3.437 \(\int \sec ^3(c+d x) (a+b \tan ^2(c+d x))^2 \, dx\)

Optimal. Leaf size=128 \[ \frac{\left (8 a^2-4 a b+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{\left (8 a^2-4 a b+b^2\right ) \tan (c+d x) \sec (c+d x)}{16 d}+\frac{b (8 a-3 b) \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac{b \tan (c+d x) \sec ^5(c+d x) \left (a-(a-b) \sin ^2(c+d x)\right )}{6 d} \]

[Out]

((8*a^2 - 4*a*b + b^2)*ArcTanh[Sin[c + d*x]])/(16*d) + ((8*a^2 - 4*a*b + b^2)*Sec[c + d*x]*Tan[c + d*x])/(16*d
) + ((8*a - 3*b)*b*Sec[c + d*x]^3*Tan[c + d*x])/(24*d) + (b*Sec[c + d*x]^5*(a - (a - b)*Sin[c + d*x]^2)*Tan[c
+ d*x])/(6*d)

________________________________________________________________________________________

Rubi [A]  time = 0.164569, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3676, 413, 385, 199, 206} \[ \frac{\left (8 a^2-4 a b+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{\left (8 a^2-4 a b+b^2\right ) \tan (c+d x) \sec (c+d x)}{16 d}+\frac{b (8 a-3 b) \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac{b \tan (c+d x) \sec ^5(c+d x) \left (a-(a-b) \sin ^2(c+d x)\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^2 - 4*a*b + b^2)*ArcTanh[Sin[c + d*x]])/(16*d) + ((8*a^2 - 4*a*b + b^2)*Sec[c + d*x]*Tan[c + d*x])/(16*d
) + ((8*a - 3*b)*b*Sec[c + d*x]^3*Tan[c + d*x])/(24*d) + (b*Sec[c + d*x]^5*(a - (a - b)*Sin[c + d*x]^2)*Tan[c
+ d*x])/(6*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 199

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

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 ^3(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 )^4} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac{b \sec ^5(c+d x) \left (a-(a-b) \sin ^2(c+d x)\right ) \tan (c+d x)}{6 d}-\frac{\operatorname{Subst}\left (\int \frac{-a (6 a-b)+3 (a-b) (2 a-b) x^2}{\left (1-x^2\right )^3} \, dx,x,\sin (c+d x)\right )}{6 d}\\ &=\frac{(8 a-3 b) b \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac{b \sec ^5(c+d x) \left (a-(a-b) \sin ^2(c+d x)\right ) \tan (c+d x)}{6 d}+\frac{\left (8 a^2-4 a b+b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{8 d}\\ &=\frac{\left (8 a^2-4 a b+b^2\right ) \sec (c+d x) \tan (c+d x)}{16 d}+\frac{(8 a-3 b) b \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac{b \sec ^5(c+d x) \left (a-(a-b) \sin ^2(c+d x)\right ) \tan (c+d x)}{6 d}+\frac{\left (8 a^2-4 a b+b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{16 d}\\ &=\frac{\left (8 a^2-4 a b+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{\left (8 a^2-4 a b+b^2\right ) \sec (c+d x) \tan (c+d x)}{16 d}+\frac{(8 a-3 b) b \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac{b \sec ^5(c+d x) \left (a-(a-b) \sin ^2(c+d x)\right ) \tan (c+d x)}{6 d}\\ \end{align*}

Mathematica [C]  time = 10.8173, size = 875, normalized size = 6.84 \[ \frac{\sin (c+d x) \left (380 (a-b)^2 \text{HypergeometricPFQ}\left (\left \{\frac{3}{2},2,2,2\right \},\left \{1,1,\frac{9}{2}\right \},\sin ^2(c+d x)\right ) \sqrt{\sin ^2(c+d x)} \sin ^{10}(c+d x)+128 (a-b)^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 ) \sqrt{\sin ^2(c+d x)} \sin ^{10}(c+d x)+16 (a-b)^2 \text{HypergeometricPFQ}\left (\left \{\frac{3}{2},2,2,2,2,2\right \},\left \{1,1,1,1,\frac{9}{2}\right \},\sin ^2(c+d x)\right ) \sqrt{\sin ^2(c+d x)} \sin ^{10}(c+d x)+525 (a-b)^2 \tanh ^{-1}\left (\sqrt{\sin ^2(c+d x)}\right ) \sin ^8(c+d x)-968 a (a-b) \text{HypergeometricPFQ}\left (\left \{\frac{3}{2},2,2,2\right \},\left \{1,1,\frac{9}{2}\right \},\sin ^2(c+d x)\right ) \sqrt{\sin ^2(c+d x)} \sin ^8(c+d x)-288 a (a-b) \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 ) \sqrt{\sin ^2(c+d x)} \sin ^8(c+d x)-32 a (a-b) \text{HypergeometricPFQ}\left (\left \{\frac{3}{2},2,2,2,2,2\right \},\left \{1,1,1,1,\frac{9}{2}\right \},\sin ^2(c+d x)\right ) \sqrt{\sin ^2(c+d x)} \sin ^8(c+d x)-19845 (a-b)^2 \tanh ^{-1}\left (\sqrt{\sin ^2(c+d x)}\right ) \sin ^6(c+d x)-1365 a (a-b) \tanh ^{-1}\left (\sqrt{\sin ^2(c+d x)}\right ) \sin ^6(c+d x)+8855 (a-b)^2 \sqrt{\sin ^2(c+d x)} \sin ^6(c+d x)+620 a^2 \text{HypergeometricPFQ}\left (\left \{\frac{3}{2},2,2,2\right \},\left \{1,1,\frac{9}{2}\right \},\sin ^2(c+d x)\right ) \sqrt{\sin ^2(c+d x)} \sin ^6(c+d x)+160 a^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 ) \sqrt{\sin ^2(c+d x)} \sin ^6(c+d x)+16 a^2 \text{HypergeometricPFQ}\left (\left \{\frac{3}{2},2,2,2,2,2\right \},\left \{1,1,1,1,\frac{9}{2}\right \},\sin ^2(c+d x)\right ) \sqrt{\sin ^2(c+d x)} \sin ^6(c+d x)+1680 a^2 \tanh ^{-1}\left (\sqrt{\sin ^2(c+d x)}\right ) \sin ^4(c+d x)+32970 (a-b)^2 \tanh ^{-1}\left (\sqrt{\sin ^2(c+d x)}\right ) \sin ^4(c+d x)+54180 a (a-b) \tanh ^{-1}\left (\sqrt{\sin ^2(c+d x)}\right ) \sin ^4(c+d x)-32970 (a-b)^2 \sqrt{\sin ^2(c+d x)} \sin ^4(c+d x)-23555 a (a-b) \sqrt{\sin ^2(c+d x)} \sin ^4(c+d x)-36855 a^2 \tanh ^{-1}\left (\sqrt{\sin ^2(c+d x)}\right ) \sin ^2(c+d x)-91875 a (a-b) \tanh ^{-1}\left (\sqrt{\sin ^2(c+d x)}\right ) \sin ^2(c+d x)+14980 a^2 \sin ^2(c+d x)^{3/2}+91875 a (a-b) \sin ^2(c+d x)^{3/2}+65625 a^2 \tanh ^{-1}\left (\sqrt{\sin ^2(c+d x)}\right )-65625 a^2 \sqrt{\sin ^2(c+d x)}\right )}{2520 d \sin ^2(c+d x)^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sin[c + d*x]*(65625*a^2*ArcTanh[Sqrt[Sin[c + d*x]^2]] - 36855*a^2*ArcTanh[Sqrt[Sin[c + d*x]^2]]*Sin[c + d*x]^
2 - 91875*a*(a - b)*ArcTanh[Sqrt[Sin[c + d*x]^2]]*Sin[c + d*x]^2 + 1680*a^2*ArcTanh[Sqrt[Sin[c + d*x]^2]]*Sin[
c + d*x]^4 + 54180*a*(a - b)*ArcTanh[Sqrt[Sin[c + d*x]^2]]*Sin[c + d*x]^4 + 32970*(a - b)^2*ArcTanh[Sqrt[Sin[c
+ d*x]^2]]*Sin[c + d*x]^4 - 1365*a*(a - b)*ArcTanh[Sqrt[Sin[c + d*x]^2]]*Sin[c + d*x]^6 - 19845*(a - b)^2*Arc
Tanh[Sqrt[Sin[c + d*x]^2]]*Sin[c + d*x]^6 + 525*(a - b)^2*ArcTanh[Sqrt[Sin[c + d*x]^2]]*Sin[c + d*x]^8 - 65625
*a^2*Sqrt[Sin[c + d*x]^2] - 23555*a*(a - b)*Sin[c + d*x]^4*Sqrt[Sin[c + d*x]^2] - 32970*(a - b)^2*Sin[c + d*x]
^4*Sqrt[Sin[c + d*x]^2] + 8855*(a - b)^2*Sin[c + d*x]^6*Sqrt[Sin[c + d*x]^2] + 620*a^2*HypergeometricPFQ[{3/2,
2, 2, 2}, {1, 1, 9/2}, Sin[c + d*x]^2]*Sin[c + d*x]^6*Sqrt[Sin[c + d*x]^2] + 160*a^2*HypergeometricPFQ[{3/2,
2, 2, 2, 2}, {1, 1, 1, 9/2}, Sin[c + d*x]^2]*Sin[c + d*x]^6*Sqrt[Sin[c + d*x]^2] + 16*a^2*HypergeometricPFQ[{3
/2, 2, 2, 2, 2, 2}, {1, 1, 1, 1, 9/2}, Sin[c + d*x]^2]*Sin[c + d*x]^6*Sqrt[Sin[c + d*x]^2] - 968*a*(a - b)*Hyp
ergeometricPFQ[{3/2, 2, 2, 2}, {1, 1, 9/2}, Sin[c + d*x]^2]*Sin[c + d*x]^8*Sqrt[Sin[c + d*x]^2] - 288*a*(a - b
)*HypergeometricPFQ[{3/2, 2, 2, 2, 2}, {1, 1, 1, 9/2}, Sin[c + d*x]^2]*Sin[c + d*x]^8*Sqrt[Sin[c + d*x]^2] - 3
2*a*(a - b)*HypergeometricPFQ[{3/2, 2, 2, 2, 2, 2}, {1, 1, 1, 1, 9/2}, Sin[c + d*x]^2]*Sin[c + d*x]^8*Sqrt[Sin
[c + d*x]^2] + 380*(a - b)^2*HypergeometricPFQ[{3/2, 2, 2, 2}, {1, 1, 9/2}, Sin[c + d*x]^2]*Sin[c + d*x]^10*Sq
rt[Sin[c + d*x]^2] + 128*(a - b)^2*HypergeometricPFQ[{3/2, 2, 2, 2, 2}, {1, 1, 1, 9/2}, Sin[c + d*x]^2]*Sin[c
+ d*x]^10*Sqrt[Sin[c + d*x]^2] + 16*(a - b)^2*HypergeometricPFQ[{3/2, 2, 2, 2, 2, 2}, {1, 1, 1, 1, 9/2}, Sin[c
+ d*x]^2]*Sin[c + d*x]^10*Sqrt[Sin[c + d*x]^2] + 14980*a^2*(Sin[c + d*x]^2)^(3/2) + 91875*a*(a - b)*(Sin[c +
d*x]^2)^(3/2)))/(2520*d*(Sin[c + d*x]^2)^(5/2))

________________________________________________________________________________________

Maple [B]  time = 0.066, size = 248, normalized size = 1.9 \begin{align*}{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{6\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{24\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{48\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{48\,d}}-{\frac{{b}^{2}\sin \left ( dx+c \right ) }{16\,d}}+{\frac{{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{16\,d}}+{\frac{ab \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{ab \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{ab\sin \left ( dx+c \right ) }{4\,d}}-{\frac{ab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{{a}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+b*tan(d*x+c)^2)^2,x)

[Out]

1/6/d*b^2*sin(d*x+c)^5/cos(d*x+c)^6+1/24/d*b^2*sin(d*x+c)^5/cos(d*x+c)^4-1/48/d*b^2*sin(d*x+c)^5/cos(d*x+c)^2-
1/48/d*b^2*sin(d*x+c)^3-1/16/d*b^2*sin(d*x+c)+1/16/d*b^2*ln(sec(d*x+c)+tan(d*x+c))+1/2/d*a*b*sin(d*x+c)^3/cos(
d*x+c)^4+1/4/d*a*b*sin(d*x+c)^3/cos(d*x+c)^2+1/4/d*a*b*sin(d*x+c)-1/4/d*a*b*ln(sec(d*x+c)+tan(d*x+c))+1/2/d*a^
2*sec(d*x+c)*tan(d*x+c)+1/2/d*a^2*ln(sec(d*x+c)+tan(d*x+c))

________________________________________________________________________________________

Maxima [A]  time = 1.12479, size = 211, normalized size = 1.65 \begin{align*} \frac{3 \,{\left (8 \, a^{2} - 4 \, a b + b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (8 \, a^{2} - 4 \, a b + b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left (3 \,{\left (8 \, a^{2} - 4 \, a b + b^{2}\right )} \sin \left (d x + c\right )^{5} - 8 \,{\left (6 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{3} + 3 \,{\left (8 \, a^{2} + 4 \, a b - b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1}}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(3*(8*a^2 - 4*a*b + b^2)*log(sin(d*x + c) + 1) - 3*(8*a^2 - 4*a*b + b^2)*log(sin(d*x + c) - 1) - 2*(3*(8*
a^2 - 4*a*b + b^2)*sin(d*x + c)^5 - 8*(6*a^2 - b^2)*sin(d*x + c)^3 + 3*(8*a^2 + 4*a*b - b^2)*sin(d*x + c))/(si
n(d*x + c)^6 - 3*sin(d*x + c)^4 + 3*sin(d*x + c)^2 - 1))/d

________________________________________________________________________________________

Fricas [A]  time = 1.77664, size = 343, normalized size = 2.68 \begin{align*} \frac{3 \,{\left (8 \, a^{2} - 4 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (8 \, a^{2} - 4 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (3 \,{\left (8 \, a^{2} - 4 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (12 \, a b - 7 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \, b^{2}\right )} \sin \left (d x + c\right )}{96 \, d \cos \left (d x + c\right )^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(3*(8*a^2 - 4*a*b + b^2)*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 3*(8*a^2 - 4*a*b + b^2)*cos(d*x + c)^6*lo
g(-sin(d*x + c) + 1) + 2*(3*(8*a^2 - 4*a*b + b^2)*cos(d*x + c)^4 + 2*(12*a*b - 7*b^2)*cos(d*x + c)^2 + 8*b^2)*
sin(d*x + c))/(d*cos(d*x + c)^6)

________________________________________________________________________________________

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 ^{3}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(a+b*tan(d*x+c)**2)**2,x)

[Out]

Integral((a + b*tan(c + d*x)**2)**2*sec(c + d*x)**3, x)

________________________________________________________________________________________

Giac [A]  time = 1.86341, size = 225, normalized size = 1.76 \begin{align*} \frac{3 \,{\left (8 \, a^{2} - 4 \, a b + b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 3 \,{\left (8 \, a^{2} - 4 \, a b + b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (24 \, a^{2} \sin \left (d x + c\right )^{5} - 12 \, a b \sin \left (d x + c\right )^{5} + 3 \, b^{2} \sin \left (d x + c\right )^{5} - 48 \, a^{2} \sin \left (d x + c\right )^{3} + 8 \, b^{2} \sin \left (d x + c\right )^{3} + 24 \, a^{2} \sin \left (d x + c\right ) + 12 \, 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 )}^{3}}}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(3*(8*a^2 - 4*a*b + b^2)*log(abs(sin(d*x + c) + 1)) - 3*(8*a^2 - 4*a*b + b^2)*log(abs(sin(d*x + c) - 1))
- 2*(24*a^2*sin(d*x + c)^5 - 12*a*b*sin(d*x + c)^5 + 3*b^2*sin(d*x + c)^5 - 48*a^2*sin(d*x + c)^3 + 8*b^2*sin(
d*x + c)^3 + 24*a^2*sin(d*x + c) + 12*a*b*sin(d*x + c) - 3*b^2*sin(d*x + c))/(sin(d*x + c)^2 - 1)^3)/d