3.703 \(\int \frac{\sec ^2(x) (a+b \tan (x))^3}{c+d \tan (x)} \, dx\)

Optimal. Leaf size=78 \[ \frac{b \tan (x) (b c-a d)^2}{d^3}-\frac{(b c-a d) (a+b \tan (x))^2}{2 d^2}-\frac{(b c-a d)^3 \log (c+d \tan (x))}{d^4}+\frac{(a+b \tan (x))^3}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.149608, antiderivative size = 78, 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 = {4342, 43} \[ \frac{b \tan (x) (b c-a d)^2}{d^3}-\frac{(b c-a d) (a+b \tan (x))^2}{2 d^2}-\frac{(b c-a d)^3 \log (c+d \tan (x))}{d^4}+\frac{(a+b \tan (x))^3}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4342

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^2, x_Symbol] :> With[{d = FreeFactors[Tan[c*(a + b*x)], x]}, Dist[d/
(b*c), Subst[Int[SubstFor[1, Tan[c*(a + b*x)]/d, u, x], x], x, Tan[c*(a + b*x)]/d], x] /; FunctionOfQ[Tan[c*(a
+ b*x)]/d, u, x, True]] /; FreeQ[{a, b, c}, x] && NonsumQ[u] && (EqQ[F, Sec] || EqQ[F, sec])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.897265, size = 133, normalized size = 1.71 \[ \frac{(a+b \tan (x))^3 (c \cos (x)+d \sin (x)) \left (b d^2 (9 a \sin (2 x) (a d-b c)+b (9 a d-3 b c+2 b d \tan (x)))+6 \cos ^2(x) (b c-a d)^3 (\log (\cos (x))-\log (c \cos (x)+d \sin (x)))+b^3 (-d) \left (d^2-3 c^2\right ) \sin (2 x)\right )}{6 d^4 (c+d \tan (x)) (a \cos (x)+b \sin (x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c*Cos[x] + d*Sin[x])*(a + b*Tan[x])^3*(6*(b*c - a*d)^3*Cos[x]^2*(Log[Cos[x]] - Log[c*Cos[x] + d*Sin[x]]) - b
^3*d*(-3*c^2 + d^2)*Sin[2*x] + b*d^2*(9*a*(-(b*c) + a*d)*Sin[2*x] + b*(-3*b*c + 9*a*d + 2*b*d*Tan[x]))))/(6*d^
4*(a*Cos[x] + b*Sin[x])^3*(c + d*Tan[x]))

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Maple [A]  time = 0.075, size = 143, normalized size = 1.8 \begin{align*}{\frac{{b}^{3} \left ( \tan \left ( x \right ) \right ) ^{3}}{3\,d}}+{\frac{3\,{b}^{2} \left ( \tan \left ( x \right ) \right ) ^{2}a}{2\,d}}-{\frac{{b}^{3} \left ( \tan \left ( x \right ) \right ) ^{2}c}{2\,{d}^{2}}}+3\,{\frac{{a}^{2}b\tan \left ( x \right ) }{d}}-3\,{\frac{a{b}^{2}c\tan \left ( x \right ) }{{d}^{2}}}+{\frac{{b}^{3}{c}^{2}\tan \left ( x \right ) }{{d}^{3}}}+{\frac{\ln \left ( c+d\tan \left ( x \right ) \right ){a}^{3}}{d}}-3\,{\frac{\ln \left ( c+d\tan \left ( x \right ) \right ){a}^{2}bc}{{d}^{2}}}+3\,{\frac{\ln \left ( c+d\tan \left ( x \right ) \right ) a{b}^{2}{c}^{2}}{{d}^{3}}}-{\frac{\ln \left ( c+d\tan \left ( x \right ) \right ){b}^{3}{c}^{3}}{{d}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.976686, size = 159, normalized size = 2.04 \begin{align*} \frac{2 \, b^{3} d^{2} \tan \left (x\right )^{3} - 3 \,{\left (b^{3} c d - 3 \, a b^{2} d^{2}\right )} \tan \left (x\right )^{2} + 6 \,{\left (b^{3} c^{2} - 3 \, a b^{2} c d + 3 \, a^{2} b d^{2}\right )} \tan \left (x\right )}{6 \, d^{3}} - \frac{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left (d \tan \left (x\right ) + c\right )}{d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.98869, size = 459, normalized size = 5.88 \begin{align*} -\frac{3 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \cos \left (x\right )^{3} \log \left (2 \, c d \cos \left (x\right ) \sin \left (x\right ) +{\left (c^{2} - d^{2}\right )} \cos \left (x\right )^{2} + d^{2}\right ) - 3 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \cos \left (x\right )^{3} \log \left (\cos \left (x\right )^{2}\right ) + 3 \,{\left (b^{3} c d^{2} - 3 \, a b^{2} d^{3}\right )} \cos \left (x\right ) - 2 \,{\left (b^{3} d^{3} +{\left (3 \, b^{3} c^{2} d - 9 \, a b^{2} c d^{2} +{\left (9 \, a^{2} b - b^{3}\right )} d^{3}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{6 \, d^{4} \cos \left (x\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*cos(x)^3*log(2*c*d*cos(x)*sin(x) + (c^2 - d^2)*cos
(x)^2 + d^2) - 3*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*cos(x)^3*log(cos(x)^2) + 3*(b^3*c*d^2 - 3
*a*b^2*d^3)*cos(x) - 2*(b^3*d^3 + (3*b^3*c^2*d - 9*a*b^2*c*d^2 + (9*a^2*b - b^3)*d^3)*cos(x)^2)*sin(x))/(d^4*c
os(x)^3)

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Sympy [A]  time = 7.21373, size = 95, normalized size = 1.22 \begin{align*} \frac{b^{3} \tan ^{3}{\left (x \right )}}{3 d} + \frac{\left (3 a b^{2} d - b^{3} c\right ) \tan ^{2}{\left (x \right )}}{2 d^{2}} + \frac{\left (a d - b c\right )^{3} \left (\begin{cases} \frac{\tan{\left (x \right )}}{c} & \text{for}\: d = 0 \\\frac{\log{\left (c + d \tan{\left (x \right )} \right )}}{d} & \text{otherwise} \end{cases}\right )}{d^{3}} + \frac{\left (3 a^{2} b d^{2} - 3 a b^{2} c d + b^{3} c^{2}\right ) \tan{\left (x \right )}}{d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b**3*tan(x)**3/(3*d) + (3*a*b**2*d - b**3*c)*tan(x)**2/(2*d**2) + (a*d - b*c)**3*Piecewise((tan(x)/c, Eq(d, 0)
), (log(c + d*tan(x))/d, True))/d**3 + (3*a**2*b*d**2 - 3*a*b**2*c*d + b**3*c**2)*tan(x)/d**3

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Giac [A]  time = 1.08839, size = 166, normalized size = 2.13 \begin{align*} \frac{2 \, b^{3} d^{2} \tan \left (x\right )^{3} - 3 \, b^{3} c d \tan \left (x\right )^{2} + 9 \, a b^{2} d^{2} \tan \left (x\right )^{2} + 6 \, b^{3} c^{2} \tan \left (x\right ) - 18 \, a b^{2} c d \tan \left (x\right ) + 18 \, a^{2} b d^{2} \tan \left (x\right )}{6 \, d^{3}} - \frac{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left ({\left | d \tan \left (x\right ) + c \right |}\right )}{d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*b^3*d^2*tan(x)^3 - 3*b^3*c*d*tan(x)^2 + 9*a*b^2*d^2*tan(x)^2 + 6*b^3*c^2*tan(x) - 18*a*b^2*c*d*tan(x) +
18*a^2*b*d^2*tan(x))/d^3 - (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*log(abs(d*tan(x) + c))/d^4