3.438 \(\int \sec (c+d x) (a+b \tan ^2(c+d x))^2 \, dx\)

Optimal. Leaf size=96 \[ \frac{\left (8 a^2-8 a b+3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{3 b (2 a-b) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{b \tan (c+d x) \sec ^3(c+d x) \left (a-(a-b) \sin ^2(c+d x)\right )}{4 d} \]

[Out]

((8*a^2 - 8*a*b + 3*b^2)*ArcTanh[Sin[c + d*x]])/(8*d) + (3*(2*a - b)*b*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (b*S
ec[c + d*x]^3*(a - (a - b)*Sin[c + d*x]^2)*Tan[c + d*x])/(4*d)

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Rubi [A]  time = 0.0859652, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3676, 413, 385, 206} \[ \frac{\left (8 a^2-8 a b+3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{3 b (2 a-b) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{b \tan (c+d x) \sec ^3(c+d x) \left (a-(a-b) \sin ^2(c+d x)\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]*(a + b*Tan[c + d*x]^2)^2,x]

[Out]

((8*a^2 - 8*a*b + 3*b^2)*ArcTanh[Sin[c + d*x]])/(8*d) + (3*(2*a - b)*b*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (b*S
ec[c + d*x]^3*(a - (a - b)*Sin[c + d*x]^2)*Tan[c + d*x])/(4*d)

Rule 3676

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> With[{ff = F
reeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 -
ff^2*x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n/2] && IntegerQ[p]

Rule 413

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((a*d - c*b)*x*(a + b*x^n)^
(p + 1)*(c + d*x^n)^(q - 1))/(a*b*n*(p + 1)), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q
, x]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \sec (c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a-(a-b) x^2\right )^2}{\left (1-x^2\right )^3} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac{b \sec ^3(c+d x) \left (a-(a-b) \sin ^2(c+d x)\right ) \tan (c+d x)}{4 d}-\frac{\operatorname{Subst}\left (\int \frac{-a (4 a-b)+(4 a-3 b) (a-b) x^2}{\left (1-x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{4 d}\\ &=\frac{3 (2 a-b) b \sec (c+d x) \tan (c+d x)}{8 d}+\frac{b \sec ^3(c+d x) \left (a-(a-b) \sin ^2(c+d x)\right ) \tan (c+d x)}{4 d}+\frac{\left (8 a^2-8 a b+3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{8 d}\\ &=\frac{\left (8 a^2-8 a b+3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{3 (2 a-b) b \sec (c+d x) \tan (c+d x)}{8 d}+\frac{b \sec ^3(c+d x) \left (a-(a-b) \sin ^2(c+d x)\right ) \tan (c+d x)}{4 d}\\ \end{align*}

Mathematica [C]  time = 8.49057, size = 347, normalized size = 3.61 \[ \frac{\csc ^3(c+d x) \left (128 \sin ^6(c+d x) \left (\frac{1}{2} a^2 (5 \cos (2 (c+d x))+9) \cos ^2(c+d x)+b \sin ^2(c+d x) \left (5 a \cos (2 (c+d x))+7 a+5 b \sin ^2(c+d x)\right )\right ) \text{HypergeometricPFQ}\left (\left \{\frac{3}{2},2,2,2\right \},\left \{1,1,\frac{9}{2}\right \},\sin ^2(c+d x)\right )+128 \sin ^6(c+d x) \left ((b-a) \sin ^2(c+d x)+a\right )^2 \text{HypergeometricPFQ}\left (\left \{\frac{3}{2},2,2,2,2\right \},\left \{1,1,1,\frac{9}{2}\right \},\sin ^2(c+d x)\right )+35 \left (\left (-3161 a^2+5108 a b-1947 b^2\right ) \sin ^4(c+d x)+\frac{3 \tanh ^{-1}\left (\sqrt{\sin ^2(c+d x)}\right ) \left (\left (-400 a^2+778 a b-378 b^2\right ) \sin ^6(c+d x)+\left (1674 a^2-2286 a b+649 b^2\right ) \sin ^4(c+d x)+1125 a^2+9 (a-b)^2 \sin ^8(c+d x)-2 a (1172 a-875 b) \sin ^2(c+d x)\right )}{\sqrt{\sin ^2(c+d x)}}-3375 a^2+485 (a-b)^2 \sin ^6(c+d x)+3 a (1969 a-1750 b) \sin ^2(c+d x)\right )\right )}{6720 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]*(a + b*Tan[c + d*x]^2)^2,x]

[Out]

(Csc[c + d*x]^3*(128*HypergeometricPFQ[{3/2, 2, 2, 2, 2}, {1, 1, 1, 9/2}, Sin[c + d*x]^2]*Sin[c + d*x]^6*(a +
(-a + b)*Sin[c + d*x]^2)^2 + 128*HypergeometricPFQ[{3/2, 2, 2, 2}, {1, 1, 9/2}, Sin[c + d*x]^2]*Sin[c + d*x]^6
*((a^2*Cos[c + d*x]^2*(9 + 5*Cos[2*(c + d*x)]))/2 + b*Sin[c + d*x]^2*(7*a + 5*a*Cos[2*(c + d*x)] + 5*b*Sin[c +
d*x]^2)) + 35*(-3375*a^2 + 3*a*(1969*a - 1750*b)*Sin[c + d*x]^2 + (-3161*a^2 + 5108*a*b - 1947*b^2)*Sin[c + d
*x]^4 + 485*(a - b)^2*Sin[c + d*x]^6 + (3*ArcTanh[Sqrt[Sin[c + d*x]^2]]*(1125*a^2 - 2*a*(1172*a - 875*b)*Sin[c
+ d*x]^2 + (1674*a^2 - 2286*a*b + 649*b^2)*Sin[c + d*x]^4 + (-400*a^2 + 778*a*b - 378*b^2)*Sin[c + d*x]^6 + 9
*(a - b)^2*Sin[c + d*x]^8))/Sqrt[Sin[c + d*x]^2])))/(6720*d)

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Maple [A]  time = 0.04, size = 178, normalized size = 1.9 \begin{align*}{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d}}-{\frac{3\,{b}^{2}\sin \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{ab \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{ab\sin \left ( dx+c \right ) }{d}}-{\frac{ab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(a+b*tan(d*x+c)^2)^2,x)

[Out]

1/4/d*b^2*sin(d*x+c)^5/cos(d*x+c)^4-1/8/d*b^2*sin(d*x+c)^5/cos(d*x+c)^2-1/8/d*b^2*sin(d*x+c)^3-3/8/d*b^2*sin(d
*x+c)+3/8/d*b^2*ln(sec(d*x+c)+tan(d*x+c))+1/d*a*b*sin(d*x+c)^3/cos(d*x+c)^2+1/d*a*b*sin(d*x+c)-1/d*a*b*ln(sec(
d*x+c)+tan(d*x+c))+1/d*a^2*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 1.04013, size = 161, normalized size = 1.68 \begin{align*} \frac{{\left (8 \, a^{2} - 8 \, a b + 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (8 \, a^{2} - 8 \, a b + 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left ({\left (8 \, a b - 5 \, b^{2}\right )} \sin \left (d x + c\right )^{3} -{\left (8 \, a b - 3 \, b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+b*tan(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

1/16*((8*a^2 - 8*a*b + 3*b^2)*log(sin(d*x + c) + 1) - (8*a^2 - 8*a*b + 3*b^2)*log(sin(d*x + c) - 1) - 2*((8*a*
b - 5*b^2)*sin(d*x + c)^3 - (8*a*b - 3*b^2)*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d

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Fricas [A]  time = 1.40442, size = 284, normalized size = 2.96 \begin{align*} \frac{{\left (8 \, a^{2} - 8 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (8 \, a^{2} - 8 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left ({\left (8 \, a b - 5 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, b^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+b*tan(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

1/16*((8*a^2 - 8*a*b + 3*b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (8*a^2 - 8*a*b + 3*b^2)*cos(d*x + c)^4*lo
g(-sin(d*x + c) + 1) + 2*((8*a*b - 5*b^2)*cos(d*x + c)^2 + 2*b^2)*sin(d*x + c))/(d*cos(d*x + c)^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan ^{2}{\left (c + d x \right )}\right )^{2} \sec{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+b*tan(d*x+c)**2)**2,x)

[Out]

Integral((a + b*tan(c + d*x)**2)**2*sec(c + d*x), x)

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Giac [A]  time = 1.82696, size = 162, normalized size = 1.69 \begin{align*} \frac{{\left (8 \, a^{2} - 8 \, a b + 3 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) -{\left (8 \, a^{2} - 8 \, a b + 3 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (8 \, a b \sin \left (d x + c\right )^{3} - 5 \, b^{2} \sin \left (d x + c\right )^{3} - 8 \, a b \sin \left (d x + c\right ) + 3 \, b^{2} \sin \left (d x + c\right )\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+b*tan(d*x+c)^2)^2,x, algorithm="giac")

[Out]

1/16*((8*a^2 - 8*a*b + 3*b^2)*log(abs(sin(d*x + c) + 1)) - (8*a^2 - 8*a*b + 3*b^2)*log(abs(sin(d*x + c) - 1))
- 2*(8*a*b*sin(d*x + c)^3 - 5*b^2*sin(d*x + c)^3 - 8*a*b*sin(d*x + c) + 3*b^2*sin(d*x + c))/(sin(d*x + c)^2 -
1)^2)/d