3.1365 \(\int \frac {\sec ^3(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=178 \[ -\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 d \left (a^2-b^2\right )}+\frac {a \sec ^2(c+d x) \left (4 a b-\left (a^2+3 b^2\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^2}-\frac {a^2 b^3 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}+\frac {a (a+3 b) \log (1-\sin (c+d x))}{16 d (a+b)^3}-\frac {a (a-3 b) \log (\sin (c+d x)+1)}{16 d (a-b)^3} \]

[Out]

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

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Rubi [A]  time = 0.38, antiderivative size = 178, 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} \[ -\frac {a^2 b^3 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 d \left (a^2-b^2\right )}+\frac {a \sec ^2(c+d x) \left (4 a b-\left (a^2+3 b^2\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^2}+\frac {a (a+3 b) \log (1-\sin (c+d x))}{16 d (a+b)^3}-\frac {a (a-3 b) \log (\sin (c+d x)+1)}{16 d (a-b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 1.24, size = 163, normalized size = 0.92 \[ \frac {-\frac {16 a^2 b^3 \log (a+b \sin (c+d x))}{(a-b)^3 (a+b)^3}+\frac {a-b}{(a+b)^2 (\sin (c+d x)-1)}+\frac {a+b}{(a-b)^2 (\sin (c+d x)+1)}+\frac {1}{(a+b) (\sin (c+d x)-1)^2}-\frac {1}{(a-b) (\sin (c+d x)+1)^2}+\frac {a (a+3 b) \log (1-\sin (c+d x))}{(a+b)^3}-\frac {a (a-3 b) \log (\sin (c+d x)+1)}{(a-b)^3}}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.21, size = 258, normalized size = 1.45 \[ -\frac {16 \, a^{2} b^{3} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left (a^{5} - 6 \, a^{3} b^{2} - 8 \, a^{2} b^{3} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{5} - 6 \, a^{3} b^{2} + 8 \, a^{2} b^{3} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a^{4} b - 8 \, a^{2} b^{3} + 4 \, b^{5} - 8 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{5} - 4 \, a^{3} b^{2} + 2 \, a b^{4} - {\left (a^{5} + 2 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/16*(16*a^2*b^3*cos(d*x + c)^4*log(b*sin(d*x + c) + a) + (a^5 - 6*a^3*b^2 - 8*a^2*b^3 - 3*a*b^4)*cos(d*x + c
)^4*log(sin(d*x + c) + 1) - (a^5 - 6*a^3*b^2 + 8*a^2*b^3 - 3*a*b^4)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 4*
a^4*b - 8*a^2*b^3 + 4*b^5 - 8*(a^4*b - a^2*b^3)*cos(d*x + c)^2 - 2*(2*a^5 - 4*a^3*b^2 + 2*a*b^4 - (a^5 + 2*a^3
*b^2 - 3*a*b^4)*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 = 0.28, size = 325, normalized size = 1.83 \[ -\frac {\frac {16 \, a^{2} b^{4} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}} + \frac {{\left (a^{2} - 3 \, a b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (a^{2} + 3 \, a b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (6 \, a^{2} b^{3} \sin \left (d x + c\right )^{4} - a^{5} \sin \left (d x + c\right )^{3} - 2 \, a^{3} b^{2} \sin \left (d x + c\right )^{3} + 3 \, a b^{4} \sin \left (d x + c\right )^{3} + 4 \, a^{4} b \sin \left (d x + c\right )^{2} - 16 \, a^{2} b^{3} \sin \left (d x + c\right )^{2} - a^{5} \sin \left (d x + c\right ) + 6 \, a^{3} b^{2} \sin \left (d x + c\right ) - 5 \, a b^{4} \sin \left (d x + c\right ) - 2 \, a^{4} b + 6 \, a^{2} b^{3} + 2 \, b^{5}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/16*(16*a^2*b^4*log(abs(b*sin(d*x + c) + a))/(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7) + (a^2 - 3*a*b)*log(abs(s
in(d*x + c) + 1))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (a^2 + 3*a*b)*log(abs(sin(d*x + c) - 1))/(a^3 + 3*a^2*b +
3*a*b^2 + b^3) + 2*(6*a^2*b^3*sin(d*x + c)^4 - a^5*sin(d*x + c)^3 - 2*a^3*b^2*sin(d*x + c)^3 + 3*a*b^4*sin(d*x
+ c)^3 + 4*a^4*b*sin(d*x + c)^2 - 16*a^2*b^3*sin(d*x + c)^2 - a^5*sin(d*x + c) + 6*a^3*b^2*sin(d*x + c) - 5*a
*b^4*sin(d*x + c) - 2*a^4*b + 6*a^2*b^3 + 2*b^5)/((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 = 0.41, size = 262, normalized size = 1.47 \[ \frac {1}{2 d \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {a}{16 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {b}{16 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\ln \left (\sin \left (d x +c \right )-1\right ) a^{2}}{16 d \left (a +b \right )^{3}}+\frac {3 \ln \left (\sin \left (d x +c \right )-1\right ) a b}{16 d \left (a +b \right )^{3}}-\frac {a^{2} b^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \left (a +b \right )^{3} \left (a -b \right )^{3}}-\frac {1}{2 d \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {a}{16 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {b}{16 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}-\frac {\ln \left (1+\sin \left (d x +c \right )\right ) a^{2}}{16 d \left (a -b \right )^{3}}+\frac {3 \ln \left (1+\sin \left (d x +c \right )\right ) a b}{16 d \left (a -b \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*sin(d*x+c)^2/(a+b*sin(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.33, size = 265, normalized size = 1.49 \[ -\frac {\frac {16 \, a^{2} b^{3} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac {{\left (a^{2} - 3 \, a b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (a^{2} + 3 \, a b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (4 \, a^{2} b \sin \left (d x + c\right )^{2} - {\left (a^{3} + 3 \, a b^{2}\right )} \sin \left (d x + c\right )^{3} - 2 \, a^{2} b - 2 \, b^{3} - {\left (a^{3} - 5 \, a b^{2}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/16*(16*a^2*b^3*log(b*sin(d*x + c) + a)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) + (a^2 - 3*a*b)*log(sin(d*x + c)
+ 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (a^2 + 3*a*b)*log(sin(d*x + c) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) +
2*(4*a^2*b*sin(d*x + c)^2 - (a^3 + 3*a*b^2)*sin(d*x + c)^3 - 2*a^2*b - 2*b^3 - (a^3 - 5*a*b^2)*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 = 12.51, size = 498, normalized size = 2.80 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {b^2}{4\,{\left (a-b\right )}^3}+\frac {b}{8\,{\left (a-b\right )}^2}-\frac {1}{8\,\left (a-b\right )}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (\frac {b}{8\,{\left (a+b\right )}^2}+\frac {1}{8\,\left (a+b\right )}-\frac {b^2}{4\,{\left (a+b\right )}^3}\right )}{d}-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,a\,b^2-7\,a^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {2\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{a^4-2\,a^2\,b^2+b^4}-\frac {2\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^4-2\,a^2\,b^2+b^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (5\,a\,b^2-a^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (3\,a\,b^2-7\,a^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {4\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{a^4-2\,a^2\,b^2+b^4}-\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-5\,b^2\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {a^2\,b^3\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^2/(cos(c + d*x)^5*(a + b*sin(c + d*x))),x)

[Out]

(log(tan(c/2 + (d*x)/2) + 1)*(b^2/(4*(a - b)^3) + b/(8*(a - b)^2) - 1/(8*(a - b))))/d + (log(tan(c/2 + (d*x)/2
) - 1)*(b/(8*(a + b)^2) + 1/(8*(a + b)) - b^2/(4*(a + b)^3)))/d - ((tan(c/2 + (d*x)/2)^3*(3*a*b^2 - 7*a^3))/(4
*(a^4 + b^4 - 2*a^2*b^2)) - (2*b^3*tan(c/2 + (d*x)/2)^6)/(a^4 + b^4 - 2*a^2*b^2) - (2*b^3*tan(c/2 + (d*x)/2)^2
)/(a^4 + b^4 - 2*a^2*b^2) + (tan(c/2 + (d*x)/2)^7*(5*a*b^2 - a^3))/(4*(a^4 + b^4 - 2*a^2*b^2)) + (tan(c/2 + (d
*x)/2)^5*(3*a*b^2 - 7*a^3))/(4*(a^4 + b^4 - 2*a^2*b^2)) + (4*a^2*b*tan(c/2 + (d*x)/2)^4)/(a^4 + b^4 - 2*a^2*b^
2) - (a*tan(c/2 + (d*x)/2)*(a^2 - 5*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)) - (a^2*b^3*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]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{2}{\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5*sin(d*x+c)**2/(a+b*sin(d*x+c)),x)

[Out]

Integral(sin(c + d*x)**2*sec(c + d*x)**5/(a + b*sin(c + d*x)), x)

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