(c+d x)}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*a^2*Log[Cos[c + d*x]] - 24*a*b*Sec[c + d*x] + 6*(a^2 - b^2)*Sec[c + d*x]^2 + 8*a*b*Sec[c + d*x]^3 + 3*b^2*
Sec[c + d*x]^4)/(12*d)

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Maple [A]  time = 0.048, size = 136, normalized size = 1.6 \begin{align*}{\frac{{a}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+{\frac{{a}^{2}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{2\,ab \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{3\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{2\,ab \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{3\,d\cos \left ( dx+c \right ) }}-{\frac{2\,ab\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{3\,d}}-{\frac{4\,ab\cos \left ( dx+c \right ) }{3\,d}}+{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*a^2*tan(d*x+c)^2+a^2*ln(cos(d*x+c))/d+2/3/d*a*b*sin(d*x+c)^4/cos(d*x+c)^3-2/3/d*a*b*sin(d*x+c)^4/cos(d*x
+c)-2/3/d*a*b*cos(d*x+c)*sin(d*x+c)^2-4/3/d*cos(d*x+c)*a*b+1/4/d*b^2*sin(d*x+c)^4/cos(d*x+c)^4

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Maxima [A]  time = 0.999052, size = 101, normalized size = 1.16 \begin{align*} \frac{12 \, a^{2} \log \left (\cos \left (d x + c\right )\right ) - \frac{24 \, a b \cos \left (d x + c\right )^{3} - 8 \, a b \cos \left (d x + c\right ) - 6 \,{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \, b^{2}}{\cos \left (d x + c\right )^{4}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*a^2*log(cos(d*x + c)) - (24*a*b*cos(d*x + c)^3 - 8*a*b*cos(d*x + c) - 6*(a^2 - b^2)*cos(d*x + c)^2 -
3*b^2)/cos(d*x + c)^4)/d

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Fricas [A]  time = 0.969567, size = 205, normalized size = 2.36 \begin{align*} \frac{12 \, a^{2} \cos \left (d x + c\right )^{4} \log \left (-\cos \left (d x + c\right )\right ) - 24 \, a b \cos \left (d x + c\right )^{3} + 8 \, a b \cos \left (d x + c\right ) + 6 \,{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, b^{2}}{12 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*a^2*cos(d*x + c)^4*log(-cos(d*x + c)) - 24*a*b*cos(d*x + c)^3 + 8*a*b*cos(d*x + c) + 6*(a^2 - b^2)*co
s(d*x + c)^2 + 3*b^2)/(d*cos(d*x + c)^4)

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Sympy [A]  time = 4.02762, size = 126, normalized size = 1.45 \begin{align*} \begin{cases} - \frac{a^{2} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac{a^{2} \tan ^{2}{\left (c + d x \right )}}{2 d} + \frac{2 a b \tan ^{2}{\left (c + d x \right )} \sec{\left (c + d x \right )}}{3 d} - \frac{4 a b \sec{\left (c + d x \right )}}{3 d} + \frac{b^{2} \tan ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{4 d} - \frac{b^{2} \sec ^{2}{\left (c + d x \right )}}{4 d} & \text{for}\: d \neq 0 \\x \left (a + b \sec{\left (c \right )}\right )^{2} \tan ^{3}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2*log(tan(c + d*x)**2 + 1)/(2*d) + a**2*tan(c + d*x)**2/(2*d) + 2*a*b*tan(c + d*x)**2*sec(c + d
*x)/(3*d) - 4*a*b*sec(c + d*x)/(3*d) + b**2*tan(c + d*x)**2*sec(c + d*x)**2/(4*d) - b**2*sec(c + d*x)**2/(4*d)
, Ne(d, 0)), (x*(a + b*sec(c))**2*tan(c)**3, True))

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Giac [B]  time = 1.93279, size = 360, normalized size = 4.14 \begin{align*} -\frac{12 \, a^{2} \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 12 \, a^{2} \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac{25 \, a^{2} + 32 \, a b + \frac{124 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{128 \, a b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{198 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{96 \, a b{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{48 \, b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{124 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{25 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{4}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(12*a^2*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 12*a^2*log(abs(-(cos(d*x + c) - 1)/(cos(d
*x + c) + 1) - 1)) + (25*a^2 + 32*a*b + 124*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 128*a*b*(cos(d*x + c)
- 1)/(cos(d*x + c) + 1) + 198*a^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 96*a*b*(cos(d*x + c) - 1)^2/(cos
(d*x + c) + 1)^2 - 48*b^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 124*a^2*(cos(d*x + c) - 1)^3/(cos(d*x +
c) + 1)^3 + 25*a^2*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4)/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^4)/d