3.591 \(\int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=145 \[ \frac {2 \left (3 a^2+2 b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 d^2 f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 \left (3 a^2+2 b^2\right ) \sin (e+f x)}{15 d f (d \sec (e+f x))^{3/2}}-\frac {2 a b}{15 f (d \sec (e+f x))^{5/2}}-\frac {2 b (a+b \tan (e+f x))}{3 f (d \sec (e+f x))^{5/2}} \]

[Out]

-2/15*a*b/f/(d*sec(f*x+e))^(5/2)+2/15*(3*a^2+2*b^2)*sin(f*x+e)/d/f/(d*sec(f*x+e))^(3/2)+2/5*(3*a^2+2*b^2)*(cos
(1/2*e+1/2*f*x)^2)^(1/2)/cos(1/2*e+1/2*f*x)*EllipticE(sin(1/2*e+1/2*f*x),2^(1/2))/d^2/f/cos(f*x+e)^(1/2)/(d*se
c(f*x+e))^(1/2)-2/3*b*(a+b*tan(f*x+e))/f/(d*sec(f*x+e))^(5/2)

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Rubi [A]  time = 0.17, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3508, 3486, 3769, 3771, 2639} \[ \frac {2 \left (3 a^2+2 b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 d^2 f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 \left (3 a^2+2 b^2\right ) \sin (e+f x)}{15 d f (d \sec (e+f x))^{3/2}}-\frac {2 a b}{15 f (d \sec (e+f x))^{5/2}}-\frac {2 b (a+b \tan (e+f x))}{3 f (d \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*b)/(15*f*(d*Sec[e + f*x])^(5/2)) + (2*(3*a^2 + 2*b^2)*EllipticE[(e + f*x)/2, 2])/(5*d^2*f*Sqrt[Cos[e + f
*x]]*Sqrt[d*Sec[e + f*x]]) + (2*(3*a^2 + 2*b^2)*Sin[e + f*x])/(15*d*f*(d*Sec[e + f*x])^(3/2)) - (2*b*(a + b*Ta
n[e + f*x]))/(3*f*(d*Sec[e + f*x])^(5/2))

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 3486

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 3508

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(b*(d*Se
c[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 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rubi steps

\begin {align*} \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{5/2}} \, dx &=-\frac {2 b (a+b \tan (e+f x))}{3 f (d \sec (e+f x))^{5/2}}-\frac {2}{3} \int \frac {-\frac {3 a^2}{2}-b^2-\frac {1}{2} a b \tan (e+f x)}{(d \sec (e+f x))^{5/2}} \, dx\\ &=-\frac {2 a b}{15 f (d \sec (e+f x))^{5/2}}-\frac {2 b (a+b \tan (e+f x))}{3 f (d \sec (e+f x))^{5/2}}-\frac {1}{3} \left (-3 a^2-2 b^2\right ) \int \frac {1}{(d \sec (e+f x))^{5/2}} \, dx\\ &=-\frac {2 a b}{15 f (d \sec (e+f x))^{5/2}}+\frac {2 \left (3 a^2+2 b^2\right ) \sin (e+f x)}{15 d f (d \sec (e+f x))^{3/2}}-\frac {2 b (a+b \tan (e+f x))}{3 f (d \sec (e+f x))^{5/2}}+\frac {\left (3 a^2+2 b^2\right ) \int \frac {1}{\sqrt {d \sec (e+f x)}} \, dx}{5 d^2}\\ &=-\frac {2 a b}{15 f (d \sec (e+f x))^{5/2}}+\frac {2 \left (3 a^2+2 b^2\right ) \sin (e+f x)}{15 d f (d \sec (e+f x))^{3/2}}-\frac {2 b (a+b \tan (e+f x))}{3 f (d \sec (e+f x))^{5/2}}+\frac {\left (3 a^2+2 b^2\right ) \int \sqrt {\cos (e+f x)} \, dx}{5 d^2 \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}\\ &=-\frac {2 a b}{15 f (d \sec (e+f x))^{5/2}}+\frac {2 \left (3 a^2+2 b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 d^2 f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 \left (3 a^2+2 b^2\right ) \sin (e+f x)}{15 d f (d \sec (e+f x))^{3/2}}-\frac {2 b (a+b \tan (e+f x))}{3 f (d \sec (e+f x))^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.92, size = 92, normalized size = 0.63 \[ \frac {\left (6 a^2+4 b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )+2 \cos ^{\frac {3}{2}}(e+f x) \left (\left (a^2-b^2\right ) \sin (e+f x)-2 a b \cos (e+f x)\right )}{5 f \cos ^{\frac {5}{2}}(e+f x) (d \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((6*a^2 + 4*b^2)*EllipticE[(e + f*x)/2, 2] + 2*Cos[e + f*x]^(3/2)*(-2*a*b*Cos[e + f*x] + (a^2 - b^2)*Sin[e + f
*x]))/(5*f*Cos[e + f*x]^(5/2)*(d*Sec[e + f*x])^(5/2))

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} \tan \left (f x + e\right )^{2} + 2 \, a b \tan \left (f x + e\right ) + a^{2}\right )} \sqrt {d \sec \left (f x + e\right )}}{d^{3} \sec \left (f x + e\right )^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.88, size = 670, normalized size = 4.62 \[ -\frac {2 \left (-3 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \cos \left (f x +e \right ) \sin \left (f x +e \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) a^{2}-2 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \cos \left (f x +e \right ) \sin \left (f x +e \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) b^{2}+3 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \cos \left (f x +e \right ) \sin \left (f x +e \right ) \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) a^{2}+2 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \cos \left (f x +e \right ) \sin \left (f x +e \right ) \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) b^{2}-3 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) a^{2}-2 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) b^{2}+3 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) a^{2}+2 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) b^{2}+\left (\cos ^{4}\left (f x +e \right )\right ) a^{2}-\left (\cos ^{4}\left (f x +e \right )\right ) b^{2}+2 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right ) a b +2 a^{2} \left (\cos ^{2}\left (f x +e \right )\right )+3 b^{2} \left (\cos ^{2}\left (f x +e \right )\right )-3 \cos \left (f x +e \right ) a^{2}-2 \cos \left (f x +e \right ) b^{2}\right )}{5 f \cos \left (f x +e \right )^{3} \sin \left (f x +e \right ) \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5/f*(-3*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*cos(f*x+e)*sin(f*x+e)*EllipticF(I*(-1+
cos(f*x+e))/sin(f*x+e),I)*a^2-2*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*cos(f*x+e)*sin(f*
x+e)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*b^2+3*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1
/2)*cos(f*x+e)*sin(f*x+e)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*a^2+2*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+
e)/(1+cos(f*x+e)))^(1/2)*cos(f*x+e)*sin(f*x+e)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*b^2-3*I*(1/(1+cos(f*x
+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*a^2-2*I*(1/
(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*b
^2+3*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)*EllipticE(I*(-1+cos(f*x+e))/sin(f
*x+e),I)*a^2+2*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)*EllipticE(I*(-1+cos(f*x
+e))/sin(f*x+e),I)*b^2+cos(f*x+e)^4*a^2-cos(f*x+e)^4*b^2+2*cos(f*x+e)^3*sin(f*x+e)*a*b+2*a^2*cos(f*x+e)^2+3*b^
2*cos(f*x+e)^2-3*cos(f*x+e)*a^2-2*cos(f*x+e)*b^2)/cos(f*x+e)^3/sin(f*x+e)/(d/cos(f*x+e))^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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