3.587 \(\int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2 \, dx\)

Optimal. Leaf size=143 \[ -\frac {2 d^2 \left (5 a^2-2 b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 d \left (5 a^2-2 b^2\right ) \sin (e+f x) \sqrt {d \sec (e+f x)}}{5 f}+\frac {14 a b (d \sec (e+f x))^{3/2}}{15 f}+\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f} \]

[Out]

14/15*a*b*(d*sec(f*x+e))^(3/2)/f-2/5*(5*a^2-2*b^2)*d^2*(cos(1/2*e+1/2*f*x)^2)^(1/2)/cos(1/2*e+1/2*f*x)*Ellipti
cE(sin(1/2*e+1/2*f*x),2^(1/2))/f/cos(f*x+e)^(1/2)/(d*sec(f*x+e))^(1/2)+2/5*(5*a^2-2*b^2)*d*sin(f*x+e)*(d*sec(f
*x+e))^(1/2)/f+2/5*b*(d*sec(f*x+e))^(3/2)*(a+b*tan(f*x+e))/f

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Rubi [A]  time = 0.16, antiderivative size = 143, 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, 3768, 3771, 2639} \[ -\frac {2 d^2 \left (5 a^2-2 b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 d \left (5 a^2-2 b^2\right ) \sin (e+f x) \sqrt {d \sec (e+f x)}}{5 f}+\frac {14 a b (d \sec (e+f x))^{3/2}}{15 f}+\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(5*a^2 - 2*b^2)*d^2*EllipticE[(e + f*x)/2, 2])/(5*f*Sqrt[Cos[e + f*x]]*Sqrt[d*Sec[e + f*x]]) + (14*a*b*(d*
Sec[e + f*x])^(3/2))/(15*f) + (2*(5*a^2 - 2*b^2)*d*Sqrt[d*Sec[e + f*x]]*Sin[e + f*x])/(5*f) + (2*b*(d*Sec[e +
f*x])^(3/2)*(a + b*Tan[e + f*x]))/(5*f)

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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[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 (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2 \, dx &=\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f}+\frac {2}{5} \int (d \sec (e+f x))^{3/2} \left (\frac {5 a^2}{2}-b^2+\frac {7}{2} a b \tan (e+f x)\right ) \, dx\\ &=\frac {14 a b (d \sec (e+f x))^{3/2}}{15 f}+\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f}+\frac {1}{5} \left (5 a^2-2 b^2\right ) \int (d \sec (e+f x))^{3/2} \, dx\\ &=\frac {14 a b (d \sec (e+f x))^{3/2}}{15 f}+\frac {2 \left (5 a^2-2 b^2\right ) d \sqrt {d \sec (e+f x)} \sin (e+f x)}{5 f}+\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f}-\frac {1}{5} \left (\left (5 a^2-2 b^2\right ) d^2\right ) \int \frac {1}{\sqrt {d \sec (e+f x)}} \, dx\\ &=\frac {14 a b (d \sec (e+f x))^{3/2}}{15 f}+\frac {2 \left (5 a^2-2 b^2\right ) d \sqrt {d \sec (e+f x)} \sin (e+f x)}{5 f}+\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f}-\frac {\left (\left (5 a^2-2 b^2\right ) d^2\right ) \int \sqrt {\cos (e+f x)} \, dx}{5 \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}\\ &=-\frac {2 \left (5 a^2-2 b^2\right ) d^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {14 a b (d \sec (e+f x))^{3/2}}{15 f}+\frac {2 \left (5 a^2-2 b^2\right ) d \sqrt {d \sec (e+f x)} \sin (e+f x)}{5 f}+\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f}\\ \end {align*}

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Mathematica [A]  time = 0.65, size = 126, normalized size = 0.88 \[ -\frac {2 d^2 (a+b \tan (e+f x))^2 \left (\left (3 b^2-\frac {15 a^2}{2}\right ) \sin (2 (e+f x))+3 \left (5 a^2-2 b^2\right ) \cos ^{\frac {3}{2}}(e+f x) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )-b (10 a+3 b \tan (e+f x))\right )}{15 f \sqrt {d \sec (e+f x)} (a \cos (e+f x)+b \sin (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d^2*(a + b*Tan[e + f*x])^2*(3*(5*a^2 - 2*b^2)*Cos[e + f*x]^(3/2)*EllipticE[(e + f*x)/2, 2] + ((-15*a^2)/2
+ 3*b^2)*Sin[2*(e + f*x)] - b*(10*a + 3*b*Tan[e + f*x])))/(15*f*Sqrt[d*Sec[e + f*x]]*(a*Cos[e + f*x] + b*Sin[e
+ f*x])^2)

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{2} d \sec \left (f x + e\right ) \tan \left (f x + e\right )^{2} + 2 \, a b d \sec \left (f x + e\right ) \tan \left (f x + e\right ) + a^{2} d \sec \left (f x + e\right )\right )} \sqrt {d \sec \left (f x + e\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.91, size = 712, normalized size = 4.98 \[ -\frac {2 \left (1+\cos \left (f x +e \right )\right )^{2} \left (-1+\cos \left (f x +e \right )\right )^{2} \left (15 i \left (\cos ^{3}\left (f x +e \right )\right ) \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 )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) a^{2}-6 i \left (\cos ^{3}\left (f x +e \right )\right ) \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 )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) b^{2}-15 i \left (\cos ^{3}\left (f x +e \right )\right ) \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 )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) a^{2}+6 i \left (\cos ^{3}\left (f x +e \right )\right ) \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 )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) b^{2}+15 i \left (\cos ^{2}\left (f x +e \right )\right ) \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 )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) a^{2}-6 i \left (\cos ^{2}\left (f x +e \right )\right ) \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 )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) b^{2}-15 i \left (\cos ^{2}\left (f x +e \right )\right ) \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 )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) a^{2}+6 i \left (\cos ^{2}\left (f x +e \right )\right ) \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 )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) b^{2}+15 \left (\cos ^{3}\left (f x +e \right )\right ) a^{2}-6 \left (\cos ^{3}\left (f x +e \right )\right ) b^{2}-15 a^{2} \left (\cos ^{2}\left (f x +e \right )\right )+9 b^{2} \left (\cos ^{2}\left (f x +e \right )\right )-10 a \cos \left (f x +e \right ) b \sin \left (f x +e \right )-3 b^{2}\right ) \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}}}{15 f \sin \left (f x +e \right )^{5} \cos \left (f x +e \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15/f*(1+cos(f*x+e))^2*(-1+cos(f*x+e))^2*(15*I*cos(f*x+e)^3*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+
e)))^(1/2)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*sin(f*x+e)*a^2-6*I*cos(f*x+e)^3*(1/(1+cos(f*x+e)))^(1/2)*
(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*sin(f*x+e)*b^2-15*I*cos(f*x+e)^3*(
1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*sin(f*x+e)
*a^2+6*I*cos(f*x+e)^3*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticE(I*(-1+cos(f*x+e))/s
in(f*x+e),I)*sin(f*x+e)*b^2+15*I*cos(f*x+e)^2*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*Ellip
ticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*sin(f*x+e)*a^2-6*I*cos(f*x+e)^2*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+c
os(f*x+e)))^(1/2)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*sin(f*x+e)*b^2-15*I*cos(f*x+e)^2*(1/(1+cos(f*x+e))
)^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*sin(f*x+e)*a^2+6*I*cos(f*x
+e)^2*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*sin
(f*x+e)*b^2+15*cos(f*x+e)^3*a^2-6*cos(f*x+e)^3*b^2-15*a^2*cos(f*x+e)^2+9*b^2*cos(f*x+e)^2-10*a*cos(f*x+e)*b*si
n(f*x+e)-3*b^2)*(d/cos(f*x+e))^(3/2)/sin(f*x+e)^5/cos(f*x+e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{3/2}\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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