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

Optimal. Leaf size=198 \[ \frac{2 b d^2 \sec ^2(e+f x) \sqrt{d \sec (e+f x)} \left (14 \left (11 a^2-2 b^2\right )+65 a b \tan (e+f x)\right )}{315 f}+\frac{2 a d^2 \left (7 a^2-6 b^2\right ) \tan (e+f x) \sqrt{d \sec (e+f x)}}{21 f}+\frac{2 a d^2 \left (7 a^2-6 b^2\right ) \sqrt{d \sec (e+f x)} F\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{21 f \sqrt [4]{\sec ^2(e+f x)}}+\frac{2 b d^2 \sec ^2(e+f x) \sqrt{d \sec (e+f x)} (a+b \tan (e+f x))^2}{9 f} \]

[Out]

(2*a*(7*a^2 - 6*b^2)*d^2*EllipticF[ArcTan[Tan[e + f*x]]/2, 2]*Sqrt[d*Sec[e + f*x]])/(21*f*(Sec[e + f*x]^2)^(1/
4)) + (2*a*(7*a^2 - 6*b^2)*d^2*Sqrt[d*Sec[e + f*x]]*Tan[e + f*x])/(21*f) + (2*b*d^2*Sec[e + f*x]^2*Sqrt[d*Sec[
e + f*x]]*(a + b*Tan[e + f*x])^2)/(9*f) + (2*b*d^2*Sec[e + f*x]^2*Sqrt[d*Sec[e + f*x]]*(14*(11*a^2 - 2*b^2) +
65*a*b*Tan[e + f*x]))/(315*f)

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Rubi [A]  time = 0.154041, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3512, 743, 780, 195, 231} \[ \frac{2 b d^2 \sec ^2(e+f x) \sqrt{d \sec (e+f x)} \left (14 \left (11 a^2-2 b^2\right )+65 a b \tan (e+f x)\right )}{315 f}+\frac{2 a d^2 \left (7 a^2-6 b^2\right ) \tan (e+f x) \sqrt{d \sec (e+f x)}}{21 f}+\frac{2 a d^2 \left (7 a^2-6 b^2\right ) \sqrt{d \sec (e+f x)} F\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{21 f \sqrt [4]{\sec ^2(e+f x)}}+\frac{2 b d^2 \sec ^2(e+f x) \sqrt{d \sec (e+f x)} (a+b \tan (e+f x))^2}{9 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(7*a^2 - 6*b^2)*d^2*EllipticF[ArcTan[Tan[e + f*x]]/2, 2]*Sqrt[d*Sec[e + f*x]])/(21*f*(Sec[e + f*x]^2)^(1/
4)) + (2*a*(7*a^2 - 6*b^2)*d^2*Sqrt[d*Sec[e + f*x]]*Tan[e + f*x])/(21*f) + (2*b*d^2*Sec[e + f*x]^2*Sqrt[d*Sec[
e + f*x]]*(a + b*Tan[e + f*x])^2)/(9*f) + (2*b*d^2*Sec[e + f*x]^2*Sqrt[d*Sec[e + f*x]]*(14*(11*a^2 - 2*b^2) +
65*a*b*Tan[e + f*x]))/(315*f)

Rule 3512

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(d^(2
*IntPart[m/2])*(d*Sec[e + f*x])^(2*FracPart[m/2]))/(b*f*(Sec[e + f*x]^2)^FracPart[m/2]), Subst[Int[(a + x)^n*(
1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 + b^2, 0] &&
!IntegerQ[m/2]

Rule 743

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - a*e^
2*(m - 1) + 2*c*d*e*(m + p)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2,
0] && If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, 0, c, d, e, m
, p, x]

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 231

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(3/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rubi steps

\begin{align*} \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3 \, dx &=\frac{\left (d^2 \sqrt{d \sec (e+f x)}\right ) \operatorname{Subst}\left (\int (a+x)^3 \sqrt [4]{1+\frac{x^2}{b^2}} \, dx,x,b \tan (e+f x)\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\\ &=\frac{2 b d^2 \sec ^2(e+f x) \sqrt{d \sec (e+f x)} (a+b \tan (e+f x))^2}{9 f}+\frac{\left (2 b d^2 \sqrt{d \sec (e+f x)}\right ) \operatorname{Subst}\left (\int (a+x) \left (\frac{1}{2} \left (-4+\frac{9 a^2}{b^2}\right )+\frac{13 a x}{2 b^2}\right ) \sqrt [4]{1+\frac{x^2}{b^2}} \, dx,x,b \tan (e+f x)\right )}{9 f \sqrt [4]{\sec ^2(e+f x)}}\\ &=\frac{2 b d^2 \sec ^2(e+f x) \sqrt{d \sec (e+f x)} (a+b \tan (e+f x))^2}{9 f}+\frac{2 b d^2 \sec ^2(e+f x) \sqrt{d \sec (e+f x)} \left (14 \left (11 a^2-2 b^2\right )+65 a b \tan (e+f x)\right )}{315 f}-\frac{\left (a \left (6-\frac{7 a^2}{b^2}\right ) b d^2 \sqrt{d \sec (e+f x)}\right ) \operatorname{Subst}\left (\int \sqrt [4]{1+\frac{x^2}{b^2}} \, dx,x,b \tan (e+f x)\right )}{7 f \sqrt [4]{\sec ^2(e+f x)}}\\ &=\frac{2 a \left (7 a^2-6 b^2\right ) d^2 \sqrt{d \sec (e+f x)} \tan (e+f x)}{21 f}+\frac{2 b d^2 \sec ^2(e+f x) \sqrt{d \sec (e+f x)} (a+b \tan (e+f x))^2}{9 f}+\frac{2 b d^2 \sec ^2(e+f x) \sqrt{d \sec (e+f x)} \left (14 \left (11 a^2-2 b^2\right )+65 a b \tan (e+f x)\right )}{315 f}-\frac{\left (a \left (6-\frac{7 a^2}{b^2}\right ) b d^2 \sqrt{d \sec (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1+\frac{x^2}{b^2}\right )^{3/4}} \, dx,x,b \tan (e+f x)\right )}{21 f \sqrt [4]{\sec ^2(e+f x)}}\\ &=\frac{2 a \left (7 a^2-6 b^2\right ) d^2 F\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) \sqrt{d \sec (e+f x)}}{21 f \sqrt [4]{\sec ^2(e+f x)}}+\frac{2 a \left (7 a^2-6 b^2\right ) d^2 \sqrt{d \sec (e+f x)} \tan (e+f x)}{21 f}+\frac{2 b d^2 \sec ^2(e+f x) \sqrt{d \sec (e+f x)} (a+b \tan (e+f x))^2}{9 f}+\frac{2 b d^2 \sec ^2(e+f x) \sqrt{d \sec (e+f x)} \left (14 \left (11 a^2-2 b^2\right )+65 a b \tan (e+f x)\right )}{315 f}\\ \end{align*}

Mathematica [A]  time = 1.69054, size = 157, normalized size = 0.79 \[ -\frac{2 d (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^3 \left (63 b \left (b^2-3 a^2\right ) \cos ^2(e+f x)-15 a \left (7 a^2-6 b^2\right ) \sin (e+f x) \cos ^3(e+f x)-15 a \left (7 a^2-6 b^2\right ) \cos ^{\frac{9}{2}}(e+f x) F\left (\left .\frac{1}{2} (e+f x)\right |2\right )-\frac{5}{2} b^2 (27 a \sin (2 (e+f x))+14 b)\right )}{315 f (a \cos (e+f x)+b \sin (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d*(d*Sec[e + f*x])^(3/2)*(63*b*(-3*a^2 + b^2)*Cos[e + f*x]^2 - 15*a*(7*a^2 - 6*b^2)*Cos[e + f*x]^(9/2)*Ell
ipticF[(e + f*x)/2, 2] - 15*a*(7*a^2 - 6*b^2)*Cos[e + f*x]^3*Sin[e + f*x] - (5*b^2*(14*b + 27*a*Sin[2*(e + f*x
)]))/2)*(a + b*Tan[e + f*x])^3)/(315*f*(a*Cos[e + f*x] + b*Sin[e + f*x])^3)

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Maple [C]  time = 0.382, size = 414, normalized size = 2.1 \begin{align*}{\frac{2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2} \left ( \cos \left ( fx+e \right ) -1 \right ) ^{2}}{315\,f \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{4}} \left ( 105\,i\sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{5}{a}^{3}-90\,i\sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{5}a{b}^{2}+105\,i\sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{3}-90\,i\sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}a{b}^{2}+105\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}{a}^{3}-90\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}a{b}^{2}+189\,{a}^{2} \left ( \cos \left ( fx+e \right ) \right ) ^{2}b-63\,{b}^{3} \left ( \cos \left ( fx+e \right ) \right ) ^{2}+135\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) a{b}^{2}+35\,{b}^{3} \right ) \left ({\frac{d}{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315/f*(cos(f*x+e)+1)^2*(cos(f*x+e)-1)^2*(105*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*El
lipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)^5*a^3-90*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+
1))^(1/2)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)^5*a*b^2+105*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+
e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)^4*a^3-90*I*(1/(cos(f*x+e)+1))^(1/
2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)^4*a*b^2+105*sin(f*x+e
)*cos(f*x+e)^3*a^3-90*sin(f*x+e)*cos(f*x+e)^3*a*b^2+189*a^2*cos(f*x+e)^2*b-63*b^3*cos(f*x+e)^2+135*sin(f*x+e)*
cos(f*x+e)*a*b^2+35*b^3)*(d/cos(f*x+e))^(5/2)/cos(f*x+e)^2/sin(f*x+e)^4

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{3} d^{2} \sec \left (f x + e\right )^{2} \tan \left (f x + e\right )^{3} + 3 \, a b^{2} d^{2} \sec \left (f x + e\right )^{2} \tan \left (f x + e\right )^{2} + 3 \, a^{2} b d^{2} \sec \left (f x + e\right )^{2} \tan \left (f x + e\right ) + a^{3} d^{2} \sec \left (f x + e\right )^{2}\right )} \sqrt{d \sec \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \sec \left (f x + e\right )\right )^{\frac{5}{2}}{\left (b \tan \left (f x + e\right ) + a\right )}^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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