3.7.31 \(\int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{5/3}} \, dx\) [631]

Optimal. Leaf size=119 \[ \frac {3 a b}{10 f (d \sec (e+f x))^{5/3}}-\frac {3 \left (2 a^2+3 b^2\right ) d \, _2F_1\left (\frac {1}{2},\frac {4}{3};\frac {7}{3};\cos ^2(e+f x)\right ) \sin (e+f x)}{16 f (d \sec (e+f x))^{8/3} \sqrt {\sin ^2(e+f x)}}-\frac {3 b (a+b \tan (e+f x))}{2 f (d \sec (e+f x))^{5/3}} \]

[Out]

3/10*a*b/f/(d*sec(f*x+e))^(5/3)-3/16*(2*a^2+3*b^2)*d*hypergeom([1/2, 4/3],[7/3],cos(f*x+e)^2)*sin(f*x+e)/f/(d*
sec(f*x+e))^(8/3)/(sin(f*x+e)^2)^(1/2)-3/2*b*(a+b*tan(f*x+e))/f/(d*sec(f*x+e))^(5/3)

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Rubi [A]
time = 0.11, 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 {4}{3};\frac {7}{3};\cos ^2(e+f x)\right )}{16 f \sqrt {\sin ^2(e+f x)} (d \sec (e+f x))^{8/3}}+\frac {3 a b}{10 f (d \sec (e+f x))^{5/3}}-\frac {3 b (a+b \tan (e+f x))}{2 f (d \sec (e+f x))^{5/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])^2/(d*Sec[e + f*x])^(5/3),x]

[Out]

(3*a*b)/(10*f*(d*Sec[e + f*x])^(5/3)) - (3*(2*a^2 + 3*b^2)*d*Hypergeometric2F1[1/2, 4/3, 7/3, Cos[e + f*x]^2]*
Sin[e + f*x])/(16*f*(d*Sec[e + f*x])^(8/3)*Sqrt[Sin[e + f*x]^2]) - (3*b*(a + b*Tan[e + f*x]))/(2*f*(d*Sec[e +
f*x])^(5/3))

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}{(d \sec (e+f x))^{5/3}} \, dx &=-\frac {3 b (a+b \tan (e+f x))}{2 f (d \sec (e+f x))^{5/3}}-\frac {3}{2} \int \frac {-\frac {2 a^2}{3}-b^2+\frac {1}{3} a b \tan (e+f x)}{(d \sec (e+f x))^{5/3}} \, dx\\ &=\frac {3 a b}{10 f (d \sec (e+f x))^{5/3}}-\frac {3 b (a+b \tan (e+f x))}{2 f (d \sec (e+f x))^{5/3}}-\frac {1}{2} \left (-2 a^2-3 b^2\right ) \int \frac {1}{(d \sec (e+f x))^{5/3}} \, dx\\ &=\frac {3 a b}{10 f (d \sec (e+f x))^{5/3}}-\frac {3 b (a+b \tan (e+f x))}{2 f (d \sec (e+f x))^{5/3}}-\frac {1}{2} \left (\left (-2 a^2-3 b^2\right ) \sqrt [3]{\frac {\cos (e+f x)}{d}} \sqrt [3]{d \sec (e+f x)}\right ) \int \left (\frac {\cos (e+f x)}{d}\right )^{5/3} \, dx\\ &=\frac {3 a b}{10 f (d \sec (e+f x))^{5/3}}-\frac {3 \left (2 a^2+3 b^2\right ) \cos ^3(e+f x) \, _2F_1\left (\frac {1}{2},\frac {4}{3};\frac {7}{3};\cos ^2(e+f x)\right ) \sqrt [3]{d \sec (e+f x)} \sin (e+f x)}{16 d^2 f \sqrt {\sin ^2(e+f x)}}-\frac {3 b (a+b \tan (e+f x))}{2 f (d \sec (e+f x))^{5/3}}\\ \end {align*}

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Mathematica [A]
time = 0.57, size = 119, normalized size = 1.00 \begin {gather*} \frac {\sec ^2(e+f x) \left (-6 a b-6 a b \cos (2 (e+f x))+3 a^2 \sin (2 (e+f x))-3 b^2 \sin (2 (e+f x))+2 \left (2 a^2+3 b^2\right ) \cos ^2(e+f x)^{2/3} \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {3}{2};\sin ^2(e+f x)\right ) \tan (e+f x)\right )}{10 f (d \sec (e+f x))^{5/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x])^2/(d*Sec[e + f*x])^(5/3),x]

[Out]

(Sec[e + f*x]^2*(-6*a*b - 6*a*b*Cos[2*(e + f*x)] + 3*a^2*Sin[2*(e + f*x)] - 3*b^2*Sin[2*(e + f*x)] + 2*(2*a^2
+ 3*b^2)*(Cos[e + f*x]^2)^(2/3)*Hypergeometric2F1[1/2, 2/3, 3/2, Sin[e + f*x]^2]*Tan[e + f*x]))/(10*f*(d*Sec[e
+ f*x])^(5/3))

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Maple [F]
time = 0.28, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \tan \left (f x +e \right )\right )^{2}}{\left (d \sec \left (f x +e \right )\right )^{\frac {5}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(5/3),x)

[Out]

int((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(5/3),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))^(5/3),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e) + a)^2/(d*sec(f*x + e))^(5/3), 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))^(5/3),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))^(1/3)/(d^2*sec(f*x + e)^2), 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}}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{3}}}\, 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))**(5/3),x)

[Out]

Integral((a + b*tan(e + f*x))**2/(d*sec(e + f*x))**(5/3), 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))^(5/3),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e) + a)^2/(d*sec(f*x + e))^(5/3), 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 )}^{5/3}} \,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))^(5/3),x)

[Out]

int((a + b*tan(e + f*x))^2/(d/cos(e + f*x))^(5/3), x)

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