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

Optimal. Leaf size=138 \[ -\frac {5 a^3 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {b \left (6 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {1}{2} a x \left (a^2-6 b^2\right )-\frac {15 a^2 b \sin (c+d x)}{2 d}+\frac {3 a \tan (c+d x) (a \cos (c+d x)+b)^2}{2 d}+\frac {\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^3}{2 d} \]

[Out]

1/2*a*(a^2-6*b^2)*x+1/2*b*(6*a^2-b^2)*arctanh(sin(d*x+c))/d-15/2*a^2*b*sin(d*x+c)/d-5/2*a^3*cos(d*x+c)*sin(d*x
+c)/d+3/2*a*(b+a*cos(d*x+c))^2*tan(d*x+c)/d+1/2*(b+a*cos(d*x+c))^3*sec(d*x+c)*tan(d*x+c)/d

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Rubi [A]  time = 0.50, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3872, 2889, 3048, 3047, 3033, 3023, 2735, 3770} \[ \frac {b \left (6 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {1}{2} a x \left (a^2-6 b^2\right )-\frac {15 a^2 b \sin (c+d x)}{2 d}-\frac {5 a^3 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {3 a \tan (c+d x) (a \cos (c+d x)+b)^2}{2 d}+\frac {\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^3}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(a^2 - 6*b^2)*x)/2 + (b*(6*a^2 - b^2)*ArcTanh[Sin[c + d*x]])/(2*d) - (15*a^2*b*Sin[c + d*x])/(2*d) - (5*a^3
*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (3*a*(b + a*Cos[c + d*x])^2*Tan[c + d*x])/(2*d) + ((b + a*Cos[c + d*x])^3*
Sec[c + d*x]*Tan[c + d*x])/(2*d)

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2889

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^m*(1 - Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f,
m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n])

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 3033

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

Rule 3047

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C - B*c*d + A*d^2)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(
c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c
*C - B*d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*
d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)
))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2,
0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3048

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + A*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m
- 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n +
2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(m + n + 2) + C*(c^2*(
m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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]

Rubi steps

\begin {align*} \int (a+b \sec (c+d x))^3 \sin ^2(c+d x) \, dx &=-\int (-b-a \cos (c+d x))^3 \sec (c+d x) \tan ^2(c+d x) \, dx\\ &=-\int (-b-a \cos (c+d x))^3 \left (1-\cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {(b+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}-\frac {1}{2} \int (-b-a \cos (c+d x))^2 \left (-3 a+b \cos (c+d x)+4 a \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {3 a (b+a \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac {(b+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}-\frac {1}{2} \int (-b-a \cos (c+d x)) \left (6 a^2-b^2-5 a b \cos (c+d x)-10 a^2 \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {5 a^3 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {3 a (b+a \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac {(b+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}-\frac {1}{4} \int \left (-2 b \left (6 a^2-b^2\right )-2 a \left (a^2-6 b^2\right ) \cos (c+d x)+30 a^2 b \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {15 a^2 b \sin (c+d x)}{2 d}-\frac {5 a^3 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {3 a (b+a \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac {(b+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}-\frac {1}{4} \int \left (-2 b \left (6 a^2-b^2\right )-2 a \left (a^2-6 b^2\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac {1}{2} a \left (a^2-6 b^2\right ) x-\frac {15 a^2 b \sin (c+d x)}{2 d}-\frac {5 a^3 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {3 a (b+a \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac {(b+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \left (b \left (6 a^2-b^2\right )\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} a \left (a^2-6 b^2\right ) x+\frac {b \left (6 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {15 a^2 b \sin (c+d x)}{2 d}-\frac {5 a^3 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {3 a (b+a \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac {(b+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end {align*}

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Mathematica [B]  time = 0.97, size = 327, normalized size = 2.37 \[ \frac {\sec ^2(c+d x) \left (-\frac {1}{2} a^3 \sin (2 (c+d x))-\frac {1}{4} a^3 \sin (4 (c+d x))+a^3 c+a^3 d x+\left (2 b^3-3 a^2 b\right ) \sin (c+d x)+\cos (2 (c+d x)) \left (\left (b^3-6 a^2 b\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+a \left (a^2-6 b^2\right ) (c+d x)-b \left (b^2-6 a^2\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-3 a^2 b \sin (3 (c+d x))-6 a^2 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 a^2 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+6 a b^2 \sin (2 (c+d x))-6 a b^2 c-6 a b^2 d x+b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-b^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^2*(a^3*c - 6*a*b^2*c + a^3*d*x - 6*a*b^2*d*x - 6*a^2*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]
+ b^3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 6*a^2*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - b^3*Log[Co
s[(c + d*x)/2] + Sin[(c + d*x)/2]] + Cos[2*(c + d*x)]*(a*(a^2 - 6*b^2)*(c + d*x) + (-6*a^2*b + b^3)*Log[Cos[(c
+ d*x)/2] - Sin[(c + d*x)/2]] - b*(-6*a^2 + b^2)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + (-3*a^2*b + 2*b^
3)*Sin[c + d*x] - (a^3*Sin[2*(c + d*x)])/2 + 6*a*b^2*Sin[2*(c + d*x)] - 3*a^2*b*Sin[3*(c + d*x)] - (a^3*Sin[4*
(c + d*x)])/4))/(4*d)

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fricas [A]  time = 0.51, size = 151, normalized size = 1.09 \[ \frac {2 \, {\left (a^{3} - 6 \, a b^{2}\right )} d x \cos \left (d x + c\right )^{2} + {\left (6 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (6 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{3} \cos \left (d x + c\right )^{3} + 6 \, a^{2} b \cos \left (d x + c\right )^{2} - 6 \, a b^{2} \cos \left (d x + c\right ) - b^{3}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*(a^3 - 6*a*b^2)*d*x*cos(d*x + c)^2 + (6*a^2*b - b^3)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (6*a^2*b -
b^3)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) - 2*(a^3*cos(d*x + c)^3 + 6*a^2*b*cos(d*x + c)^2 - 6*a*b^2*cos(d*x
+ c) - b^3)*sin(d*x + c))/(d*cos(d*x + c)^2)

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giac [B]  time = 0.38, size = 346, normalized size = 2.51 \[ \frac {{\left (a^{3} - 6 \, a b^{2}\right )} {\left (d x + c\right )} + {\left (6 \, a^{2} b - b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (6 \, a^{2} b - b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 6 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a^3 - 6*a*b^2)*(d*x + c) + (6*a^2*b - b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (6*a^2*b - b^3)*log(abs(
tan(1/2*d*x + 1/2*c) - 1)) + 2*(a^3*tan(1/2*d*x + 1/2*c)^7 - 6*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 6*a*b^2*tan(1/2*
d*x + 1/2*c)^7 + b^3*tan(1/2*d*x + 1/2*c)^7 - 3*a^3*tan(1/2*d*x + 1/2*c)^5 + 6*a^2*b*tan(1/2*d*x + 1/2*c)^5 -
6*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 3*b^3*tan(1/2*d*x + 1/2*c)^5 + 3*a^3*tan(1/2*d*x + 1/2*c)^3 + 6*a^2*b*tan(1/2
*d*x + 1/2*c)^3 + 6*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 3*b^3*tan(1/2*d*x + 1/2*c)^3 - a^3*tan(1/2*d*x + 1/2*c) - 6
*a^2*b*tan(1/2*d*x + 1/2*c) + 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)^4
- 1)^2)/d

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maple [A]  time = 0.42, size = 167, normalized size = 1.21 \[ -\frac {a^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {a^{3} x}{2}+\frac {a^{3} c}{2 d}-\frac {3 a^{2} b \sin \left (d x +c \right )}{d}+\frac {3 a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}-3 a \,b^{2} x +\frac {3 a \,b^{2} \tan \left (d x +c \right )}{d}-\frac {3 a \,b^{2} c}{d}+\frac {b^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}+\frac {b^{3} \sin \left (d x +c \right )}{2 d}-\frac {b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.04, size = 129, normalized size = 0.93 \[ \frac {{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 12 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a b^{2} - b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, a^{2} b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((2*d*x + 2*c - sin(2*d*x + 2*c))*a^3 - 12*(d*x + c - tan(d*x + c))*a*b^2 - b^3*(2*sin(d*x + c)/(sin(d*x +
c)^2 - 1) + log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 6*a^2*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c
) - 1) - 2*sin(d*x + c)))/d

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mupad [B]  time = 1.27, size = 202, normalized size = 1.46 \[ \frac {a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {b^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {b^3\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}-\frac {a^3\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d}-\frac {3\,a^2\,b\,\sin \left (c+d\,x\right )}{d}-\frac {6\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {6\,a^2\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,a\,b^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^2*(a + b/cos(c + d*x))^3,x)

[Out]

(a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d - (b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (
b^3*sin(c + d*x))/(2*d*cos(c + d*x)^2) - (a^3*cos(c + d*x)*sin(c + d*x))/(2*d) - (3*a^2*b*sin(c + d*x))/d - (6
*a*b^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (6*a^2*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))
/d + (3*a*b^2*sin(c + d*x))/(d*cos(c + d*x))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sec {\left (c + d x \right )}\right )^{3} \sin ^{2}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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