\(\int \sin ^3(c+d x) (a+b \tan (c+d x))^4 \, dx\) [41]

Optimal result

Integrand size = 21, antiderivative size = 275 \[ \int \sin ^3(c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {4 a^3 b \text {arctanh}(\sin (c+d x))}{d}-\frac {10 a b^3 \text {arctanh}(\sin (c+d x))}{d}-\frac {a^4 \cos (c+d x)}{d}+\frac {12 a^2 b^2 \cos (c+d x)}{d}-\frac {3 b^4 \cos (c+d x)}{d}+\frac {a^4 \cos ^3(c+d x)}{3 d}-\frac {2 a^2 b^2 \cos ^3(c+d x)}{d}+\frac {b^4 \cos ^3(c+d x)}{3 d}+\frac {6 a^2 b^2 \sec (c+d x)}{d}-\frac {3 b^4 \sec (c+d x)}{d}+\frac {b^4 \sec ^3(c+d x)}{3 d}-\frac {4 a^3 b \sin (c+d x)}{d}+\frac {10 a b^3 \sin (c+d x)}{d}-\frac {4 a^3 b \sin ^3(c+d x)}{3 d}+\frac {10 a b^3 \sin ^3(c+d x)}{3 d}+\frac {2 a b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{d} \]

[Out]

4*a^3*b*arctanh(sin(d*x+c))/d-10*a*b^3*arctanh(sin(d*x+c))/d-a^4*cos(d*x+c)/d+12*a^2*b^2*cos(d*x+c)/d-3*b^4*co
s(d*x+c)/d+1/3*a^4*cos(d*x+c)^3/d-2*a^2*b^2*cos(d*x+c)^3/d+1/3*b^4*cos(d*x+c)^3/d+6*a^2*b^2*sec(d*x+c)/d-3*b^4
*sec(d*x+c)/d+1/3*b^4*sec(d*x+c)^3/d-4*a^3*b*sin(d*x+c)/d+10*a*b^3*sin(d*x+c)/d-4/3*a^3*b*sin(d*x+c)^3/d+10/3*
a*b^3*sin(d*x+c)^3/d+2*a*b^3*sin(d*x+c)^3*tan(d*x+c)^2/d

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3598, 2713, 2672, 308, 212, 2670, 276, 294} \[ \int \sin ^3(c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {a^4 \cos ^3(c+d x)}{3 d}-\frac {a^4 \cos (c+d x)}{d}+\frac {4 a^3 b \text {arctanh}(\sin (c+d x))}{d}-\frac {4 a^3 b \sin ^3(c+d x)}{3 d}-\frac {4 a^3 b \sin (c+d x)}{d}-\frac {2 a^2 b^2 \cos ^3(c+d x)}{d}+\frac {12 a^2 b^2 \cos (c+d x)}{d}+\frac {6 a^2 b^2 \sec (c+d x)}{d}-\frac {10 a b^3 \text {arctanh}(\sin (c+d x))}{d}+\frac {10 a b^3 \sin ^3(c+d x)}{3 d}+\frac {10 a b^3 \sin (c+d x)}{d}+\frac {2 a b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{d}+\frac {b^4 \cos ^3(c+d x)}{3 d}-\frac {3 b^4 \cos (c+d x)}{d}+\frac {b^4 \sec ^3(c+d x)}{3 d}-\frac {3 b^4 \sec (c+d x)}{d} \]

[In]

Int[Sin[c + d*x]^3*(a + b*Tan[c + d*x])^4,x]

[Out]

(4*a^3*b*ArcTanh[Sin[c + d*x]])/d - (10*a*b^3*ArcTanh[Sin[c + d*x]])/d - (a^4*Cos[c + d*x])/d + (12*a^2*b^2*Co
s[c + d*x])/d - (3*b^4*Cos[c + d*x])/d + (a^4*Cos[c + d*x]^3)/(3*d) - (2*a^2*b^2*Cos[c + d*x]^3)/d + (b^4*Cos[
c + d*x]^3)/(3*d) + (6*a^2*b^2*Sec[c + d*x])/d - (3*b^4*Sec[c + d*x])/d + (b^4*Sec[c + d*x]^3)/(3*d) - (4*a^3*
b*Sin[c + d*x])/d + (10*a*b^3*Sin[c + d*x])/d - (4*a^3*b*Sin[c + d*x]^3)/(3*d) + (10*a*b^3*Sin[c + d*x]^3)/(3*
d) + (2*a*b^3*Sin[c + d*x]^3*Tan[c + d*x]^2)/d

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 276

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

Rule 294

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 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 2670

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

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 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 3598

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

Rubi steps \begin{align*} \text {integral}& = \int \left (a^4 \sin ^3(c+d x)+4 a^3 b \sin ^3(c+d x) \tan (c+d x)+6 a^2 b^2 \sin ^3(c+d x) \tan ^2(c+d x)+4 a b^3 \sin ^3(c+d x) \tan ^3(c+d x)+b^4 \sin ^3(c+d x) \tan ^4(c+d x)\right ) \, dx \\ & = a^4 \int \sin ^3(c+d x) \, dx+\left (4 a^3 b\right ) \int \sin ^3(c+d x) \tan (c+d x) \, dx+\left (6 a^2 b^2\right ) \int \sin ^3(c+d x) \tan ^2(c+d x) \, dx+\left (4 a b^3\right ) \int \sin ^3(c+d x) \tan ^3(c+d x) \, dx+b^4 \int \sin ^3(c+d x) \tan ^4(c+d x) \, dx \\ & = -\frac {a^4 \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac {\left (4 a^3 b\right ) \text {Subst}\left (\int \frac {x^4}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}-\frac {\left (6 a^2 b^2\right ) \text {Subst}\left (\int \frac {\left (1-x^2\right )^2}{x^2} \, dx,x,\cos (c+d x)\right )}{d}+\frac {\left (4 a b^3\right ) \text {Subst}\left (\int \frac {x^6}{\left (1-x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{d}-\frac {b^4 \text {Subst}\left (\int \frac {\left (1-x^2\right )^3}{x^4} \, dx,x,\cos (c+d x)\right )}{d} \\ & = -\frac {a^4 \cos (c+d x)}{d}+\frac {a^4 \cos ^3(c+d x)}{3 d}+\frac {2 a b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{d}+\frac {\left (4 a^3 b\right ) \text {Subst}\left (\int \left (-1-x^2+\frac {1}{1-x^2}\right ) \, dx,x,\sin (c+d x)\right )}{d}-\frac {\left (6 a^2 b^2\right ) \text {Subst}\left (\int \left (-2+\frac {1}{x^2}+x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac {\left (10 a b^3\right ) \text {Subst}\left (\int \frac {x^4}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}-\frac {b^4 \text {Subst}\left (\int \left (3+\frac {1}{x^4}-\frac {3}{x^2}-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d} \\ & = -\frac {a^4 \cos (c+d x)}{d}+\frac {12 a^2 b^2 \cos (c+d x)}{d}-\frac {3 b^4 \cos (c+d x)}{d}+\frac {a^4 \cos ^3(c+d x)}{3 d}-\frac {2 a^2 b^2 \cos ^3(c+d x)}{d}+\frac {b^4 \cos ^3(c+d x)}{3 d}+\frac {6 a^2 b^2 \sec (c+d x)}{d}-\frac {3 b^4 \sec (c+d x)}{d}+\frac {b^4 \sec ^3(c+d x)}{3 d}-\frac {4 a^3 b \sin (c+d x)}{d}-\frac {4 a^3 b \sin ^3(c+d x)}{3 d}+\frac {2 a b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{d}+\frac {\left (4 a^3 b\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}-\frac {\left (10 a b^3\right ) \text {Subst}\left (\int \left (-1-x^2+\frac {1}{1-x^2}\right ) \, dx,x,\sin (c+d x)\right )}{d} \\ & = \frac {4 a^3 b \text {arctanh}(\sin (c+d x))}{d}-\frac {a^4 \cos (c+d x)}{d}+\frac {12 a^2 b^2 \cos (c+d x)}{d}-\frac {3 b^4 \cos (c+d x)}{d}+\frac {a^4 \cos ^3(c+d x)}{3 d}-\frac {2 a^2 b^2 \cos ^3(c+d x)}{d}+\frac {b^4 \cos ^3(c+d x)}{3 d}+\frac {6 a^2 b^2 \sec (c+d x)}{d}-\frac {3 b^4 \sec (c+d x)}{d}+\frac {b^4 \sec ^3(c+d x)}{3 d}-\frac {4 a^3 b \sin (c+d x)}{d}+\frac {10 a b^3 \sin (c+d x)}{d}-\frac {4 a^3 b \sin ^3(c+d x)}{3 d}+\frac {10 a b^3 \sin ^3(c+d x)}{3 d}+\frac {2 a b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{d}-\frac {\left (10 a b^3\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d} \\ & = \frac {4 a^3 b \text {arctanh}(\sin (c+d x))}{d}-\frac {10 a b^3 \text {arctanh}(\sin (c+d x))}{d}-\frac {a^4 \cos (c+d x)}{d}+\frac {12 a^2 b^2 \cos (c+d x)}{d}-\frac {3 b^4 \cos (c+d x)}{d}+\frac {a^4 \cos ^3(c+d x)}{3 d}-\frac {2 a^2 b^2 \cos ^3(c+d x)}{d}+\frac {b^4 \cos ^3(c+d x)}{3 d}+\frac {6 a^2 b^2 \sec (c+d x)}{d}-\frac {3 b^4 \sec (c+d x)}{d}+\frac {b^4 \sec ^3(c+d x)}{3 d}-\frac {4 a^3 b \sin (c+d x)}{d}+\frac {10 a b^3 \sin (c+d x)}{d}-\frac {4 a^3 b \sin ^3(c+d x)}{3 d}+\frac {10 a b^3 \sin ^3(c+d x)}{3 d}+\frac {2 a b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(1017\) vs. \(2(275)=550\).

Time = 7.87 (sec) , antiderivative size = 1017, normalized size of antiderivative = 3.70 \[ \int \sin ^3(c+d x) (a+b \tan (c+d x))^4 \, dx=-\frac {b^2 \left (-36 a^2+17 b^2\right ) \cos ^4(c+d x) (a+b \tan (c+d x))^4}{6 d (a \cos (c+d x)+b \sin (c+d x))^4}-\frac {\left (3 a^4-42 a^2 b^2+11 b^4\right ) \cos ^5(c+d x) (a+b \tan (c+d x))^4}{4 d (a \cos (c+d x)+b \sin (c+d x))^4}+\frac {\left (a^4-6 a^2 b^2+b^4\right ) \cos ^4(c+d x) \cos (3 (c+d x)) (a+b \tan (c+d x))^4}{12 d (a \cos (c+d x)+b \sin (c+d x))^4}-\frac {2 \left (2 a^3 b-5 a b^3\right ) \cos ^4(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 \tan (c+d x))^4}{d (a \cos (c+d x)+b \sin (c+d x))^4}+\frac {2 \left (2 a^3 b-5 a b^3\right ) \cos ^4(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 \tan (c+d x))^4}{d (a \cos (c+d x)+b \sin (c+d x))^4}+\frac {\left (12 a b^3+b^4\right ) \cos ^4(c+d x) (a+b \tan (c+d x))^4}{12 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2 (a \cos (c+d x)+b \sin (c+d x))^4}+\frac {b^4 \cos ^4(c+d x) \sin \left (\frac {1}{2} (c+d x)\right ) (a+b \tan (c+d x))^4}{6 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 (a \cos (c+d x)+b \sin (c+d x))^4}-\frac {b^4 \cos ^4(c+d x) \sin \left (\frac {1}{2} (c+d x)\right ) (a+b \tan (c+d x))^4}{6 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 (a \cos (c+d x)+b \sin (c+d x))^4}+\frac {\left (-12 a b^3+b^4\right ) \cos ^4(c+d x) (a+b \tan (c+d x))^4}{12 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2 (a \cos (c+d x)+b \sin (c+d x))^4}+\frac {\cos ^4(c+d x) \left (36 a^2 b^2 \sin \left (\frac {1}{2} (c+d x)\right )-17 b^4 \sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^4}{6 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^4}+\frac {\cos ^4(c+d x) \left (-36 a^2 b^2 \sin \left (\frac {1}{2} (c+d x)\right )+17 b^4 \sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^4}{6 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^4}-\frac {a b \left (5 a^2-9 b^2\right ) \cos ^4(c+d x) \sin (c+d x) (a+b \tan (c+d x))^4}{d (a \cos (c+d x)+b \sin (c+d x))^4}+\frac {a b \left (a^2-b^2\right ) \cos ^4(c+d x) \sin (3 (c+d x)) (a+b \tan (c+d x))^4}{3 d (a \cos (c+d x)+b \sin (c+d x))^4} \]

[In]

Integrate[Sin[c + d*x]^3*(a + b*Tan[c + d*x])^4,x]

[Out]

-1/6*(b^2*(-36*a^2 + 17*b^2)*Cos[c + d*x]^4*(a + b*Tan[c + d*x])^4)/(d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) -
((3*a^4 - 42*a^2*b^2 + 11*b^4)*Cos[c + d*x]^5*(a + b*Tan[c + d*x])^4)/(4*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4
) + ((a^4 - 6*a^2*b^2 + b^4)*Cos[c + d*x]^4*Cos[3*(c + d*x)]*(a + b*Tan[c + d*x])^4)/(12*d*(a*Cos[c + d*x] + b
*Sin[c + d*x])^4) - (2*(2*a^3*b - 5*a*b^3)*Cos[c + d*x]^4*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a + b*Tan[
c + d*x])^4)/(d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + (2*(2*a^3*b - 5*a*b^3)*Cos[c + d*x]^4*Log[Cos[(c + d*x)
/2] + Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^4)/(d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + ((12*a*b^3 + b^4)*Co
s[c + d*x]^4*(a + b*Tan[c + d*x])^4)/(12*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2*(a*Cos[c + d*x] + b*Sin[c +
d*x])^4) + (b^4*Cos[c + d*x]^4*Sin[(c + d*x)/2]*(a + b*Tan[c + d*x])^4)/(6*d*(Cos[(c + d*x)/2] - Sin[(c + d*x
)/2])^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) - (b^4*Cos[c + d*x]^4*Sin[(c + d*x)/2]*(a + b*Tan[c + d*x])^4)/(6
*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + ((-12*a*b^3 + b^4)*Cos[c + d
*x]^4*(a + b*Tan[c + d*x])^4)/(12*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^
4) + (Cos[c + d*x]^4*(36*a^2*b^2*Sin[(c + d*x)/2] - 17*b^4*Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^4)/(6*d*(Cos
[(c + d*x)/2] - Sin[(c + d*x)/2])*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + (Cos[c + d*x]^4*(-36*a^2*b^2*Sin[(c +
d*x)/2] + 17*b^4*Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^4)/(6*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(a*Cos[
c + d*x] + b*Sin[c + d*x])^4) - (a*b*(5*a^2 - 9*b^2)*Cos[c + d*x]^4*Sin[c + d*x]*(a + b*Tan[c + d*x])^4)/(d*(a
*Cos[c + d*x] + b*Sin[c + d*x])^4) + (a*b*(a^2 - b^2)*Cos[c + d*x]^4*Sin[3*(c + d*x)]*(a + b*Tan[c + d*x])^4)/
(3*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4)

Maple [A] (verified)

Time = 4.71 (sec) , antiderivative size = 267, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(\frac {-\frac {a^{4} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}+4 a^{3} b \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+6 a^{2} b^{2} \left (\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+4 a \,b^{3} \left (\frac {\sin ^{7}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{2}+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{6}+\frac {5 \sin \left (d x +c \right )}{2}-\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+b^{4} \left (\frac {\sin ^{8}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {5 \left (\sin ^{8}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )}-\frac {5 \left (\frac {16}{5}+\sin ^{6}\left (d x +c \right )+\frac {6 \left (\sin ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\sin ^{2}\left (d x +c \right )\right )}{5}\right ) \cos \left (d x +c \right )}{3}\right )}{d}\)	\(267\)
default	\(\frac {-\frac {a^{4} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}+4 a^{3} b \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+6 a^{2} b^{2} \left (\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+4 a \,b^{3} \left (\frac {\sin ^{7}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{2}+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{6}+\frac {5 \sin \left (d x +c \right )}{2}-\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+b^{4} \left (\frac {\sin ^{8}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {5 \left (\sin ^{8}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )}-\frac {5 \left (\frac {16}{5}+\sin ^{6}\left (d x +c \right )+\frac {6 \left (\sin ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\sin ^{2}\left (d x +c \right )\right )}{5}\right ) \cos \left (d x +c \right )}{3}\right )}{d}\)	\(267\)
risch	\(-\frac {3 \,{\mathrm e}^{i \left (d x +c \right )} a^{4}}{8 d}-\frac {11 \,{\mathrm e}^{i \left (d x +c \right )} b^{4}}{8 d}+\frac {{\mathrm e}^{-3 i \left (d x +c \right )} a^{4}}{24 d}+\frac {{\mathrm e}^{-3 i \left (d x +c \right )} b^{4}}{24 d}-\frac {3 \,{\mathrm e}^{-i \left (d x +c \right )} a^{4}}{8 d}-\frac {11 \,{\mathrm e}^{-i \left (d x +c \right )} b^{4}}{8 d}+\frac {{\mathrm e}^{3 i \left (d x +c \right )} a^{4}}{24 d}+\frac {{\mathrm e}^{3 i \left (d x +c \right )} b^{4}}{24 d}+\frac {4 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {10 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {4 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {10 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {21 \,{\mathrm e}^{i \left (d x +c \right )} a^{2} b^{2}}{4 d}+\frac {21 \,{\mathrm e}^{-i \left (d x +c \right )} a^{2} b^{2}}{4 d}-\frac {{\mathrm e}^{-3 i \left (d x +c \right )} a^{2} b^{2}}{4 d}-\frac {{\mathrm e}^{3 i \left (d x +c \right )} a^{2} b^{2}}{4 d}+\frac {5 i {\mathrm e}^{i \left (d x +c \right )} a^{3} b}{2 d}-\frac {9 i {\mathrm e}^{i \left (d x +c \right )} a \,b^{3}}{2 d}-\frac {2 b^{2} {\mathrm e}^{i \left (d x +c \right )} \left (-18 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+9 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+6 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}-36 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+14 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-18 a^{2}+9 b^{2}-6 i a b \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {5 i {\mathrm e}^{-i \left (d x +c \right )} a^{3} b}{2 d}+\frac {9 i {\mathrm e}^{-i \left (d x +c \right )} a \,b^{3}}{2 d}+\frac {i {\mathrm e}^{-3 i \left (d x +c \right )} a^{3} b}{6 d}-\frac {i {\mathrm e}^{-3 i \left (d x +c \right )} a \,b^{3}}{6 d}-\frac {i {\mathrm e}^{3 i \left (d x +c \right )} a^{3} b}{6 d}+\frac {i {\mathrm e}^{3 i \left (d x +c \right )} a \,b^{3}}{6 d}\)	\(616\)

[In]

int(sin(d*x+c)^3*(a+b*tan(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.81 \[ \int \sin ^3(c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {{\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{6} - 3 \, {\left (a^{4} - 12 \, a^{2} b^{2} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (2 \, a^{3} b - 5 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, a^{3} b - 5 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + b^{4} + 9 \, {\left (2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{5} + 3 \, a b^{3} \cos \left (d x + c\right ) - 2 \, {\left (4 \, a^{3} b - 7 \, a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{3}} \]

[In]

integrate(sin(d*x+c)^3*(a+b*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

1/3*((a^4 - 6*a^2*b^2 + b^4)*cos(d*x + c)^6 - 3*(a^4 - 12*a^2*b^2 + 3*b^4)*cos(d*x + c)^4 + 3*(2*a^3*b - 5*a*b
^3)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(2*a^3*b - 5*a*b^3)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + b^4 +
9*(2*a^2*b^2 - b^4)*cos(d*x + c)^2 + 2*(2*(a^3*b - a*b^3)*cos(d*x + c)^5 + 3*a*b^3*cos(d*x + c) - 2*(4*a^3*b
- 7*a*b^3)*cos(d*x + c)^3)*sin(d*x + c))/(d*cos(d*x + c)^3)

Sympy [F]

\[ \int \sin ^3(c+d x) (a+b \tan (c+d x))^4 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{4} \sin ^{3}{\left (c + d x \right )}\, dx \]

[In]

integrate(sin(d*x+c)**3*(a+b*tan(d*x+c))**4,x)

[Out]

Integral((a + b*tan(c + d*x))**4*sin(c + d*x)**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.79 \[ \int \sin ^3(c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {{\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a^{4} - 2 \, {\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} a^{3} b - 6 \, {\left (\cos \left (d x + c\right )^{3} - \frac {3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} a^{2} b^{2} + {\left (4 \, \sin \left (d x + c\right )^{3} - \frac {6 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 24 \, \sin \left (d x + c\right )\right )} a b^{3} + {\left (\cos \left (d x + c\right )^{3} - \frac {9 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} - 9 \, \cos \left (d x + c\right )\right )} b^{4}}{3 \, d} \]

[In]

integrate(sin(d*x+c)^3*(a+b*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

1/3*((cos(d*x + c)^3 - 3*cos(d*x + c))*a^4 - 2*(2*sin(d*x + c)^3 - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c
) - 1) + 6*sin(d*x + c))*a^3*b - 6*(cos(d*x + c)^3 - 3/cos(d*x + c) - 6*cos(d*x + c))*a^2*b^2 + (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))*a*b^3 + (cos(d*x + c)^3 - (9*cos(d*x + c)^2 - 1)/cos(d*x + c)^3 - 9*cos(d*x + c))*b^4)/d

Giac [F(-1)]

Timed out. \[ \int \sin ^3(c+d x) (a+b \tan (c+d x))^4 \, dx=\text {Timed out} \]

[In]

integrate(sin(d*x+c)^3*(a+b*tan(d*x+c))^4,x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 8.57 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.16 \[ \int \sin ^3(c+d x) (a+b \tan (c+d x))^4 \, dx=-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (8\,a^4-96\,a^2\,b^2+32\,b^4\right )+4\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (20\,a\,b^3-8\,a^3\,b\right )-\frac {4\,a^4}{3}-\frac {32\,b^4}{3}+32\,a^2\,b^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {32\,a^4}{3}-64\,a^2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {20\,a\,b^3}{3}-\frac {8\,a^3\,b}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (20\,a\,b^3-8\,a^3\,b\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {20\,a\,b^3}{3}-\frac {8\,a^3\,b}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (56\,a\,b^3-48\,a^3\,b\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (56\,a\,b^3-48\,a^3\,b\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-1\right )}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (20\,a\,b^3-8\,a^3\,b\right )}{d} \]

[In]

int(sin(c + d*x)^3*(a + b*tan(c + d*x))^4,x)

[Out]

- (tan(c/2 + (d*x)/2)^4*(8*a^4 + 32*b^4 - 96*a^2*b^2) + 4*a^4*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)*(20*a*
b^3 - 8*a^3*b) - (4*a^4)/3 - (32*b^4)/3 + 32*a^2*b^2 - tan(c/2 + (d*x)/2)^6*((32*a^4)/3 - 64*a^2*b^2) + tan(c/
2 + (d*x)/2)^3*((20*a*b^3)/3 - (8*a^3*b)/3) - tan(c/2 + (d*x)/2)^11*(20*a*b^3 - 8*a^3*b) - tan(c/2 + (d*x)/2)^
9*((20*a*b^3)/3 - (8*a^3*b)/3) - tan(c/2 + (d*x)/2)^5*(56*a*b^3 - 48*a^3*b) + tan(c/2 + (d*x)/2)^7*(56*a*b^3 -
48*a^3*b))/(d*(3*tan(c/2 + (d*x)/2)^4 - 3*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^12 - 1)) - (atanh(tan(c/2
+ (d*x)/2))*(20*a*b^3 - 8*a^3*b))/d