3.1364 \(\int \frac {\sec ^2(c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=182 \[ \frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 d \left (a^2-b^2\right )}+\frac {a^3 b^2 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}-\frac {\sec ^2(c+d x) \left (4 a^3-b \left (5 a^2-b^2\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^2}+\frac {b (3 a+b) \log (1-\sin (c+d x))}{16 d (a+b)^3}-\frac {b (3 a-b) \log (\sin (c+d x)+1)}{16 d (a-b)^3} \]

[Out]

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

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Rubi [A]  time = 0.37, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2837, 12, 1647, 823, 801} \[ \frac {a^3 b^2 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}+\frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 d \left (a^2-b^2\right )}-\frac {\sec ^2(c+d x) \left (4 a^3-b \left (5 a^2-b^2\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^2}+\frac {b (3 a+b) \log (1-\sin (c+d x))}{16 d (a+b)^3}-\frac {b (3 a-b) \log (\sin (c+d x)+1)}{16 d (a-b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(3*a + b)*Log[1 - Sin[c + d*x]])/(16*(a + b)^3*d) - ((3*a - b)*b*Log[1 + Sin[c + d*x]])/(16*(a - b)^3*d) +
(a^3*b^2*Log[a + b*Sin[c + d*x]])/((a^2 - b^2)^3*d) + (Sec[c + d*x]^4*(a - b*Sin[c + d*x]))/(4*(a^2 - b^2)*d)
- (Sec[c + d*x]^2*(4*a^3 - b*(5*a^2 - b^2)*Sin[c + d*x]))/(8*(a^2 - b^2)^2*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rule 2837

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

Rubi steps

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

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Mathematica [A]  time = 1.25, size = 166, normalized size = 0.91 \[ \frac {\frac {16 a^3 b^2 \log (a+b \sin (c+d x))}{(a-b)^3 (a+b)^3}+\frac {3 a+b}{(a+b)^2 (\sin (c+d x)-1)}+\frac {b-3 a}{(a-b)^2 (\sin (c+d x)+1)}+\frac {1}{(a+b) (\sin (c+d x)-1)^2}+\frac {1}{(a-b) (\sin (c+d x)+1)^2}+\frac {b (3 a+b) \log (1-\sin (c+d x))}{(a+b)^3}-\frac {b (3 a-b) \log (\sin (c+d x)+1)}{(a-b)^3}}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*(3*a + b)*Log[1 - Sin[c + d*x]])/(a + b)^3 - ((3*a - b)*b*Log[1 + Sin[c + d*x]])/(a - b)^3 + (16*a^3*b^2*L
og[a + b*Sin[c + d*x]])/((a - b)^3*(a + b)^3) + 1/((a + b)*(-1 + Sin[c + d*x])^2) + (3*a + b)/((a + b)^2*(-1 +
Sin[c + d*x])) + 1/((a - b)*(1 + Sin[c + d*x])^2) + (-3*a + b)/((a - b)^2*(1 + Sin[c + d*x])))/(16*d)

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fricas [A]  time = 1.14, size = 260, normalized size = 1.43 \[ \frac {16 \, a^{3} b^{2} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (3 \, a^{4} b + 8 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (3 \, a^{4} b - 8 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a^{5} - 8 \, a^{3} b^{2} + 4 \, a b^{4} - 8 \, {\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{4} b - 4 \, a^{2} b^{3} + 2 \, b^{5} - {\left (5 \, a^{4} b - 6 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.29, size = 326, normalized size = 1.79 \[ \frac {\frac {16 \, a^{3} b^{3} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}} - \frac {{\left (3 \, a b - b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (3 \, a b + b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (6 \, a^{3} b^{2} \sin \left (d x + c\right )^{4} - 5 \, a^{4} b \sin \left (d x + c\right )^{3} + 6 \, a^{2} b^{3} \sin \left (d x + c\right )^{3} - b^{5} \sin \left (d x + c\right )^{3} + 4 \, a^{5} \sin \left (d x + c\right )^{2} - 16 \, a^{3} b^{2} \sin \left (d x + c\right )^{2} + 3 \, a^{4} b \sin \left (d x + c\right ) - 2 \, a^{2} b^{3} \sin \left (d x + c\right ) - b^{5} \sin \left (d x + c\right ) - 2 \, a^{5} + 6 \, a^{3} b^{2} + 2 \, a b^{4}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/16*(16*a^3*b^3*log(abs(b*sin(d*x + c) + a))/(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7) - (3*a*b - b^2)*log(abs(si
n(d*x + c) + 1))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (3*a*b + b^2)*log(abs(sin(d*x + c) - 1))/(a^3 + 3*a^2*b + 3
*a*b^2 + b^3) + 2*(6*a^3*b^2*sin(d*x + c)^4 - 5*a^4*b*sin(d*x + c)^3 + 6*a^2*b^3*sin(d*x + c)^3 - b^5*sin(d*x
+ c)^3 + 4*a^5*sin(d*x + c)^2 - 16*a^3*b^2*sin(d*x + c)^2 + 3*a^4*b*sin(d*x + c) - 2*a^2*b^3*sin(d*x + c) - b^
5*sin(d*x + c) - 2*a^5 + 6*a^3*b^2 + 2*a*b^4)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(sin(d*x + c)^2 - 1)^2))/d

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maple [A]  time = 0.42, size = 261, normalized size = 1.43 \[ \frac {1}{2 d \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {3 a}{16 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {b}{16 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {3 \ln \left (\sin \left (d x +c \right )-1\right ) a b}{16 d \left (a +b \right )^{3}}+\frac {\ln \left (\sin \left (d x +c \right )-1\right ) b^{2}}{16 d \left (a +b \right )^{3}}+\frac {a^{3} b^{2} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {1}{2 d \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3 a}{16 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {b}{16 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}-\frac {3 \ln \left (1+\sin \left (d x +c \right )\right ) a b}{16 d \left (a -b \right )^{3}}+\frac {\ln \left (1+\sin \left (d x +c \right )\right ) b^{2}}{16 d \left (a -b \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*sin(d*x+c)^3/(a+b*sin(d*x+c)),x)

[Out]

1/2/d/(8*a+8*b)/(sin(d*x+c)-1)^2+3/16/d/(a+b)^2/(sin(d*x+c)-1)*a+1/16/d/(a+b)^2/(sin(d*x+c)-1)*b+3/16/d/(a+b)^
3*ln(sin(d*x+c)-1)*a*b+1/16/d/(a+b)^3*ln(sin(d*x+c)-1)*b^2+1/d*a^3/(a+b)^3*b^2/(a-b)^3*ln(a+b*sin(d*x+c))+1/2/
d/(8*a-8*b)/(1+sin(d*x+c))^2-3/16/d/(a-b)^2/(1+sin(d*x+c))*a+1/16/d/(a-b)^2/(1+sin(d*x+c))*b-3/16/d/(a-b)^3*ln
(1+sin(d*x+c))*a*b+1/16/d/(a-b)^3*ln(1+sin(d*x+c))*b^2

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maxima [A]  time = 0.32, size = 267, normalized size = 1.47 \[ \frac {\frac {16 \, a^{3} b^{2} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac {{\left (3 \, a b - b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (3 \, a b + b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (4 \, a^{3} \sin \left (d x + c\right )^{2} - {\left (5 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )^{3} - 2 \, a^{3} - 2 \, a b^{2} + {\left (3 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/16*(16*a^3*b^2*log(b*sin(d*x + c) + a)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) - (3*a*b - b^2)*log(sin(d*x + c)
+ 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (3*a*b + b^2)*log(sin(d*x + c) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + 2
*(4*a^3*sin(d*x + c)^2 - (5*a^2*b - b^3)*sin(d*x + c)^3 - 2*a^3 - 2*a*b^2 + (3*a^2*b + b^3)*sin(d*x + c))/((a^
4 - 2*a^2*b^2 + b^4)*sin(d*x + c)^4 + a^4 - 2*a^2*b^2 + b^4 - 2*(a^4 - 2*a^2*b^2 + b^4)*sin(d*x + c)^2))/d

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mupad [B]  time = 12.45, size = 471, normalized size = 2.59 \[ \frac {b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (3\,a+b\right )}{8\,d\,{\left (a+b\right )}^3}-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (11\,a^2\,b-7\,b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{a^4-2\,a^2\,b^2+b^4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (3\,a^2\,b+b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (11\,a^2\,b-7\,b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {2\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^4-2\,a^2\,b^2+b^4}+\frac {2\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{a^4-2\,a^2\,b^2+b^4}-\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^2+b^2\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {b^2}{4\,{\left (a-b\right )}^3}+\frac {3\,b}{8\,{\left (a-b\right )}^2}\right )}{d}+\frac {a^3\,b^2\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^3/(cos(c + d*x)^5*(a + b*sin(c + d*x))),x)

[Out]

(b*log(tan(c/2 + (d*x)/2) - 1)*(3*a + b))/(8*d*(a + b)^3) - ((tan(c/2 + (d*x)/2)^3*(11*a^2*b - 7*b^3))/(4*(a^4
+ b^4 - 2*a^2*b^2)) - (4*a^3*tan(c/2 + (d*x)/2)^4)/(a^4 + b^4 - 2*a^2*b^2) - (tan(c/2 + (d*x)/2)^7*(3*a^2*b +
b^3))/(4*(a^4 + b^4 - 2*a^2*b^2)) + (tan(c/2 + (d*x)/2)^5*(11*a^2*b - 7*b^3))/(4*(a^4 + b^4 - 2*a^2*b^2)) + (
2*a*b^2*tan(c/2 + (d*x)/2)^2)/(a^4 + b^4 - 2*a^2*b^2) + (2*a*b^2*tan(c/2 + (d*x)/2)^6)/(a^4 + b^4 - 2*a^2*b^2)
- (b*tan(c/2 + (d*x)/2)*(3*a^2 + b^2))/(4*(a^4 + b^4 - 2*a^2*b^2)))/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 +
(d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) - (log(tan(c/2 + (d*x)/2) + 1)*(b^2/(4*(a - b
)^3) + (3*b)/(8*(a - b)^2)))/d + (a^3*b^2*log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^2))/(d*(a^6 -
b^6 + 3*a^2*b^4 - 3*a^4*b^2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5*sin(d*x+c)**3/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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