3.2.79 \(\int (a+b \sec (c+d x))^2 \sin ^6(c+d x) \, dx\) [179]

Optimal. Leaf size=175 \[ \frac {5}{16} \left (a^2-6 b^2\right ) x+\frac {2 a b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {2 a b \sin (c+d x)}{d}-\frac {\left (11 a^2-18 b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {\left (13 a^2-6 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {2 a b \sin ^3(c+d x)}{3 d}-\frac {2 a b \sin ^5(c+d x)}{5 d}+\frac {b^2 \tan (c+d x)}{d} \]

[Out]

5/16*(a^2-6*b^2)*x+2*a*b*arctanh(sin(d*x+c))/d-2*a*b*sin(d*x+c)/d-1/16*(11*a^2-18*b^2)*cos(d*x+c)*sin(d*x+c)/d
+1/24*(13*a^2-6*b^2)*cos(d*x+c)^3*sin(d*x+c)/d-1/6*a^2*cos(d*x+c)^5*sin(d*x+c)/d-2/3*a*b*sin(d*x+c)^3/d-2/5*a*
b*sin(d*x+c)^5/d+b^2*tan(d*x+c)/d

________________________________________________________________________________________

Rubi [A]
time = 0.34, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3957, 2990, 2672, 308, 212, 466, 1828, 1171, 396, 209} \begin {gather*} \frac {\left (13 a^2-6 b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{24 d}-\frac {\left (11 a^2-18 b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {5}{16} x \left (a^2-6 b^2\right )-\frac {a^2 \sin (c+d x) \cos ^5(c+d x)}{6 d}-\frac {2 a b \sin ^5(c+d x)}{5 d}-\frac {2 a b \sin ^3(c+d x)}{3 d}-\frac {2 a b \sin (c+d x)}{d}+\frac {2 a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b^2 \tan (c+d x)}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^2*Sin[c + d*x]^6,x]

[Out]

(5*(a^2 - 6*b^2)*x)/16 + (2*a*b*ArcTanh[Sin[c + d*x]])/d - (2*a*b*Sin[c + d*x])/d - ((11*a^2 - 18*b^2)*Cos[c +
d*x]*Sin[c + d*x])/(16*d) + ((13*a^2 - 6*b^2)*Cos[c + d*x]^3*Sin[c + d*x])/(24*d) - (a^2*Cos[c + d*x]^5*Sin[c
+ d*x])/(6*d) - (2*a*b*Sin[c + d*x]^3)/(3*d) - (2*a*b*Sin[c + d*x]^5)/(5*d) + (b^2*Tan[c + d*x])/d

Rule 209

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

Rule 212

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

Rule 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 396

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

Rule 466

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x
*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1171

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1
)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] &&
NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 1828

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*
g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rule 2672

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 2990

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^2, x_Symbol] :> Dist[2*a*(b/d), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n + 1), x], x] + Int[(g*Cos[e
+ f*x])^p*(d*Sin[e + f*x])^n*(a^2 + b^2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 -
b^2, 0]

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

\begin {align*} \int (a+b \sec (c+d x))^2 \sin ^6(c+d x) \, dx &=\int (-b-a \cos (c+d x))^2 \sin ^4(c+d x) \tan ^2(c+d x) \, dx\\ &=(2 a b) \int \sin ^5(c+d x) \tan (c+d x) \, dx+\int \left (b^2+a^2 \cos ^2(c+d x)\right ) \sin ^4(c+d x) \tan ^2(c+d x) \, dx\\ &=\frac {\text {Subst}\left (\int \frac {x^6 \left (a^2+b^2+b^2 x^2\right )}{\left (1+x^2\right )^4} \, dx,x,\tan (c+d x)\right )}{d}+\frac {(2 a b) \text {Subst}\left (\int \frac {x^6}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {\text {Subst}\left (\int \frac {-a^2+6 a^2 x^2-6 a^2 x^4-6 b^2 x^6}{\left (1+x^2\right )^3} \, dx,x,\tan (c+d x)\right )}{6 d}+\frac {(2 a b) \text {Subst}\left (\int \left (-1-x^2-x^4+\frac {1}{1-x^2}\right ) \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac {2 a b \sin (c+d x)}{d}+\frac {\left (13 a^2-6 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {2 a b \sin ^3(c+d x)}{3 d}-\frac {2 a b \sin ^5(c+d x)}{5 d}+\frac {\text {Subst}\left (\int \frac {-3 \left (3 a^2-2 b^2\right )+24 \left (a^2-b^2\right ) x^2+24 b^2 x^4}{\left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{24 d}+\frac {(2 a b) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {2 a b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {2 a b \sin (c+d x)}{d}-\frac {\left (11 a^2-18 b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {\left (13 a^2-6 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {2 a b \sin ^3(c+d x)}{3 d}-\frac {2 a b \sin ^5(c+d x)}{5 d}-\frac {\text {Subst}\left (\int \frac {-3 \left (5 a^2-14 b^2\right )-48 b^2 x^2}{1+x^2} \, dx,x,\tan (c+d x)\right )}{48 d}\\ &=\frac {2 a b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {2 a b \sin (c+d x)}{d}-\frac {\left (11 a^2-18 b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {\left (13 a^2-6 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {2 a b \sin ^3(c+d x)}{3 d}-\frac {2 a b \sin ^5(c+d x)}{5 d}+\frac {b^2 \tan (c+d x)}{d}+\frac {\left (5 \left (a^2-6 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{16 d}\\ &=\frac {5}{16} \left (a^2-6 b^2\right ) x+\frac {2 a b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {2 a b \sin (c+d x)}{d}-\frac {\left (11 a^2-18 b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {\left (13 a^2-6 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {2 a b \sin ^3(c+d x)}{3 d}-\frac {2 a b \sin ^5(c+d x)}{5 d}+\frac {b^2 \tan (c+d x)}{d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 1.09, size = 193, normalized size = 1.10 \begin {gather*} \frac {60 \left (5 \left (a^2-6 b^2\right ) (c+d x)-32 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+32 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-2128 a b \sin (c+d x)+\left (-185 a^2+1410 b^2-5 \left (29 a^2-84 b^2\right ) \cos (2 (c+d x))+232 a b \cos (3 (c+d x))+35 a^2 \cos (4 (c+d x))-30 b^2 \cos (4 (c+d x))-24 a b \cos (5 (c+d x))-5 a^2 \cos (6 (c+d x))\right ) \tan (c+d x)}{960 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x])^2*Sin[c + d*x]^6,x]

[Out]

(60*(5*(a^2 - 6*b^2)*(c + d*x) - 32*a*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 32*a*b*Log[Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2]]) - 2128*a*b*Sin[c + d*x] + (-185*a^2 + 1410*b^2 - 5*(29*a^2 - 84*b^2)*Cos[2*(c + d*x)] +
232*a*b*Cos[3*(c + d*x)] + 35*a^2*Cos[4*(c + d*x)] - 30*b^2*Cos[4*(c + d*x)] - 24*a*b*Cos[5*(c + d*x)] - 5*a^2
*Cos[6*(c + d*x)])*Tan[c + d*x])/(960*d)

________________________________________________________________________________________

Maple [A]
time = 0.12, size = 163, normalized size = 0.93

method	result	size
derivativedivides	\(\frac {b^{2} \left (\frac {\sin ^{7}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )+2 b a \left (-\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a^{2} \left (-\frac {\left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}\)	\(163\)
default	\(\frac {b^{2} \left (\frac {\sin ^{7}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )+2 b a \left (-\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a^{2} \left (-\frac {\left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}\)	\(163\)
risch	\(\frac {5 a^{2} x}{16}-\frac {15 b^{2} x}{8}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} b^{2}}{4 d}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} b^{2}}{4 d}+\frac {2 i b^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {15 i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{128 d}-\frac {11 i b a \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {11 i b a \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {15 i a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{128 d}+\frac {2 b a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {2 b a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {a^{2} \sin \left (6 d x +6 c \right )}{192 d}-\frac {b a \sin \left (5 d x +5 c \right )}{40 d}+\frac {3 a^{2} \sin \left (4 d x +4 c \right )}{64 d}-\frac {\sin \left (4 d x +4 c \right ) b^{2}}{32 d}+\frac {7 b a \sin \left (3 d x +3 c \right )}{24 d}\)	\(265\)
norman	\(\frac {\left (-\frac {5 a^{2}}{16}+\frac {15 b^{2}}{8}\right ) x +\left (-\frac {45 a^{2}}{16}+\frac {135 b^{2}}{8}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {25 a^{2}}{16}+\frac {75 b^{2}}{8}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {25 a^{2}}{16}+\frac {75 b^{2}}{8}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {5 a^{2}}{16}-\frac {15 b^{2}}{8}\right ) x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {25 a^{2}}{16}-\frac {75 b^{2}}{8}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {25 a^{2}}{16}-\frac {75 b^{2}}{8}\right ) x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {45 a^{2}}{16}-\frac {135 b^{2}}{8}\right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {\left (33 a^{2}+58 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {\left (5 a^{2}-32 b a -30 b^{2}\right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {\left (5 a^{2}+32 b a -30 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {\left (35 a^{2}-256 b a -210 b^{2}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {\left (35 a^{2}+256 b a -210 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {\left (565 a^{2}-5216 b a -3390 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{120 d}+\frac {\left (565 a^{2}+5216 b a -3390 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{120 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}-\frac {2 b a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {2 b a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\)	\(464\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^2*sin(d*x+c)^6,x,method=_RETURNVERBOSE)

[Out]

1/d*(b^2*(sin(d*x+c)^7/cos(d*x+c)+(sin(d*x+c)^5+5/4*sin(d*x+c)^3+15/8*sin(d*x+c))*cos(d*x+c)-15/8*d*x-15/8*c)+
2*b*a*(-1/5*sin(d*x+c)^5-1/3*sin(d*x+c)^3-sin(d*x+c)+ln(sec(d*x+c)+tan(d*x+c)))+a^2*(-1/6*(sin(d*x+c)^5+5/4*si
n(d*x+c)^3+15/8*sin(d*x+c))*cos(d*x+c)+5/16*d*x+5/16*c))

________________________________________________________________________________________

Maxima [A]
time = 0.50, size = 173, normalized size = 0.99 \begin {gather*} \frac {5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 60 \, d x + 60 \, c + 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - 64 \, {\left (6 \, \sin \left (d x + c\right )^{5} + 10 \, \sin \left (d x + c\right )^{3} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 30 \, \sin \left (d x + c\right )\right )} a b - 120 \, {\left (15 \, d x + 15 \, c - \frac {9 \, \tan \left (d x + c\right )^{3} + 7 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1} - 8 \, \tan \left (d x + c\right )\right )} b^{2}}{960 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^2*sin(d*x+c)^6,x, algorithm="maxima")

[Out]

1/960*(5*(4*sin(2*d*x + 2*c)^3 + 60*d*x + 60*c + 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c))*a^2 - 64*(6*sin(d*x
+ c)^5 + 10*sin(d*x + c)^3 - 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1) + 30*sin(d*x + c))*a*b - 120
*(15*d*x + 15*c - (9*tan(d*x + c)^3 + 7*tan(d*x + c))/(tan(d*x + c)^4 + 2*tan(d*x + c)^2 + 1) - 8*tan(d*x + c)
)*b^2)/d

________________________________________________________________________________________

Fricas [A]
time = 4.02, size = 176, normalized size = 1.01 \begin {gather*} \frac {75 \, {\left (a^{2} - 6 \, b^{2}\right )} d x \cos \left (d x + c\right ) + 240 \, a b \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 240 \, a b \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (40 \, a^{2} \cos \left (d x + c\right )^{6} + 96 \, a b \cos \left (d x + c\right )^{5} - 352 \, a b \cos \left (d x + c\right )^{3} - 10 \, {\left (13 \, a^{2} - 6 \, b^{2}\right )} \cos \left (d x + c\right )^{4} + 736 \, a b \cos \left (d x + c\right ) + 15 \, {\left (11 \, a^{2} - 18 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 240 \, b^{2}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^2*sin(d*x+c)^6,x, algorithm="fricas")

[Out]

1/240*(75*(a^2 - 6*b^2)*d*x*cos(d*x + c) + 240*a*b*cos(d*x + c)*log(sin(d*x + c) + 1) - 240*a*b*cos(d*x + c)*l
og(-sin(d*x + c) + 1) - (40*a^2*cos(d*x + c)^6 + 96*a*b*cos(d*x + c)^5 - 352*a*b*cos(d*x + c)^3 - 10*(13*a^2 -
6*b^2)*cos(d*x + c)^4 + 736*a*b*cos(d*x + c) + 15*(11*a^2 - 18*b^2)*cos(d*x + c)^2 - 240*b^2)*sin(d*x + c))/(
d*cos(d*x + c))

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))**2*sin(d*x+c)**6,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 379 vs. \(2 (163) = 326\).
time = 0.50, size = 379, normalized size = 2.17 \begin {gather*} \frac {480 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 480 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 75 \, {\left (a^{2} - 6 \, b^{2}\right )} {\left (d x + c\right )} - \frac {480 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \frac {2 \, {\left (75 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 480 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 210 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 425 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 3040 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 870 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 990 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 8256 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 660 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 990 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 8256 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 660 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 425 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3040 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 870 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 75 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 480 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 210 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6}}}{240 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^2*sin(d*x+c)^6,x, algorithm="giac")

[Out]

1/240*(480*a*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 480*a*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 75*(a^2 - 6*b
^2)*(d*x + c) - 480*b^2*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) + 2*(75*a^2*tan(1/2*d*x + 1/2*c)^11
- 480*a*b*tan(1/2*d*x + 1/2*c)^11 - 210*b^2*tan(1/2*d*x + 1/2*c)^11 + 425*a^2*tan(1/2*d*x + 1/2*c)^9 - 3040*a*
b*tan(1/2*d*x + 1/2*c)^9 - 870*b^2*tan(1/2*d*x + 1/2*c)^9 + 990*a^2*tan(1/2*d*x + 1/2*c)^7 - 8256*a*b*tan(1/2*
d*x + 1/2*c)^7 - 660*b^2*tan(1/2*d*x + 1/2*c)^7 - 990*a^2*tan(1/2*d*x + 1/2*c)^5 - 8256*a*b*tan(1/2*d*x + 1/2*
c)^5 + 660*b^2*tan(1/2*d*x + 1/2*c)^5 - 425*a^2*tan(1/2*d*x + 1/2*c)^3 - 3040*a*b*tan(1/2*d*x + 1/2*c)^3 + 870
*b^2*tan(1/2*d*x + 1/2*c)^3 - 75*a^2*tan(1/2*d*x + 1/2*c) - 480*a*b*tan(1/2*d*x + 1/2*c) + 210*b^2*tan(1/2*d*x
+ 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^6)/d

________________________________________________________________________________________

Mupad [B]
time = 3.10, size = 231, normalized size = 1.32 \begin {gather*} \frac {\frac {5\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,a^2}{8}+4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,a\,b-\frac {15\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,b^2}{4}}{d}-\frac {\frac {15\,a^2\,\sin \left (c+d\,x\right )}{128}-\frac {5\,b^2\,\sin \left (c+d\,x\right )}{4}+\frac {3\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{32}-\frac {a^2\,\sin \left (5\,c+5\,d\,x\right )}{48}+\frac {a^2\,\sin \left (7\,c+7\,d\,x\right )}{384}-\frac {15\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{64}+\frac {b^2\,\sin \left (5\,c+5\,d\,x\right )}{64}+\frac {59\,a\,b\,\sin \left (2\,c+2\,d\,x\right )}{48}-\frac {2\,a\,b\,\sin \left (4\,c+4\,d\,x\right )}{15}+\frac {a\,b\,\sin \left (6\,c+6\,d\,x\right )}{80}}{d\,\cos \left (c+d\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^6*(a + b/cos(c + d*x))^2,x)

[Out]

((5*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/8 - (15*b^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/
4 + 4*a*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d - ((15*a^2*sin(c + d*x))/128 - (5*b^2*sin(c + d*x))/
4 + (3*a^2*sin(3*c + 3*d*x))/32 - (a^2*sin(5*c + 5*d*x))/48 + (a^2*sin(7*c + 7*d*x))/384 - (15*b^2*sin(3*c + 3
*d*x))/64 + (b^2*sin(5*c + 5*d*x))/64 + (59*a*b*sin(2*c + 2*d*x))/48 - (2*a*b*sin(4*c + 4*d*x))/15 + (a*b*sin(
6*c + 6*d*x))/80)/(d*cos(c + d*x))

________________________________________________________________________________________