3.7.30 \(\int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx\) [630]
Optimal. Leaf size=119 \[ -\frac {15 a b}{2 f \sqrt [3]{d \sec (e+f x)}}-\frac {3 \left (2 a^2-3 b^2\right ) d \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};\cos ^2(e+f x)\right ) \sin (e+f x)}{8 f (d \sec (e+f x))^{4/3} \sqrt {\sin ^2(e+f x)}}+\frac {3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}} \]
[Out]
-15/2*a*b/f/(d*sec(f*x+e))^(1/3)-3/8*(2*a^2-3*b^2)*d*hypergeom([1/2, 2/3],[5/3],cos(f*x+e)^2)*sin(f*x+e)/f/(d*
sec(f*x+e))^(4/3)/(sin(f*x+e)^2)^(1/2)+3/2*b*(a+b*tan(f*x+e))/f/(d*sec(f*x+e))^(1/3)
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Rubi [A]
time = 0.10, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3589, 3567,
3857, 2722} \begin {gather*} -\frac {3 d \left (2 a^2-3 b^2\right ) \sin (e+f x) \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};\cos ^2(e+f x)\right )}{8 f \sqrt {\sin ^2(e+f x)} (d \sec (e+f x))^{4/3}}-\frac {15 a b}{2 f \sqrt [3]{d \sec (e+f x)}}+\frac {3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(a + b*Tan[e + f*x])^2/(d*Sec[e + f*x])^(1/3),x]
[Out]
(-15*a*b)/(2*f*(d*Sec[e + f*x])^(1/3)) - (3*(2*a^2 - 3*b^2)*d*Hypergeometric2F1[1/2, 2/3, 5/3, Cos[e + f*x]^2]
*Sin[e + f*x])/(8*f*(d*Sec[e + f*x])^(4/3)*Sqrt[Sin[e + f*x]^2]) + (3*b*(a + b*Tan[e + f*x]))/(2*f*(d*Sec[e +
f*x])^(1/3))
Rule 2722
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] && !IntegerQ[2*n]
Rule 3567
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*((d*Sec[
e + f*x])^m/(f*m)), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])
Rule 3589
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[b*(d*Sec
[e + f*x])^m*((a + b*Tan[e + f*x])/(f*(m + 1))), x] + Dist[1/(m + 1), Int[(d*Sec[e + f*x])^m*(a^2*(m + 1) - b^
2 + a*b*(m + 2)*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 + b^2, 0] && NeQ[m, -1]
Rule 3857
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x])^(n - 1)*((Sin[c + d*x]/b)^(n - 1)
*Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[n]
Rubi steps
\begin {align*} \int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx &=\frac {3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}}+\frac {3}{2} \int \frac {\frac {2 a^2}{3}-b^2+\frac {5}{3} a b \tan (e+f x)}{\sqrt [3]{d \sec (e+f x)}} \, dx\\ &=-\frac {15 a b}{2 f \sqrt [3]{d \sec (e+f x)}}+\frac {3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}}+\frac {1}{2} \left (2 a^2-3 b^2\right ) \int \frac {1}{\sqrt [3]{d \sec (e+f x)}} \, dx\\ &=-\frac {15 a b}{2 f \sqrt [3]{d \sec (e+f x)}}+\frac {3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}}+\frac {1}{2} \left (\left (2 a^2-3 b^2\right ) \left (\frac {\cos (e+f x)}{d}\right )^{2/3} (d \sec (e+f x))^{2/3}\right ) \int \sqrt [3]{\frac {\cos (e+f x)}{d}} \, dx\\ &=-\frac {15 a b}{2 f \sqrt [3]{d \sec (e+f x)}}-\frac {3 \left (2 a^2-3 b^2\right ) \cos ^2(e+f x) \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};\cos ^2(e+f x)\right ) (d \sec (e+f x))^{2/3} \sin (e+f x)}{8 d f \sqrt {\sin ^2(e+f x)}}+\frac {3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}}\\ \end {align*}
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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(4138\) vs. \(2(119)=238\).
time = 31.29, size = 4138, normalized size = 34.77 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + b*Tan[e + f*x])^2/(d*Sec[e + f*x])^(1/3),x]
[Out]
(3*b^2*Cos[e + f*x]*Sin[e + f*x]*(a + b*Tan[e + f*x])^2)/(2*f*(d*Sec[e + f*x])^(1/3)*(a*Cos[e + f*x] + b*Sin[e
+ f*x])^2) + (3*Cos[e + f*x]^2*((-1)^(1/3) + Sec[e + f*x]^(2/3) + Sqrt[3]*Sec[e + f*x]^(2/3))^14*(-1 + Sec[e
+ f*x]^2)*Sqrt[Cos[e + f*x]^2*(-1 + Sec[e + f*x]^2)]*(-4*a*b*Sec[e + f*x] + (2*a^2 - 3*b^2)*Sqrt[1 - Cos[e + f
*x]^2]*Sec[e + f*x]^2 - ((2*a^2 - 3*b^2)*((-1)^(1/3)*3^(1/4)*(-6*EllipticE[ArcCos[((-1)^(1/3) - (-1 + Sqrt[3])
*Sec[e + f*x]^(2/3))/((-1)^(1/3) + (1 + Sqrt[3])*Sec[e + f*x]^(2/3))], (2 + Sqrt[3])/4] - (-3 + Sqrt[3])*Ellip
ticF[ArcCos[((-1)^(1/3) - (-1 + Sqrt[3])*Sec[e + f*x]^(2/3))/((-1)^(1/3) + (1 + Sqrt[3])*Sec[e + f*x]^(2/3))],
(2 + Sqrt[3])/4])*((-1)^(1/3) + (1 + Sqrt[3])*Sec[e + f*x]^(2/3))^3*Sqrt[(((-1)^(1/3) + Sec[e + f*x]^(2/3))*S
ec[e + f*x]^(2/3))/((-1)^(1/3) + (1 + Sqrt[3])*Sec[e + f*x]^(2/3))^2]*Sqrt[((-1)^(2/3) - (-1)^(1/3)*Sec[e + f*
x]^(2/3) + Sec[e + f*x]^(4/3))/((-1)^(1/3) + (1 + Sqrt[3])*Sec[e + f*x]^(2/3))^2] + 6*(1 + Sqrt[3])*Sec[e + f*
x]^(2/3)*(-1 + Sec[e + f*x]^2)))/(6*Sqrt[1 - Cos[e + f*x]^2]*((-1)^(1/3) + (1 + Sqrt[3])*Sec[e + f*x]^(2/3))))
*(2*a^2*Cos[e + f*x] - 3*b^2*Cos[e + f*x] + 4*a*b*Sin[e + f*x])*(a + b*Tan[e + f*x])^2)/(2*f*(d*Sec[e + f*x])^
(1/3)*(a*Cos[e + f*x] + b*Sin[e + f*x])^2*(-2*(-1)^(2/3)*a^2*Sin[e + f*x] + 3*(-1)^(2/3)*b^2*Sin[e + f*x] - 28
*(-1)^(1/3)*a^2*Sec[e + f*x]^(2/3)*Sin[e + f*x] - 28*(-1)^(1/3)*Sqrt[3]*a^2*Sec[e + f*x]^(2/3)*Sin[e + f*x] +
42*(-1)^(1/3)*b^2*Sec[e + f*x]^(2/3)*Sin[e + f*x] + 42*(-1)^(1/3)*Sqrt[3]*b^2*Sec[e + f*x]^(2/3)*Sin[e + f*x]
- 728*a^2*Sec[e + f*x]^(4/3)*Sin[e + f*x] - 364*Sqrt[3]*a^2*Sec[e + f*x]^(4/3)*Sin[e + f*x] + 1092*b^2*Sec[e +
f*x]^(4/3)*Sin[e + f*x] + 546*Sqrt[3]*b^2*Sec[e + f*x]^(4/3)*Sin[e + f*x] + 56084*(-1)^(1/3)*a^2*Sec[e + f*x]
^(8/3)*Sin[e + f*x] + 32060*(-1)^(1/3)*Sqrt[3]*a^2*Sec[e + f*x]^(8/3)*Sin[e + f*x] - 84126*(-1)^(1/3)*b^2*Sec[
e + f*x]^(8/3)*Sin[e + f*x] - 48090*(-1)^(1/3)*Sqrt[3]*b^2*Sec[e + f*x]^(8/3)*Sin[e + f*x] + 305032*a^2*Sec[e
+ f*x]^(10/3)*Sin[e + f*x] + 176540*Sqrt[3]*a^2*Sec[e + f*x]^(10/3)*Sin[e + f*x] - 457548*b^2*Sec[e + f*x]^(10
/3)*Sin[e + f*x] - 264810*Sqrt[3]*b^2*Sec[e + f*x]^(10/3)*Sin[e + f*x] - 3954808*(-1)^(1/3)*a^2*Sec[e + f*x]^(
14/3)*Sin[e + f*x] - 2283424*(-1)^(1/3)*Sqrt[3]*a^2*Sec[e + f*x]^(14/3)*Sin[e + f*x] + 5932212*(-1)^(1/3)*b^2*
Sec[e + f*x]^(14/3)*Sin[e + f*x] + 3425136*(-1)^(1/3)*Sqrt[3]*b^2*Sec[e + f*x]^(14/3)*Sin[e + f*x] - 9625616*a
^2*Sec[e + f*x]^(16/3)*Sin[e + f*x] - 5557552*Sqrt[3]*a^2*Sec[e + f*x]^(16/3)*Sin[e + f*x] + 14438424*b^2*Sec[
e + f*x]^(16/3)*Sin[e + f*x] + 8336328*Sqrt[3]*b^2*Sec[e + f*x]^(16/3)*Sin[e + f*x] + 27089920*(-1)^(1/3)*a^2*
Sec[e + f*x]^(20/3)*Sin[e + f*x] + 15640768*(-1)^(1/3)*Sqrt[3]*a^2*Sec[e + f*x]^(20/3)*Sin[e + f*x] - 40634880
*(-1)^(1/3)*b^2*Sec[e + f*x]^(20/3)*Sin[e + f*x] - 23461152*(-1)^(1/3)*Sqrt[3]*b^2*Sec[e + f*x]^(20/3)*Sin[e +
f*x] + 32361056*a^2*Sec[e + f*x]^(22/3)*Sin[e + f*x] + 18683392*Sqrt[3]*a^2*Sec[e + f*x]^(22/3)*Sin[e + f*x]
- 48541584*b^2*Sec[e + f*x]^(22/3)*Sin[e + f*x] - 28025088*Sqrt[3]*b^2*Sec[e + f*x]^(22/3)*Sin[e + f*x] - 2980
5440*(-1)^(1/3)*a^2*Sec[e + f*x]^(26/3)*Sin[e + f*x] - 17208128*(-1)^(1/3)*Sqrt[3]*a^2*Sec[e + f*x]^(26/3)*Sin
[e + f*x] + 44708160*(-1)^(1/3)*b^2*Sec[e + f*x]^(26/3)*Sin[e + f*x] + 25812192*(-1)^(1/3)*Sqrt[3]*b^2*Sec[e +
f*x]^(26/3)*Sin[e + f*x] - 24330496*a^2*Sec[e + f*x]^(28/3)*Sin[e + f*x] - 14047232*Sqrt[3]*a^2*Sec[e + f*x]^
(28/3)*Sin[e + f*x] + 36495744*b^2*Sec[e + f*x]^(28/3)*Sin[e + f*x] + 21070848*Sqrt[3]*b^2*Sec[e + f*x]^(28/3)
*Sin[e + f*x] + 6614272*(-1)^(1/3)*a^2*Sec[e + f*x]^(32/3)*Sin[e + f*x] + 3818752*(-1)^(1/3)*Sqrt[3]*a^2*Sec[e
+ f*x]^(32/3)*Sin[e + f*x] - 9921408*(-1)^(1/3)*b^2*Sec[e + f*x]^(32/3)*Sin[e + f*x] - 5728128*(-1)^(1/3)*Sqr
t[3]*b^2*Sec[e + f*x]^(32/3)*Sin[e + f*x] + 1290752*a^2*Sec[e + f*x]^(34/3)*Sin[e + f*x] + 745216*Sqrt[3]*a^2*
Sec[e + f*x]^(34/3)*Sin[e + f*x] - 1936128*b^2*Sec[e + f*x]^(34/3)*Sin[e + f*x] - 1117824*Sqrt[3]*b^2*Sec[e +
f*x]^(34/3)*Sin[e + f*x] - 56*(-1)^(1/3)*a*b*Sec[e + f*x]^(5/3)*Sqrt[Cos[e + f*x]^2*(-1 + Sec[e + f*x]^2)]*Sin
[e + f*x] - 56*(-1)^(1/3)*Sqrt[3]*a*b*Sec[e + f*x]^(5/3)*Sqrt[Cos[e + f*x]^2*(-1 + Sec[e + f*x]^2)]*Sin[e + f*
x] - 1456*a*b*Sec[e + f*x]^(7/3)*Sqrt[Cos[e + f*x]^2*(-1 + Sec[e + f*x]^2)]*Sin[e + f*x] - 728*Sqrt[3]*a*b*Sec
[e + f*x]^(7/3)*Sqrt[Cos[e + f*x]^2*(-1 + Sec[e + f*x]^2)]*Sin[e + f*x] + 112168*(-1)^(1/3)*a*b*Sec[e + f*x]^(
11/3)*Sqrt[Cos[e + f*x]^2*(-1 + Sec[e + f*x]^2)]*Sin[e + f*x] + 64120*(-1)^(1/3)*Sqrt[3]*a*b*Sec[e + f*x]^(11/
3)*Sqrt[Cos[e + f*x]^2*(-1 + Sec[e + f*x]^2)]*Sin[e + f*x] + 610064*a*b*Sec[e + f*x]^(13/3)*Sqrt[Cos[e + f*x]^
2*(-1 + Sec[e + f*x]^2)]*Sin[e + f*x] + 353080*Sqrt[3]*a*b*Sec[e + f*x]^(13/3)*Sqrt[Cos[e + f*x]^2*(-1 + Sec[e
+ f*x]^2)]*Sin[e + f*x] - 7909616*(-1)^(1/3)*a*b*Sec[e + f*x]^(17/3)*Sqrt[Cos[e + f*x]^2*(-1 + Sec[e + f*x]^2
)]*Sin[e + f*x] - 4566848*(-1)^(1/3)*Sqrt[3]*a*b*Sec[e + f*x]^(17/3)*Sqrt[Cos[e + f*x]^2*(-1 + Sec[e + f*x]^2)
]*Sin[e + f*x] - 19251232*a*b*Sec[e + f*x]^(19/...
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Maple [F]
time = 0.28, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \tan \left (f x +e \right )\right )^{2}}{\left (d \sec \left (f x +e \right )\right )^{\frac {1}{3}}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(1/3),x)
[Out]
int((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(1/3),x)
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(1/3),x, algorithm="maxima")
[Out]
integrate((b*tan(f*x + e) + a)^2/(d*sec(f*x + e))^(1/3), x)
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(1/3),x, algorithm="fricas")
[Out]
integral((b^2*tan(f*x + e)^2 + 2*a*b*tan(f*x + e) + a^2)*(d*sec(f*x + e))^(2/3)/(d*sec(f*x + e)), x)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{2}}{\sqrt [3]{d \sec {\left (e + f x \right )}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*tan(f*x+e))**2/(d*sec(f*x+e))**(1/3),x)
[Out]
Integral((a + b*tan(e + f*x))**2/(d*sec(e + f*x))**(1/3), x)
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(1/3),x, algorithm="giac")
[Out]
integrate((b*tan(f*x + e) + a)^2/(d*sec(f*x + e))^(1/3), x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{1/3}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b*tan(e + f*x))^2/(d/cos(e + f*x))^(1/3),x)
[Out]
int((a + b*tan(e + f*x))^2/(d/cos(e + f*x))^(1/3), x)
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