3.640 \(\int (d \sec (e+f x))^m (a+b \tan (e+f x))^3 \, dx\)

Optimal. Leaf size=173 \[ -\frac{a \left (3 b^2-a^2 (m+1)\right ) \tan (e+f x) \sec ^2(e+f x)^{-m/2} (d \sec (e+f x))^m \, _2F_1\left (\frac{1}{2},1-\frac{m}{2};\frac{3}{2};-\tan ^2(e+f x)\right )}{f (m+1)}-\frac{b (d \sec (e+f x))^m \left (2 (m+1) \left (b^2-a^2 (m+3)\right )-a b m (m+4) \tan (e+f x)\right )}{f m \left (m^2+3 m+2\right )}+\frac{b (a+b \tan (e+f x))^2 (d \sec (e+f x))^m}{f (m+2)} \]

[Out]

-((a*(3*b^2 - a^2*(1 + m))*Hypergeometric2F1[1/2, 1 - m/2, 3/2, -Tan[e + f*x]^2]*(d*Sec[e + f*x])^m*Tan[e + f*
x])/(f*(1 + m)*(Sec[e + f*x]^2)^(m/2))) + (b*(d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^2)/(f*(2 + m)) - (b*(d*Se
c[e + f*x])^m*(2*(1 + m)*(b^2 - a^2*(3 + m)) - a*b*m*(4 + m)*Tan[e + f*x]))/(f*m*(2 + 3*m + m^2))

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Rubi [A]  time = 0.19662, antiderivative size = 167, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3512, 743, 780, 245} \[ \frac{a \left (a^2-\frac{3 b^2}{m+1}\right ) \tan (e+f x) \sec ^2(e+f x)^{-m/2} (d \sec (e+f x))^m \, _2F_1\left (\frac{1}{2},1-\frac{m}{2};\frac{3}{2};-\tan ^2(e+f x)\right )}{f}-\frac{b (d \sec (e+f x))^m \left (2 (m+1) \left (b^2-a^2 (m+3)\right )-a b m (m+4) \tan (e+f x)\right )}{f m \left (m^2+3 m+2\right )}+\frac{b (a+b \tan (e+f x))^2 (d \sec (e+f x))^m}{f (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(a^2 - (3*b^2)/(1 + m))*Hypergeometric2F1[1/2, 1 - m/2, 3/2, -Tan[e + f*x]^2]*(d*Sec[e + f*x])^m*Tan[e + f*
x])/(f*(Sec[e + f*x]^2)^(m/2)) + (b*(d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^2)/(f*(2 + m)) - (b*(d*Sec[e + f*x
])^m*(2*(1 + m)*(b^2 - a^2*(3 + m)) - a*b*m*(4 + m)*Tan[e + f*x]))/(f*m*(2 + 3*m + m^2))

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 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, -((b*x^n)/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rubi steps

\begin{align*} \int (d \sec (e+f x))^m (a+b \tan (e+f x))^3 \, dx &=\frac{\left ((d \sec (e+f x))^m \sec ^2(e+f x)^{-m/2}\right ) \operatorname{Subst}\left (\int (a+x)^3 \left (1+\frac{x^2}{b^2}\right )^{-1+\frac{m}{2}} \, dx,x,b \tan (e+f x)\right )}{b f}\\ &=\frac{b (d \sec (e+f x))^m (a+b \tan (e+f x))^2}{f (2+m)}+\frac{\left (b (d \sec (e+f x))^m \sec ^2(e+f x)^{-m/2}\right ) \operatorname{Subst}\left (\int (a+x) \left (-2+\frac{a^2 (2+m)}{b^2}+\frac{a (4+m) x}{b^2}\right ) \left (1+\frac{x^2}{b^2}\right )^{-1+\frac{m}{2}} \, dx,x,b \tan (e+f x)\right )}{f (2+m)}\\ &=\frac{b (d \sec (e+f x))^m (a+b \tan (e+f x))^2}{f (2+m)}-\frac{b (d \sec (e+f x))^m \left (2 (1+m) \left (b^2-a^2 (3+m)\right )-a b m (4+m) \tan (e+f x)\right )}{f m \left (2+3 m+m^2\right )}-\frac{\left (a \left (3 b^2-a^2 (1+m)\right ) (d \sec (e+f x))^m \sec ^2(e+f x)^{-m/2}\right ) \operatorname{Subst}\left (\int \left (1+\frac{x^2}{b^2}\right )^{-1+\frac{m}{2}} \, dx,x,b \tan (e+f x)\right )}{b f (1+m)}\\ &=-\frac{a \left (3 b^2-a^2 (1+m)\right ) \, _2F_1\left (\frac{1}{2},1-\frac{m}{2};\frac{3}{2};-\tan ^2(e+f x)\right ) (d \sec (e+f x))^m \sec ^2(e+f x)^{-m/2} \tan (e+f x)}{f (1+m)}+\frac{b (d \sec (e+f x))^m (a+b \tan (e+f x))^2}{f (2+m)}-\frac{b (d \sec (e+f x))^m \left (2 (1+m) \left (b^2-a^2 (3+m)\right )-a b m (4+m) \tan (e+f x)\right )}{f m \left (2+3 m+m^2\right )}\\ \end{align*}

Mathematica [A]  time = 57.2557, size = 175, normalized size = 1.01 \[ \frac{\sec ^2(e+f x)^{-m/2} (d \sec (e+f x))^m \left (b \left (\left (3 a^2 (m+2)-2 b^2\right ) \left (\sec ^2(e+f x)^{m/2}-1\right )+a b m (m+2) \tan ^3(e+f x) \, _2F_1\left (\frac{3}{2},1-\frac{m}{2};\frac{5}{2};-\tan ^2(e+f x)\right )+b^2 m \tan ^2(e+f x) \sec ^2(e+f x)^{m/2}\right )+a^3 m (m+2) \tan (e+f x) \, _2F_1\left (\frac{1}{2},1-\frac{m}{2};\frac{3}{2};-\tan ^2(e+f x)\right )\right )}{f m (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.543, size = 0, normalized size = 0. \begin{align*} \int \left ( d\sec \left ( fx+e \right ) \right ) ^{m} \left ( a+b\tan \left ( fx+e \right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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