Optimal. Leaf size=383 \[ -\frac{2 \left (4 B n \left (2 n^2+8 n+9\right )+i A \left (8 n^3+32 n^2+36 n+15\right )\right ) \sqrt{\tan (c+d x)} (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \text{Hypergeometric2F1}\left (\frac{1}{2},1-n,\frac{3}{2},-i \tan (c+d x)\right )}{d (2 n+1) (2 n+3) (2 n+5)}+\frac{2 (B+i A) \sqrt{\tan (c+d x)} (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n F_1\left (\frac{1}{2};1-n,1;\frac{3}{2};-i \tan (c+d x),i \tan (c+d x)\right )}{d}-\frac{2 \left (B \left (4 n^2+10 n+15\right )+2 i A n (2 n+5)\right ) \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (2 n+1) (2 n+3) (2 n+5)}-\frac{2 (-A (2 n+5)+2 i B n) \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+3) (2 n+5)}+\frac{2 B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+5)} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 1.13789, antiderivative size = 383, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3597, 3601, 3564, 130, 430, 429, 3599, 66, 64} \[ \frac{2 (B+i A) \sqrt{\tan (c+d x)} (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n F_1\left (\frac{1}{2};1-n,1;\frac{3}{2};-i \tan (c+d x),i \tan (c+d x)\right )}{d}-\frac{2 \left (4 B n \left (2 n^2+8 n+9\right )+i A \left (8 n^3+32 n^2+36 n+15\right )\right ) \sqrt{\tan (c+d x)} (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \, _2F_1\left (\frac{1}{2},1-n;\frac{3}{2};-i \tan (c+d x)\right )}{d (2 n+1) (2 n+3) (2 n+5)}-\frac{2 \left (B \left (4 n^2+10 n+15\right )+2 i A n (2 n+5)\right ) \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (2 n+1) (2 n+3) (2 n+5)}-\frac{2 (-A (2 n+5)+2 i B n) \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+3) (2 n+5)}+\frac{2 B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+5)} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 3597
Rule 3601
Rule 3564
Rule 130
Rule 430
Rule 429
Rule 3599
Rule 66
Rule 64
Rubi steps
\begin{align*} \int \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx &=\frac{2 B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (5+2 n)}+\frac{2 \int \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^n \left (-\frac{5 a B}{2}-\frac{1}{2} a (2 i B n-A (5+2 n)) \tan (c+d x)\right ) \, dx}{a (5+2 n)}\\ &=-\frac{2 (2 i B n-A (5+2 n)) \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (3+2 n) (5+2 n)}+\frac{2 B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (5+2 n)}+\frac{4 \int \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n \left (\frac{3}{4} a^2 (2 i B n-A (5+2 n))-\frac{1}{4} a^2 \left (2 i A n (5+2 n)+B \left (15+10 n+4 n^2\right )\right ) \tan (c+d x)\right ) \, dx}{a^2 (3+2 n) (5+2 n)}\\ &=-\frac{2 \left (2 i A n (5+2 n)+B \left (15+10 n+4 n^2\right )\right ) \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (1+2 n) (3+2 n) (5+2 n)}-\frac{2 (2 i B n-A (5+2 n)) \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (3+2 n) (5+2 n)}+\frac{2 B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (5+2 n)}+\frac{8 \int \frac{(a+i a \tan (c+d x))^n \left (\frac{1}{8} a^3 \left (2 i A n (5+2 n)+B \left (15+10 n+4 n^2\right )\right )+\frac{1}{8} a^3 \left (4 i B n \left (9+8 n+2 n^2\right )-A \left (15+36 n+32 n^2+8 n^3\right )\right ) \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx}{a^3 (1+2 n) (3+2 n) (5+2 n)}\\ &=-\frac{2 \left (2 i A n (5+2 n)+B \left (15+10 n+4 n^2\right )\right ) \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (1+2 n) (3+2 n) (5+2 n)}-\frac{2 (2 i B n-A (5+2 n)) \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (3+2 n) (5+2 n)}+\frac{2 B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (5+2 n)}+(i A+B) \int \frac{(a+i a \tan (c+d x))^n}{\sqrt{\tan (c+d x)}} \, dx-\frac{\left (4 B n \left (9+8 n+2 n^2\right )+i A \left (15+36 n+32 n^2+8 n^3\right )\right ) \int \frac{(a-i a \tan (c+d x)) (a+i a \tan (c+d x))^n}{\sqrt{\tan (c+d x)}} \, dx}{a (1+2 n) (3+2 n) (5+2 n)}\\ &=-\frac{2 \left (2 i A n (5+2 n)+B \left (15+10 n+4 n^2\right )\right ) \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (1+2 n) (3+2 n) (5+2 n)}-\frac{2 (2 i B n-A (5+2 n)) \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (3+2 n) (5+2 n)}+\frac{2 B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (5+2 n)}-\frac{\left (a^2 (A-i B)\right ) \operatorname{Subst}\left (\int \frac{(a+x)^{-1+n}}{\sqrt{-\frac{i x}{a}} \left (-a^2+a x\right )} \, dx,x,i a \tan (c+d x)\right )}{d}-\frac{\left (a \left (4 B n \left (9+8 n+2 n^2\right )+i A \left (15+36 n+32 n^2+8 n^3\right )\right )\right ) \operatorname{Subst}\left (\int \frac{(a+i a x)^{-1+n}}{\sqrt{x}} \, dx,x,\tan (c+d x)\right )}{d (1+2 n) (3+2 n) (5+2 n)}\\ &=-\frac{2 \left (2 i A n (5+2 n)+B \left (15+10 n+4 n^2\right )\right ) \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (1+2 n) (3+2 n) (5+2 n)}-\frac{2 (2 i B n-A (5+2 n)) \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (3+2 n) (5+2 n)}+\frac{2 B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (5+2 n)}-\frac{\left (2 a^3 (i A+B)\right ) \operatorname{Subst}\left (\int \frac{\left (a+i a x^2\right )^{-1+n}}{-a^2+i a^2 x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}-\frac{\left (\left (4 B n \left (9+8 n+2 n^2\right )+i A \left (15+36 n+32 n^2+8 n^3\right )\right ) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n\right ) \operatorname{Subst}\left (\int \frac{(1+i x)^{-1+n}}{\sqrt{x}} \, dx,x,\tan (c+d x)\right )}{d (1+2 n) (3+2 n) (5+2 n)}\\ &=-\frac{2 \left (2 i A n (5+2 n)+B \left (15+10 n+4 n^2\right )\right ) \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (1+2 n) (3+2 n) (5+2 n)}-\frac{2 \left (4 B n \left (9+8 n+2 n^2\right )+i A \left (15+36 n+32 n^2+8 n^3\right )\right ) \, _2F_1\left (\frac{1}{2},1-n;\frac{3}{2};-i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (1+2 n) (3+2 n) (5+2 n)}-\frac{2 (2 i B n-A (5+2 n)) \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (3+2 n) (5+2 n)}+\frac{2 B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (5+2 n)}-\frac{\left (2 a^2 (i A+B) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n\right ) \operatorname{Subst}\left (\int \frac{\left (1+i x^2\right )^{-1+n}}{-a^2+i a^2 x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=-\frac{2 \left (2 i A n (5+2 n)+B \left (15+10 n+4 n^2\right )\right ) \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (1+2 n) (3+2 n) (5+2 n)}+\frac{2 (i A+B) F_1\left (\frac{1}{2};1-n,1;\frac{3}{2};-i \tan (c+d x),i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n}{d}-\frac{2 \left (4 B n \left (9+8 n+2 n^2\right )+i A \left (15+36 n+32 n^2+8 n^3\right )\right ) \, _2F_1\left (\frac{1}{2},1-n;\frac{3}{2};-i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (1+2 n) (3+2 n) (5+2 n)}-\frac{2 (2 i B n-A (5+2 n)) \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (3+2 n) (5+2 n)}+\frac{2 B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (5+2 n)}\\ \end{align*}
Mathematica [F] time = 15.2754, size = 0, normalized size = 0. \[ \int \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx \]
Verification is Not applicable to the result.
[In]
[Out]
________________________________________________________________________________________
Maple [F] time = 0.336, size = 0, normalized size = 0. \begin{align*} \int \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{n} \left ( A+B\tan \left ( dx+c \right ) \right ) \, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{\frac{5}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left ({\left (A - i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} -{\left (A - 3 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} -{\left (A + 3 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + A + i \, B\right )} \left (\frac{2 \, a e^{\left (2 i \, d x + 2 i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{n} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
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]
[Out]
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{\frac{5}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]