3.4.0 ∫ sec 2 (x)(−∘ _______ 4−3 tan(x)+3 tan(x)) ___________________________________________

Integrand size = 32, antiderivative size = 40 \[ \int \frac {\sec ^2(x) \left (-\sqrt {4-3 \tan (x)}+3 \tan (x)\right )}{(4-3 \tan (x))^{3/2}} \, dx=\frac {1}{3} \log (4-3 \tan (x))+\frac {8}{3 \sqrt {4-3 \tan (x)}}+\frac {2}{3} \sqrt {4-3 \tan (x)} \]

Rubi [A] (veriﬁed)

Time = 0.10 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {4427, 45} \[ \int \frac {\sec ^2(x) \left (-\sqrt {4-3 \tan (x)}+3 \tan (x)\right )}{(4-3 \tan (x))^{3/2}} \, dx=\frac {2}{3} \sqrt {4-3 \tan (x)}+\frac {8}{3 \sqrt {4-3 \tan (x)}}+\frac {1}{3} \log (4-3 \tan (x)) \]

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])

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])

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \left (\frac {3 x}{(4-3 x)^{3/2}}+\frac {1}{-4+3 x}\right ) \, dx,x,\tan (x)\right ) \\ & = \frac {1}{3} \log (4-3 \tan (x))+3 \text {Subst}\left (\int \frac {x}{(4-3 x)^{3/2}} \, dx,x,\tan (x)\right ) \\ & = \frac {1}{3} \log (4-3 \tan (x))+3 \text {Subst}\left (\int \left (\frac {4}{3 (4-3 x)^{3/2}}-\frac {1}{3 \sqrt {4-3 x}}\right ) \, dx,x,\tan (x)\right ) \\ & = \frac {1}{3} \log (4-3 \tan (x))+\frac {8}{3 \sqrt {4-3 \tan (x)}}+\frac {2}{3} \sqrt {4-3 \tan (x)} \\ \end{align*}

Mathematica [A] (veriﬁed)

Time = 5.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.75 \[ \int \frac {\sec ^2(x) \left (-\sqrt {4-3 \tan (x)}+3 \tan (x)\right )}{(4-3 \tan (x))^{3/2}} \, dx=\frac {1}{3} \left (\log (4-3 \tan (x))+\frac {2 (8-3 \tan (x))}{\sqrt {4-3 \tan (x)}}\right ) \]

Integrate[(Sec[x]^2*(-Sqrt[4 - 3*Tan[x]] + 3*Tan[x]))/(4 - 3*Tan[x])^(3/2),x]

Maple [A] (veriﬁed)

Time = 0.45 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.78


method	result	size

derivativedivides	\(\frac {\ln \left (4-3 \tan \left (x \right )\right )}{3}+\frac {8}{3 \sqrt {4-3 \tan \left (x \right )}}+\frac {2 \sqrt {4-3 \tan \left (x \right )}}{3}\)	\(31\)
default	\(\frac {\ln \left (4-3 \tan \left (x \right )\right )}{3}+\frac {8}{3 \sqrt {4-3 \tan \left (x \right )}}+\frac {2 \sqrt {4-3 \tan \left (x \right )}}{3}\)	\(31\)

int((-(4-3*tan(x))^(1/2)+3*tan(x))/cos(x)^2/(4-3*tan(x))^(3/2),x,method=_RETURNVERBOSE)

Fricas [B] (veriﬁcation not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (30) = 60\).

Time = 0.26 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.05 \[ \int \frac {\sec ^2(x) \left (-\sqrt {4-3 \tan (x)}+3 \tan (x)\right )}{(4-3 \tan (x))^{3/2}} \, dx=\frac {{\left (4 \, \cos \left (x\right ) - 3 \, \sin \left (x\right )\right )} \log \left (\frac {7}{4} \, \cos \left (x\right )^{2} - 6 \, \cos \left (x\right ) \sin \left (x\right ) + \frac {9}{4}\right ) - {\left (4 \, \cos \left (x\right ) - 3 \, \sin \left (x\right )\right )} \log \left (\cos \left (x\right )^{2}\right ) + 4 \, \sqrt {\frac {4 \, \cos \left (x\right ) - 3 \, \sin \left (x\right )}{\cos \left (x\right )}} {\left (8 \, \cos \left (x\right ) - 3 \, \sin \left (x\right )\right )}}{6 \, {\left (4 \, \cos \left (x\right ) - 3 \, \sin \left (x\right )\right )}} \]

integrate((-(4-3*tan(x))^(1/2)+3*tan(x))/cos(x)^2/(4-3*tan(x))^(3/2),x, algorithm="fricas")

1/6*((4*cos(x) - 3*sin(x))*log(7/4*cos(x)^2 - 6*cos(x)*sin(x) + 9/4) - (4*cos(x) - 3*sin(x))*log(cos(x)^2) + 4

*sqrt((4*cos(x) - 3*sin(x))/cos(x))*(8*cos(x) - 3*sin(x)))/(4*cos(x) - 3*sin(x))

Sympy [F]

\[ \int \frac {\sec ^2(x) \left (-\sqrt {4-3 \tan (x)}+3 \tan (x)\right )}{(4-3 \tan (x))^{3/2}} \, dx=- \int \frac {\sqrt {4 - 3 \tan {\left (x \right )}}}{- 3 \sqrt {4 - 3 \tan {\left (x \right )}} \cos ^{2}{\left (x \right )} \tan {\left (x \right )} + 4 \sqrt {4 - 3 \tan {\left (x \right )}} \cos ^{2}{\left (x \right )}}\, dx - \int \left (- \frac {3 \tan {\left (x \right )}}{- 3 \sqrt {4 - 3 \tan {\left (x \right )}} \cos ^{2}{\left (x \right )} \tan {\left (x \right )} + 4 \sqrt {4 - 3 \tan {\left (x \right )}} \cos ^{2}{\left (x \right )}}\right )\, dx \]

integrate((-(4-3*tan(x))**(1/2)+3*tan(x))/cos(x)**2/(4-3*tan(x))**(3/2),x)

-Integral(sqrt(4 - 3*tan(x))/(-3*sqrt(4 - 3*tan(x))*cos(x)**2*tan(x) + 4*sqrt(4 - 3*tan(x))*cos(x)**2), x) - I

ntegral(-3*tan(x)/(-3*sqrt(4 - 3*tan(x))*cos(x)**2*tan(x) + 4*sqrt(4 - 3*tan(x))*cos(x)**2), x)

Maxima [A] (veriﬁcation not implemented)

Time = 0.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.75 \[ \int \frac {\sec ^2(x) \left (-\sqrt {4-3 \tan (x)}+3 \tan (x)\right )}{(4-3 \tan (x))^{3/2}} \, dx=\frac {2}{3} \, \sqrt {-3 \, \tan \left (x\right ) + 4} + \frac {8}{3 \, \sqrt {-3 \, \tan \left (x\right ) + 4}} + \frac {1}{3} \, \log \left (-3 \, \tan \left (x\right ) + 4\right ) \]

integrate((-(4-3*tan(x))^(1/2)+3*tan(x))/cos(x)^2/(4-3*tan(x))^(3/2),x, algorithm="maxima")

2/3*sqrt(-3*tan(x) + 4) + 8/3/sqrt(-3*tan(x) + 4) + 1/3*log(-3*tan(x) + 4)

Giac [A] (veriﬁcation not implemented)

Time = 0.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.78 \[ \int \frac {\sec ^2(x) \left (-\sqrt {4-3 \tan (x)}+3 \tan (x)\right )}{(4-3 \tan (x))^{3/2}} \, dx=\frac {2}{3} \, \sqrt {-3 \, \tan \left (x\right ) + 4} + \frac {8}{3 \, \sqrt {-3 \, \tan \left (x\right ) + 4}} + \frac {1}{3} \, \log \left ({\left | -3 \, \tan \left (x\right ) + 4 \right |}\right ) \]

integrate((-(4-3*tan(x))^(1/2)+3*tan(x))/cos(x)^2/(4-3*tan(x))^(3/2),x, algorithm="giac")

2/3*sqrt(-3*tan(x) + 4) + 8/3/sqrt(-3*tan(x) + 4) + 1/3*log(abs(-3*tan(x) + 4))

Mupad [B] (veriﬁcation not implemented)

Time = 1.47 (sec) , antiderivative size = 105, normalized size of antiderivative = 2.62 \[ \int \frac {\sec ^2(x) \left (-\sqrt {4-3 \tan (x)}+3 \tan (x)\right )}{(4-3 \tan (x))^{3/2}} \, dx=\frac {\ln \left ({\mathrm {e}}^{x\,2{}\mathrm {i}}\,\left (-\frac {16}{3}-4{}\mathrm {i}\right )-\frac {16}{3}+4{}\mathrm {i}\right )}{3}-\frac {\ln \left ({\mathrm {e}}^{x\,2{}\mathrm {i}}\,\left (\frac {16}{3}-4{}\mathrm {i}\right )+\frac {16}{3}-4{}\mathrm {i}\right )}{3}+\frac {2\,{\mathrm {e}}^{x\,1{}\mathrm {i}}\,\cos \left (x\right )\,\left (\frac {32\,{\mathrm {e}}^{x\,1{}\mathrm {i}}\,\cos \left (x\right )}{3}-4\,{\mathrm {e}}^{x\,1{}\mathrm {i}}\,\sin \left (x\right )\right )\,\sqrt {4-\frac {3\,\sin \left (x\right )}{\cos \left (x\right )}}}{8\,{\mathrm {e}}^{x\,2{}\mathrm {i}}+8\,\cos \left (2\,x\right )\,{\mathrm {e}}^{x\,2{}\mathrm {i}}-6\,\sin \left (2\,x\right )\,{\mathrm {e}}^{x\,2{}\mathrm {i}}} \]

log(- exp(x*2i)*(16/3 + 4i) - (16/3 - 4i))/3 - log(exp(x*2i)*(16/3 - 4i) + (16/3 - 4i))/3 + (2*exp(x*1i)*cos(x

)*((32*exp(x*1i)*cos(x))/3 - 4*exp(x*1i)*sin(x))*(4 - (3*sin(x))/cos(x))^(1/2))/(8*exp(x*2i) + 8*cos(2*x)*exp(

\(\int \frac {\sec ^2(x) (-\sqrt {4-3 \tan (x)}+3 \tan (x))}{(4-3 \tan (x))^{3/2}} \, dx\) [400]

Optimal result