3.190 \(\int (a+b \sec (c+d x))^3 \sin ^6(c+d x) \, dx\)

Optimal. Leaf size=299 \[ -\frac{3 a^2 b \sin ^5(c+d x)}{5 d}-\frac{a^2 b \sin ^3(c+d x)}{d}-\frac{3 a^2 b \sin (c+d x)}{d}+\frac{3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a^3 \sin ^5(c+d x) \cos (c+d x)}{6 d}-\frac{5 a^3 \sin ^3(c+d x) \cos (c+d x)}{24 d}-\frac{5 a^3 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{5 a^3 x}{16}+\frac{45 a b^2 \tan (c+d x)}{8 d}-\frac{3 a b^2 \sin ^4(c+d x) \tan (c+d x)}{4 d}-\frac{15 a b^2 \sin ^2(c+d x) \tan (c+d x)}{8 d}-\frac{45}{8} a b^2 x+\frac{5 b^3 \sin ^3(c+d x)}{6 d}+\frac{5 b^3 \sin (c+d x)}{2 d}+\frac{b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{2 d}-\frac{5 b^3 \tanh ^{-1}(\sin (c+d x))}{2 d} \]

[Out]

(5*a^3*x)/16 - (45*a*b^2*x)/8 + (3*a^2*b*ArcTanh[Sin[c + d*x]])/d - (5*b^3*ArcTanh[Sin[c + d*x]])/(2*d) - (3*a
^2*b*Sin[c + d*x])/d + (5*b^3*Sin[c + d*x])/(2*d) - (5*a^3*Cos[c + d*x]*Sin[c + d*x])/(16*d) - (a^2*b*Sin[c +
d*x]^3)/d + (5*b^3*Sin[c + d*x]^3)/(6*d) - (5*a^3*Cos[c + d*x]*Sin[c + d*x]^3)/(24*d) - (3*a^2*b*Sin[c + d*x]^
5)/(5*d) - (a^3*Cos[c + d*x]*Sin[c + d*x]^5)/(6*d) + (45*a*b^2*Tan[c + d*x])/(8*d) - (15*a*b^2*Sin[c + d*x]^2*
Tan[c + d*x])/(8*d) - (3*a*b^2*Sin[c + d*x]^4*Tan[c + d*x])/(4*d) + (b^3*Sin[c + d*x]^3*Tan[c + d*x]^2)/(2*d)

________________________________________________________________________________________

Rubi [A]  time = 0.335932, antiderivative size = 299, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 11, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.524, Rules used = {3872, 2912, 2635, 8, 2592, 302, 206, 2591, 288, 321, 203} \[ -\frac{3 a^2 b \sin ^5(c+d x)}{5 d}-\frac{a^2 b \sin ^3(c+d x)}{d}-\frac{3 a^2 b \sin (c+d x)}{d}+\frac{3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a^3 \sin ^5(c+d x) \cos (c+d x)}{6 d}-\frac{5 a^3 \sin ^3(c+d x) \cos (c+d x)}{24 d}-\frac{5 a^3 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{5 a^3 x}{16}+\frac{45 a b^2 \tan (c+d x)}{8 d}-\frac{3 a b^2 \sin ^4(c+d x) \tan (c+d x)}{4 d}-\frac{15 a b^2 \sin ^2(c+d x) \tan (c+d x)}{8 d}-\frac{45}{8} a b^2 x+\frac{5 b^3 \sin ^3(c+d x)}{6 d}+\frac{5 b^3 \sin (c+d x)}{2 d}+\frac{b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{2 d}-\frac{5 b^3 \tanh ^{-1}(\sin (c+d x))}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^3*Sin[c + d*x]^6,x]

[Out]

(5*a^3*x)/16 - (45*a*b^2*x)/8 + (3*a^2*b*ArcTanh[Sin[c + d*x]])/d - (5*b^3*ArcTanh[Sin[c + d*x]])/(2*d) - (3*a
^2*b*Sin[c + d*x])/d + (5*b^3*Sin[c + d*x])/(2*d) - (5*a^3*Cos[c + d*x]*Sin[c + d*x])/(16*d) - (a^2*b*Sin[c +
d*x]^3)/d + (5*b^3*Sin[c + d*x]^3)/(6*d) - (5*a^3*Cos[c + d*x]*Sin[c + d*x]^3)/(24*d) - (3*a^2*b*Sin[c + d*x]^
5)/(5*d) - (a^3*Cos[c + d*x]*Sin[c + d*x]^5)/(6*d) + (45*a*b^2*Tan[c + d*x])/(8*d) - (15*a*b^2*Sin[c + d*x]^2*
Tan[c + d*x])/(8*d) - (3*a*b^2*Sin[c + d*x]^4*Tan[c + d*x])/(4*d) + (b^3*Sin[c + d*x]^3*Tan[c + d*x]^2)/(2*d)

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[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]

Rule 2912

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

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2592

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 302

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 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])

Rule 2591

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

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

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

Rule 203

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

Rubi steps

\begin{align*} \int (a+b \sec (c+d x))^3 \sin ^6(c+d x) \, dx &=-\int (-b-a \cos (c+d x))^3 \sin ^3(c+d x) \tan ^3(c+d x) \, dx\\ &=-\int \left (-a^3 \sin ^6(c+d x)-3 a^2 b \sin ^5(c+d x) \tan (c+d x)-3 a b^2 \sin ^4(c+d x) \tan ^2(c+d x)-b^3 \sin ^3(c+d x) \tan ^3(c+d x)\right ) \, dx\\ &=a^3 \int \sin ^6(c+d x) \, dx+\left (3 a^2 b\right ) \int \sin ^5(c+d x) \tan (c+d x) \, dx+\left (3 a b^2\right ) \int \sin ^4(c+d x) \tan ^2(c+d x) \, dx+b^3 \int \sin ^3(c+d x) \tan ^3(c+d x) \, dx\\ &=-\frac{a^3 \cos (c+d x) \sin ^5(c+d x)}{6 d}+\frac{1}{6} \left (5 a^3\right ) \int \sin ^4(c+d x) \, dx+\frac{\left (3 a^2 b\right ) \operatorname{Subst}\left (\int \frac{x^6}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}+\frac{\left (3 a b^2\right ) \operatorname{Subst}\left (\int \frac{x^6}{\left (1+x^2\right )^3} \, dx,x,\tan (c+d x)\right )}{d}+\frac{b^3 \operatorname{Subst}\left (\int \frac{x^6}{\left (1-x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac{5 a^3 \cos (c+d x) \sin ^3(c+d x)}{24 d}-\frac{a^3 \cos (c+d x) \sin ^5(c+d x)}{6 d}-\frac{3 a b^2 \sin ^4(c+d x) \tan (c+d x)}{4 d}+\frac{b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{2 d}+\frac{1}{8} \left (5 a^3\right ) \int \sin ^2(c+d x) \, dx+\frac{\left (3 a^2 b\right ) \operatorname{Subst}\left (\int \left (-1-x^2-x^4+\frac{1}{1-x^2}\right ) \, dx,x,\sin (c+d x)\right )}{d}+\frac{\left (15 a b^2\right ) \operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{4 d}-\frac{\left (5 b^3\right ) \operatorname{Subst}\left (\int \frac{x^4}{1-x^2} \, dx,x,\sin (c+d x)\right )}{2 d}\\ &=-\frac{3 a^2 b \sin (c+d x)}{d}-\frac{5 a^3 \cos (c+d x) \sin (c+d x)}{16 d}-\frac{a^2 b \sin ^3(c+d x)}{d}-\frac{5 a^3 \cos (c+d x) \sin ^3(c+d x)}{24 d}-\frac{3 a^2 b \sin ^5(c+d x)}{5 d}-\frac{a^3 \cos (c+d x) \sin ^5(c+d x)}{6 d}-\frac{15 a b^2 \sin ^2(c+d x) \tan (c+d x)}{8 d}-\frac{3 a b^2 \sin ^4(c+d x) \tan (c+d x)}{4 d}+\frac{b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{2 d}+\frac{1}{16} \left (5 a^3\right ) \int 1 \, dx+\frac{\left (3 a^2 b\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}+\frac{\left (45 a b^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{1+x^2} \, dx,x,\tan (c+d x)\right )}{8 d}-\frac{\left (5 b^3\right ) \operatorname{Subst}\left (\int \left (-1-x^2+\frac{1}{1-x^2}\right ) \, dx,x,\sin (c+d x)\right )}{2 d}\\ &=\frac{5 a^3 x}{16}+\frac{3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{3 a^2 b \sin (c+d x)}{d}+\frac{5 b^3 \sin (c+d x)}{2 d}-\frac{5 a^3 \cos (c+d x) \sin (c+d x)}{16 d}-\frac{a^2 b \sin ^3(c+d x)}{d}+\frac{5 b^3 \sin ^3(c+d x)}{6 d}-\frac{5 a^3 \cos (c+d x) \sin ^3(c+d x)}{24 d}-\frac{3 a^2 b \sin ^5(c+d x)}{5 d}-\frac{a^3 \cos (c+d x) \sin ^5(c+d x)}{6 d}+\frac{45 a b^2 \tan (c+d x)}{8 d}-\frac{15 a b^2 \sin ^2(c+d x) \tan (c+d x)}{8 d}-\frac{3 a b^2 \sin ^4(c+d x) \tan (c+d x)}{4 d}+\frac{b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{2 d}-\frac{\left (45 a b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{8 d}-\frac{\left (5 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{2 d}\\ &=\frac{5 a^3 x}{16}-\frac{45}{8} a b^2 x+\frac{3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{5 b^3 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{3 a^2 b \sin (c+d x)}{d}+\frac{5 b^3 \sin (c+d x)}{2 d}-\frac{5 a^3 \cos (c+d x) \sin (c+d x)}{16 d}-\frac{a^2 b \sin ^3(c+d x)}{d}+\frac{5 b^3 \sin ^3(c+d x)}{6 d}-\frac{5 a^3 \cos (c+d x) \sin ^3(c+d x)}{24 d}-\frac{3 a^2 b \sin ^5(c+d x)}{5 d}-\frac{a^3 \cos (c+d x) \sin ^5(c+d x)}{6 d}+\frac{45 a b^2 \tan (c+d x)}{8 d}-\frac{15 a b^2 \sin ^2(c+d x) \tan (c+d x)}{8 d}-\frac{3 a b^2 \sin ^4(c+d x) \tan (c+d x)}{4 d}+\frac{b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{2 d}\\ \end{align*}

Mathematica [B]  time = 6.24289, size = 818, normalized size = 2.74 \[ \frac{\left (5 b^3-6 a^2 b\right ) \cos ^3(c+d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \sec (c+d x))^3}{2 d (b+a \cos (c+d x))^3}+\frac{\left (6 a^2 b-5 b^3\right ) \cos ^3(c+d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \sec (c+d x))^3}{2 d (b+a \cos (c+d x))^3}+\frac{3 a b^2 \cos ^3(c+d x) \sin \left (\frac{1}{2} (c+d x)\right ) (a+b \sec (c+d x))^3}{d (b+a \cos (c+d x))^3 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{3 b \left (6 b^2-11 a^2\right ) \cos ^3(c+d x) \sin (c+d x) (a+b \sec (c+d x))^3}{8 d (b+a \cos (c+d x))^3}-\frac{3 a \left (5 a^2-32 b^2\right ) \cos ^3(c+d x) \sin (2 (c+d x)) (a+b \sec (c+d x))^3}{64 d (b+a \cos (c+d x))^3}-\frac{b \left (4 b^2-21 a^2\right ) \cos ^3(c+d x) \sin (3 (c+d x)) (a+b \sec (c+d x))^3}{48 d (b+a \cos (c+d x))^3}+\frac{3 a \left (a^2-2 b^2\right ) \cos ^3(c+d x) \sin (4 (c+d x)) (a+b \sec (c+d x))^3}{64 d (b+a \cos (c+d x))^3}-\frac{3 a^2 b \cos ^3(c+d x) \sin (5 (c+d x)) (a+b \sec (c+d x))^3}{80 d (b+a \cos (c+d x))^3}-\frac{a^3 \cos ^3(c+d x) \sin (6 (c+d x)) (a+b \sec (c+d x))^3}{192 d (b+a \cos (c+d x))^3}+\frac{3 a b^2 \cos ^3(c+d x) \sin \left (\frac{1}{2} (c+d x)\right ) (a+b \sec (c+d x))^3}{d (b+a \cos (c+d x))^3 \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{b^3 \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d (b+a \cos (c+d x))^3 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{b^3 \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d (b+a \cos (c+d x))^3 \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{5 a \left (a^2-18 b^2\right ) (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{16 d (b+a \cos (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x])^3*Sin[c + d*x]^6,x]

[Out]

(5*a*(a^2 - 18*b^2)*(c + d*x)*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3)/(16*d*(b + a*Cos[c + d*x])^3) + ((-6*a^2*
b + 5*b^3)*Cos[c + d*x]^3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a + b*Sec[c + d*x])^3)/(2*d*(b + a*Cos[c +
d*x])^3) + ((6*a^2*b - 5*b^3)*Cos[c + d*x]^3*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*(a + b*Sec[c + d*x])^3)
/(2*d*(b + a*Cos[c + d*x])^3) + (b^3*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3)/(4*d*(b + a*Cos[c + d*x])^3*(Cos[(
c + d*x)/2] - Sin[(c + d*x)/2])^2) + (3*a*b^2*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3*Sin[(c + d*x)/2])/(d*(b +
a*Cos[c + d*x])^3*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - (b^3*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3)/(4*d*(b
+ a*Cos[c + d*x])^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (3*a*b^2*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3
*Sin[(c + d*x)/2])/(d*(b + a*Cos[c + d*x])^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (3*b*(-11*a^2 + 6*b^2)*C
os[c + d*x]^3*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(8*d*(b + a*Cos[c + d*x])^3) - (3*a*(5*a^2 - 32*b^2)*Cos[c
+ d*x]^3*(a + b*Sec[c + d*x])^3*Sin[2*(c + d*x)])/(64*d*(b + a*Cos[c + d*x])^3) - (b*(-21*a^2 + 4*b^2)*Cos[c +
d*x]^3*(a + b*Sec[c + d*x])^3*Sin[3*(c + d*x)])/(48*d*(b + a*Cos[c + d*x])^3) + (3*a*(a^2 - 2*b^2)*Cos[c + d*
x]^3*(a + b*Sec[c + d*x])^3*Sin[4*(c + d*x)])/(64*d*(b + a*Cos[c + d*x])^3) - (3*a^2*b*Cos[c + d*x]^3*(a + b*S
ec[c + d*x])^3*Sin[5*(c + d*x)])/(80*d*(b + a*Cos[c + d*x])^3) - (a^3*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3*Si
n[6*(c + d*x)])/(192*d*(b + a*Cos[c + d*x])^3)

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Maple [A]  time = 0.051, size = 354, normalized size = 1.2 \begin{align*} -{\frac{{a}^{3}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{6\,d}}-{\frac{5\,{a}^{3}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{24\,d}}-{\frac{5\,{a}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{16\,d}}+{\frac{5\,{a}^{3}x}{16}}+{\frac{5\,{a}^{3}c}{16\,d}}-{\frac{3\,{a}^{2}b \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{5\,d}}-{\frac{{a}^{2}b \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{d}}-3\,{\frac{{a}^{2}b\sin \left ( dx+c \right ) }{d}}+3\,{\frac{{a}^{2}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{d\cos \left ( dx+c \right ) }}+3\,{\frac{a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}\cos \left ( dx+c \right ) }{d}}+{\frac{15\,\cos \left ( dx+c \right ) a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{45\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) a{b}^{2}}{8\,d}}-{\frac{45\,a{b}^{2}x}{8}}-{\frac{45\,a{b}^{2}c}{8\,d}}+{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{2\,d}}+{\frac{5\,{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{6\,d}}+{\frac{5\,{b}^{3}\sin \left ( dx+c \right ) }{2\,d}}-{\frac{5\,{b}^{3}\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((a+b*sec(d*x+c))^3*sin(d*x+c)^6,x)

[Out]

-1/6*a^3*cos(d*x+c)*sin(d*x+c)^5/d-5/24*a^3*cos(d*x+c)*sin(d*x+c)^3/d-5/16*a^3*cos(d*x+c)*sin(d*x+c)/d+5/16*a^
3*x+5/16/d*a^3*c-3/5*a^2*b*sin(d*x+c)^5/d-a^2*b*sin(d*x+c)^3/d-3*a^2*b*sin(d*x+c)/d+3/d*a^2*b*ln(sec(d*x+c)+ta
n(d*x+c))+3/d*a*b^2*sin(d*x+c)^7/cos(d*x+c)+3/d*a*b^2*sin(d*x+c)^5*cos(d*x+c)+15/4/d*a*b^2*cos(d*x+c)*sin(d*x+
c)^3+45/8/d*cos(d*x+c)*sin(d*x+c)*a*b^2-45/8*a*b^2*x-45/8/d*a*b^2*c+1/2/d*b^3*sin(d*x+c)^7/cos(d*x+c)^2+1/2/d*
b^3*sin(d*x+c)^5+5/6*b^3*sin(d*x+c)^3/d+5/2*b^3*sin(d*x+c)/d-5/2/d*b^3*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 1.53197, size = 327, normalized size = 1.09 \begin{align*} \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^{3} - 96 \,{\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^{2} b - 360 \,{\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 )} a b^{2} + 80 \,{\left (4 \, \sin \left (d x + c\right )^{3} - \frac{6 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 24 \, \sin \left (d x + c\right )\right )} b^{3}}{960 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^3*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^3 - 96*(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^2*b - 3
60*(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))*a*b^2 + 80*(4*sin(d*x + c)^3 - 6*sin(d*x + c)/(sin(d*x + c)^2 - 1) - 15*log(sin(d*x + c) + 1) + 15*log(sin
(d*x + c) - 1) + 24*sin(d*x + c))*b^3)/d

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Fricas [A]  time = 2.01143, size = 605, normalized size = 2.02 \begin{align*} \frac{75 \,{\left (a^{3} - 18 \, a b^{2}\right )} d x \cos \left (d x + c\right )^{2} + 60 \,{\left (6 \, a^{2} b - 5 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 60 \,{\left (6 \, a^{2} b - 5 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) -{\left (40 \, a^{3} \cos \left (d x + c\right )^{7} + 144 \, a^{2} b \cos \left (d x + c\right )^{6} - 10 \,{\left (13 \, a^{3} - 18 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} - 16 \,{\left (33 \, a^{2} b - 5 \, b^{3}\right )} \cos \left (d x + c\right )^{4} - 720 \, a b^{2} \cos \left (d x + c\right ) + 15 \,{\left (11 \, a^{3} - 54 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} - 120 \, b^{3} + 16 \,{\left (69 \, a^{2} b - 35 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^3*sin(d*x+c)^6,x, algorithm="fricas")

[Out]

1/240*(75*(a^3 - 18*a*b^2)*d*x*cos(d*x + c)^2 + 60*(6*a^2*b - 5*b^3)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - 60
*(6*a^2*b - 5*b^3)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) - (40*a^3*cos(d*x + c)^7 + 144*a^2*b*cos(d*x + c)^6 -
10*(13*a^3 - 18*a*b^2)*cos(d*x + c)^5 - 16*(33*a^2*b - 5*b^3)*cos(d*x + c)^4 - 720*a*b^2*cos(d*x + c) + 15*(1
1*a^3 - 54*a*b^2)*cos(d*x + c)^3 - 120*b^3 + 16*(69*a^2*b - 35*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x +
c)^2)

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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((a+b*sec(d*x+c))**3*sin(d*x+c)**6,x)

[Out]

Timed out

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Giac [B]  time = 1.54781, size = 760, normalized size = 2.54 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^3*sin(d*x+c)^6,x, algorithm="giac")

[Out]

1/240*(75*(a^3 - 18*a*b^2)*(d*x + c) + 120*(6*a^2*b - 5*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 120*(6*a^2*b
- 5*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 240*(6*a*b^2*tan(1/2*d*x + 1/2*c)^3 - b^3*tan(1/2*d*x + 1/2*c)^
3 - 6*a*b^2*tan(1/2*d*x + 1/2*c) - b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2 + 2*(75*a^3*tan(1/
2*d*x + 1/2*c)^11 - 720*a^2*b*tan(1/2*d*x + 1/2*c)^11 - 630*a*b^2*tan(1/2*d*x + 1/2*c)^11 + 480*b^3*tan(1/2*d*
x + 1/2*c)^11 + 425*a^3*tan(1/2*d*x + 1/2*c)^9 - 4560*a^2*b*tan(1/2*d*x + 1/2*c)^9 - 2610*a*b^2*tan(1/2*d*x +
1/2*c)^9 + 2720*b^3*tan(1/2*d*x + 1/2*c)^9 + 990*a^3*tan(1/2*d*x + 1/2*c)^7 - 12384*a^2*b*tan(1/2*d*x + 1/2*c)
^7 - 1980*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 5760*b^3*tan(1/2*d*x + 1/2*c)^7 - 990*a^3*tan(1/2*d*x + 1/2*c)^5 - 12
384*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 1980*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 5760*b^3*tan(1/2*d*x + 1/2*c)^5 - 425*a
^3*tan(1/2*d*x + 1/2*c)^3 - 4560*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 2610*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 2720*b^3*t
an(1/2*d*x + 1/2*c)^3 - 75*a^3*tan(1/2*d*x + 1/2*c) - 720*a^2*b*tan(1/2*d*x + 1/2*c) + 630*a*b^2*tan(1/2*d*x +
1/2*c) + 480*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^6)/d