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

Optimal. Leaf size=40 \[ \frac {1}{3} \log (4-3 \tan (x))+\frac {8}{3 \sqrt {4-3 \tan (x)}}+\frac {2}{3} \sqrt {4-3 \tan (x)} \]

[Out]

1/3*ln(4-3*tan(x))+8/3/(4-3*tan(x))^(1/2)+2/3*(4-3*tan(x))^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {4427, 45} \begin {gather*} \frac {2}{3} \sqrt {4-3 \tan (x)}+\frac {8}{3 \sqrt {4-3 \tan (x)}}+\frac {1}{3} \log (4-3 \tan (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[4 - 3*Tan[x]]/3 + 8/(3*Sqrt[4 - 3*Tan[x]]) + (2*Sqrt[4 - 3*Tan[x]])/3

Rule 45

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])

Rule 4427

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

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Mathematica [A]
time = 0.86, size = 38, normalized size = 0.95 \begin {gather*} \frac {16+\log (4-3 \tan (x)) \sqrt {4-3 \tan (x)}-6 \tan (x)}{3 \sqrt {4-3 \tan (x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(16 + Log[4 - 3*Tan[x]]*Sqrt[4 - 3*Tan[x]] - 6*Tan[x])/(3*Sqrt[4 - 3*Tan[x]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(218\) vs. \(2(30)=60\).
time = 0.47, size = 219, normalized size = 5.48

method	result	size
default	\(\frac {\left (\cos \left (x \right )-1\right )^{2} \left (1+\cos \left (x \right )\right )^{2} \left (16 \sqrt {\frac {4 \cos \left (x \right )-3 \sin \left (x \right )}{\cos \left (x \right )}}\, \cos \left (x \right )-4 \cos \left (x \right ) \ln \left (-\frac {-1+\cos \left (x \right )-\sin \left (x \right )}{\sin \left (x \right )}\right )+4 \cos \left (x \right ) \ln \left (-\frac {\sin \left (x \right )-2+2 \cos \left (x \right )}{\sin \left (x \right )}\right )-4 \cos \left (x \right ) \ln \left (-\frac {-1+\cos \left (x \right )+\sin \left (x \right )}{\sin \left (x \right )}\right )+4 \cos \left (x \right ) \ln \left (-\frac {-2 \sin \left (x \right )-1+\cos \left (x \right )}{\sin \left (x \right )}\right )-6 \sin \left (x \right ) \sqrt {\frac {4 \cos \left (x \right )-3 \sin \left (x \right )}{\cos \left (x \right )}}+3 \sin \left (x \right ) \ln \left (-\frac {-1+\cos \left (x \right )-\sin \left (x \right )}{\sin \left (x \right )}\right )-3 \sin \left (x \right ) \ln \left (-\frac {\sin \left (x \right )-2+2 \cos \left (x \right )}{\sin \left (x \right )}\right )+3 \sin \left (x \right ) \ln \left (-\frac {-1+\cos \left (x \right )+\sin \left (x \right )}{\sin \left (x \right )}\right )-3 \sin \left (x \right ) \ln \left (-\frac {-2 \sin \left (x \right )-1+\cos \left (x \right )}{\sin \left (x \right )}\right )\right )}{3 \left (4 \cos \left (x \right )-3 \sin \left (x \right )\right ) \sin \left (x \right )^{4}}\)	\(219\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(cos(x)-1)^2*(1+cos(x))^2*(16*((4*cos(x)-3*sin(x))/cos(x))^(1/2)*cos(x)-4*cos(x)*ln(-(-1+cos(x)-sin(x))/si
n(x))+4*cos(x)*ln(-(sin(x)-2+2*cos(x))/sin(x))-4*cos(x)*ln(-(-1+cos(x)+sin(x))/sin(x))+4*cos(x)*ln(-(-2*sin(x)
-1+cos(x))/sin(x))-6*sin(x)*((4*cos(x)-3*sin(x))/cos(x))^(1/2)+3*sin(x)*ln(-(-1+cos(x)-sin(x))/sin(x))-3*sin(x
)*ln(-(sin(x)-2+2*cos(x))/sin(x))+3*sin(x)*ln(-(-1+cos(x)+sin(x))/sin(x))-3*sin(x)*ln(-(-2*sin(x)-1+cos(x))/si
n(x)))/(4*cos(x)-3*sin(x))/sin(x)^4

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Maxima [A]
time = 2.16, size = 30, normalized size = 0.75 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (30) = 60\).
time = 0.86, size = 82, normalized size = 2.05 \begin {gather*} \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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)

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Giac [A]
time = 1.26, size = 31, normalized size = 0.78 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.42, size = 105, normalized size = 2.62 \begin {gather*} \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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*tan(x) - (4 - 3*tan(x))^(1/2))/(cos(x)^2*(4 - 3*tan(x))^(3/2)),x)

[Out]

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(
x*2i) - 6*sin(2*x)*exp(x*2i))

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